11.13 Using the FPU

Strangely enough, most of assembly language literature does not even mention the existence of the FPU, or floating point unit, let alone discuss programming it.

Yet, never does assembly language shine more than when we create highly optimized FPU code by doing things that can be done only in assembly language.

11.13.1 Organization of the FPU

The FPU consists of 8 80–bit floating–point registers. These are organized in a stack fashion—you can push a value on TOS (top of stack) and you can pop it.

That said, the assembly language op codes are not push and pop because those are already taken.

You can push a value on TOS by using fld, fild, and fbld. Several other op codes let you push many common constants—such as pi—on the TOS.

Similarly, you can pop a value by using fst, fstp, fist, fistp, and fbstp. Actually, only the op codes that end with a p will literally pop the value, the rest will store it somewhere else without removing it from the TOS.

We can transfer the data between the TOS and the computer memory either as a 32–bit, 64–bit, or 80–bit real, a 16–bit, 32–bit, or 64–bit integer, or an 80–bit packed decimal.

The 80–bit packed decimal is a special case of binary coded decimal which is very convenient when converting between the ASCII representation of data and the internal data of the FPU. It allows us to use 18 significant digits.

No matter how we represent data in the memory, the FPU always stores it in the 80–bit real format in its registers.

Its internal precision is at least 19 decimal digits, so even if we choose to display results as ASCII in the full 18–digit precision, we are still showing correct results.

We can perform mathematical operations on the TOS: We can calculate its sine, we can scale it (i.e., we can multiply or divide it by a power of 2), we can calculate its base–2 logarithm, and many other things.

We can also multiply or divide it by, add it to, or subtract it from, any of the FPU registers (including itself).

The official Intel op code for the TOS is st, and for the registers st(0)st(7). st and st(0), then, refer to the same register.

For whatever reasons, the original author of nasm has decided to use different op codes, namely st0st7. In other words, there are no parentheses, and the TOS is always st0, never just st. The Packed Decimal Format

The packed decimal format uses 10 bytes (80 bits) of memory to represent 18 digits. The number represented there is always an integer.

Tip: You can use it to get decimal places by multiplying the TOS by a power of 10 first.

The highest bit of the highest byte (byte 9) is the sign bit: If it is set, the number is negative, otherwise, it is positive. The rest of the bits of this byte are unused/ignored.

The remaining 9 bytes store the 18 digits of the number: 2 digits per byte.

The more significant digit is stored in the high nibble (4 bits), the less significant digit in the low nibble.

That said, you might think that -1234567 would be stored in the memory like this (using hexadecimal notation):

80 00 00 00 00 00 01 23 45 67

Alas it is not! As with everything else of Intel make, even the packed decimal is little–endian.

That means our -1234567 is stored like this:

67 45 23 01 00 00 00 00 00 80

Remember that, or you will be pulling your hair out in desperation!

Note: The book to read—if you can find it—is Richard Startz' 8087/80287/80387 for the IBM PC & Compatibles. Though it does seem to take the fact about the little–endian storage of the packed decimal for granted. I kid you not about the desperation of trying to figure out what was wrong with the filter I show below before it occurred to me I should try the little–endian order even for this type of data.

11.13.2 Excursion to Pinhole Photography

To write meaningful software, we must not only understand our programming tools, but also the field we are creating software for.

Our next filter will help us whenever we want to build a pinhole camera, so, we need some background in pinhole photography before we can continue. The Camera

The easiest way to describe any camera ever built is as some empty space enclosed in some lightproof material, with a small hole in the enclosure.

The enclosure is usually sturdy (e.g., a box), though sometimes it is flexible (the bellows). It is quite dark inside the camera. However, the hole lets light rays in through a single point (though in some cases there may be several). These light rays form an image, a representation of whatever is outside the camera, in front of the hole.

If some light sensitive material (such as film) is placed inside the camera, it can capture the image.

The hole often contains a lens, or a lens assembly, often called the objective. The Pinhole

But, strictly speaking, the lens is not necessary: The original cameras did not use a lens but a pinhole. Even today, pinholes are used, both as a tool to study how cameras work, and to achieve a special kind of image.

The image produced by the pinhole is all equally sharp. Or blurred. There is an ideal size for a pinhole: If it is either larger or smaller, the image loses its sharpness. Focal Length

This ideal pinhole diameter is a function of the square root of focal length, which is the distance of the pinhole from the film.

	D = PC * sqrt(FL)

In here, D is the ideal diameter of the pinhole, FL is the focal length, and PC is a pinhole constant. According to Jay Bender, its value is 0.04, while Kenneth Connors has determined it to be 0.037. Others have proposed other values. Plus, this value is for the daylight only: Other types of light will require a different constant, whose value can only be determined by experimentation. The F–Number

The f–number is a very useful measure of how much light reaches the film. A light meter can determine that, for example, to expose a film of specific sensitivity with f5.6 may require the exposure to last 1/1000 sec.

It does not matter whether it is a 35–mm camera, or a 6x9cm camera, etc. As long as we know the f–number, we can determine the proper exposure.

The f–number is easy to calculate:

	F = FL / D

In other words, the f–number equals the focal length divided by the diameter of the pinhole. It also means a higher f–number either implies a smaller pinhole or a larger focal distance, or both. That, in turn, implies, the higher the f–number, the longer the exposure has to be.

Furthermore, while pinhole diameter and focal distance are one–dimensional measurements, both, the film and the pinhole, are two–dimensional. That means that if you have measured the exposure at f–number A as t, then the exposure at f–number B is:

	t * (B / A) Normalized F–Number

While many modern cameras can change the diameter of their pinhole, and thus their f–number, quite smoothly and gradually, such was not always the case.

To allow for different f–numbers, cameras typically contained a metal plate with several holes of different sizes drilled to them.

Their sizes were chosen according to the above formula in such a way that the resultant f–number was one of standard f–numbers used on all cameras everywhere. For example, a very old Kodak Duaflex IV camera in my possession has three such holes for f–numbers 8, 11, and 16.

A more recently made camera may offer f–numbers of 2.8, 4, 5.6, 8, 11, 16, 22, and 32 (as well as others). These numbers were not chosen arbitrarily: They all are powers of the square root of 2, though they may be rounded somewhat. The F–Stop

A typical camera is designed in such a way that setting any of the normalized f–numbers changes the feel of the dial. It will naturally stop in that position. Because of that, these positions of the dial are called f–stops.

Since the f–numbers at each stop are powers of the square root of 2, moving the dial by 1 stop will double the amount of light required for proper exposure. Moving it by 2 stops will quadruple the required exposure. Moving the dial by 3 stops will require the increase in exposure 8 times, etc.

11.13.3 Designing the Pinhole Software

We are now ready to decide what exactly we want our pinhole software to do. Processing Program Input

Since its main purpose is to help us design a working pinhole camera, we will use the focal length as the input to the program. This is something we can determine without software: Proper focal length is determined by the size of the film and by the need to shoot "regular" pictures, wide angle pictures, or telephoto pictures.

Most of the programs we have written so far worked with individual characters, or bytes, as their input: The hex program converted individual bytes into a hexadecimal number, the csv program either let a character through, or deleted it, or changed it to a different character, etc.

One program, ftuc used the state machine to consider at most two input bytes at a time.

But our pinhole program cannot just work with individual characters, it has to deal with larger syntactic units.

For example, if we want the program to calculate the pinhole diameter (and other values we will discuss later) at the focal lengths of 100 mm, 150 mm, and 210 mm, we may want to enter something like this:

100, 150, 210

Our program needs to consider more than a single byte of input at a time. When it sees the first 1, it must understand it is seeing the first digit of a decimal number. When it sees the 0 and the other 0, it must know it is seeing more digits of the same number.

When it encounters the first comma, it must know it is no longer receiving the digits of the first number. It must be able to convert the digits of the first number into the value of 100. And the digits of the second number into the value of 150. And, of course, the digits of the third number into the numeric value of 210.

We need to decide what delimiters to accept: Do the input numbers have to be separated by a comma? If so, how do we treat two numbers separated by something else?

Personally, I like to keep it simple. Something either is a number, so I process it. Or it is not a number, so I discard it. I do not like the computer complaining about me typing in an extra character when it is obvious that it is an extra character. Duh!

Plus, it allows me to break up the monotony of computing and type in a query instead of just a number:

What is the best pinhole diameter for the focal length of 150?

There is no reason for the computer to spit out a number of complaints:

Syntax error: What
Syntax error: is
Syntax error: the
Syntax error: best

Et cetera, et cetera, et cetera.

Secondly, I like the # character to denote the start of a comment which extends to the end of the line. This does not take too much effort to code, and lets me treat input files for my software as executable scripts.

In our case, we also need to decide what units the input should come in: We choose millimeters because that is how most photographers measure the focus length.

Finally, we need to decide whether to allow the use of the decimal point (in which case we must also consider the fact that much of the world uses a decimal comma).

In our case allowing for the decimal point/comma would offer a false sense of precision: There is little if any noticeable difference between the focus lengths of 50 and 51, so allowing the user to input something like 50.5 is not a good idea. This is my opinion, mind you, but I am the one writing this program. You can make other choices in yours, of course. Offering Options

The most important thing we need to know when building a pinhole camera is the diameter of the pinhole. Since we want to shoot sharp images, we will use the above formula to calculate the pinhole diameter from focal length. As experts are offering several different values for the PC constant, we will need to have the choice.

It is traditional in UNIX® programming to have two main ways of choosing program parameters, plus to have a default for the time the user does not make a choice.

Why have two ways of choosing?

One is to allow a (relatively) permanent choice that applies automatically each time the software is run without us having to tell it over and over what we want it to do.

The permanent choices may be stored in a configuration file, typically found in the user's home directory. The file usually has the same name as the application but is started with a dot. Often "rc" is added to the file name. So, ours could be ~/.pinhole or ~/.pinholerc. (The ~/ means current user's home directory.)

The configuration file is used mostly by programs that have many configurable parameters. Those that have only one (or a few) often use a different method: They expect to find the parameter in an environment variable. In our case, we might look at an environment variable named PINHOLE.

Usually, a program uses one or the other of the above methods. Otherwise, if a configuration file said one thing, but an environment variable another, the program might get confused (or just too complicated).

Because we only need to choose one such parameter, we will go with the second method and search the environment for a variable named PINHOLE.

The other way allows us to make ad hoc decisions: "Though I usually want you to use 0.039, this time I want 0.03872." In other words, it allows us to override the permanent choice.

This type of choice is usually done with command line parameters.

Finally, a program always needs a default. The user may not make any choices. Perhaps he does not know what to choose. Perhaps he is "just browsing." Preferably, the default will be the value most users would choose anyway. That way they do not need to choose. Or, rather, they can choose the default without an additional effort.

Given this system, the program may find conflicting options, and handle them this way:

  1. If it finds an ad hoc choice (e.g., command line parameter), it should accept that choice. It must ignore any permanent choice and any default.

  2. Otherwise, if it finds a permanent option (e.g., an environment variable), it should accept it, and ignore the default.

  3. Otherwise, it should use the default.

We also need to decide what format our PC option should have.

At first site, it seems obvious to use the PINHOLE=0.04 format for the environment variable, and -p0.04 for the command line.

Allowing that is actually a security risk. The PC constant is a very small number. Naturally, we will test our software using various small values of PC. But what will happen if someone runs the program choosing a huge value?

It may crash the program because we have not designed it to handle huge numbers.

Or, we may spend more time on the program so it can handle huge numbers. We might do that if we were writing commercial software for computer illiterate audience.

Or, we might say, "Tough! The user should know better.""

Or, we just may make it impossible for the user to enter a huge number. This is the approach we will take: We will use an implied 0. prefix.

In other words, if the user wants 0.04, we will expect him to type -p04, or set PINHOLE=04 in his environment. So, if he says -p9999999, we will interpret it as 0.9999999—still ridiculous but at least safer.

Secondly, many users will just want to go with either Bender's constant or Connors' constant. To make it easier on them, we will interpret -b as identical to -p04, and -c as identical to -p037. The Output

We need to decide what we want our software to send to the output, and in what format.

Since our input allows for an unspecified number of focal length entries, it makes sense to use a traditional database–style output of showing the result of the calculation for each focal length on a separate line, while separating all values on one line by a tab character.

Optionally, we should also allow the user to specify the use of the CSV format we have studied earlier. In this case, we will print out a line of comma–separated names describing each field of every line, then show our results as before, but substituting a comma for the tab.

We need a command line option for the CSV format. We cannot use -c because that already means use Connors' constant. For some strange reason, many web sites refer to CSV files as "Excel spreadsheet" (though the CSV format predates Excel). We will, therefore, use the -e switch to inform our software we want the output in the CSV format.

We will start each line of the output with the focal length. This may sound repetitious at first, especially in the interactive mode: The user types in the focal length, and we are repeating it.

But the user can type several focal lengths on one line. The input can also come in from a file or from the output of another program. In that case the user does not see the input at all.

By the same token, the output can go to a file which we will want to examine later, or it could go to the printer, or become the input of another program.

So, it makes perfect sense to start each line with the focal length as entered by the user.

No, wait! Not as entered by the user. What if the user types in something like this:


Clearly, we need to strip those leading zeros.

So, we might consider reading the user input as is, converting it to binary inside the FPU, and printing it out from there.


What if the user types something like this:


Ha! The packed decimal FPU format lets us input 18–digit numbers. But the user has entered more than 18 digits. How do we handle that?

Well, we could modify our code to read the first 18 digits, enter it to the FPU, then read more, multiply what we already have on the TOS by 10 raised to the number of additional digits, then add to it.

Yes, we could do that. But in this program it would be ridiculous (in a different one it may be just the thing to do): Even the circumference of the Earth expressed in millimeters only takes 11 digits. Clearly, we cannot build a camera that large (not yet, anyway).

So, if the user enters such a huge number, he is either bored, or testing us, or trying to break into the system, or playing games—doing anything but designing a pinhole camera.

What will we do?

We will slap him in the face, in a manner of speaking:

17459765723452353453534535353530530534563507309676764423	???	???	???	???	???

To achieve that, we will simply ignore any leading zeros. Once we find a non–zero digit, we will initialize a counter to 0 and start taking three steps:

  1. Send the digit to the output.

  2. Append the digit to a buffer we will use later to produce the packed decimal we can send to the FPU.

  3. Increase the counter.

Now, while we are taking these three steps, we also need to watch out for one of two conditions:

  • If the counter grows above 18, we stop appending to the buffer. We continue reading the digits and sending them to the output.

  • If, or rather when, the next input character is not a digit, we are done inputting for now.

    Incidentally, we can simply discard the non–digit, unless it is a #, which we must return to the input stream. It starts a comment, so we must see it after we are done producing output and start looking for more input.

That still leaves one possibility uncovered: If all the user enters is a zero (or several zeros), we will never find a non–zero to display.

We can determine this has happened whenever our counter stays at 0. In that case we need to send 0 to the output, and perform another "slap in the face":

0	???	???	???	???	???

Once we have displayed the focal length and determined it is valid (greater than 0 but not exceeding 18 digits), we can calculate the pinhole diameter.

It is not by coincidence that pinhole contains the word pin. Indeed, many a pinhole literally is a pin hole, a hole carefully punched with the tip of a pin.

That is because a typical pinhole is very small. Our formula gets the result in millimeters. We will multiply it by 1000, so we can output the result in microns.

At this point we have yet another trap to face: Too much precision.

Yes, the FPU was designed for high precision mathematics. But we are not dealing with high precision mathematics. We are dealing with physics (optics, specifically).

Suppose we want to convert a truck into a pinhole camera (we would not be the first ones to do that!). Suppose its box is 12 meters long, so we have the focal length of 12000. Well, using Bender's constant, it gives us square root of 12000 multiplied by 0.04, which is 4.381780460 millimeters, or 4381.780460 microns.

Put either way, the result is absurdly precise. Our truck is not exactly 12000 millimeters long. We did not measure its length with such a precision, so stating we need a pinhole with the diameter of 4.381780460 millimeters is, well, deceiving. 4.4 millimeters would do just fine.

Note: I "only" used ten digits in the above example. Imagine the absurdity of going for all 18!

We need to limit the number of significant digits of our result. One way of doing it is by using an integer representing microns. So, our truck would need a pinhole with the diameter of 4382 microns. Looking at that number, we still decide that 4400 microns, or 4.4 millimeters is close enough.

Additionally, we can decide that no matter how big a result we get, we only want to display four significant digits (or any other number of them, of course). Alas, the FPU does not offer rounding to a specific number of digits (after all, it does not view the numbers as decimal but as binary).

We, therefore, must devise an algorithm to reduce the number of significant digits.

Here is mine (I think it is awkward—if you know a better one, please, let me know):

  1. Initialize a counter to 0.

  2. While the number is greater than or equal to 10000, divide it by 10 and increase the counter.

  3. Output the result.

  4. While the counter is greater than 0, output 0 and decrease the counter.

Note: The 10000 is only good if you want four significant digits. For any other number of significant digits, replace 10000 with 10 raised to the number of significant digits.

We will, then, output the pinhole diameter in microns, rounded off to four significant digits.

At this point, we know the focal length and the pinhole diameter. That means we have enough information to also calculate the f–number.

We will display the f–number, rounded to four significant digits. Chances are the f–number will tell us very little. To make it more meaningful, we can find the nearest normalized f–number, i.e., the nearest power of the square root of 2.

We do that by multiplying the actual f–number by itself, which, of course, will give us its square. We will then calculate its base–2 logarithm, which is much easier to do than calculating the base–square–root–of–2 logarithm! We will round the result to the nearest integer. Next, we will raise 2 to the result. Actually, the FPU gives us a good shortcut to do that: We can use the fscale op code to "scale" 1, which is analogous to shifting an integer left. Finally, we calculate the square root of it all, and we have the nearest normalized f–number.

If all that sounds overwhelming—or too much work, perhaps—it may become much clearer if you see the code. It takes 9 op codes altogether:

	fmul	st0, st0
	fld	st1
	fstp	st1

The first line, fmul st0, st0, squares the contents of the TOS (top of the stack, same as st, called st0 by nasm). The fld1 pushes 1 on the TOS.

The next line, fld st1, pushes the square back to the TOS. At this point the square is both in st and st(2) (it will become clear why we leave a second copy on the stack in a moment). st(1) contains 1.

Next, fyl2x calculates base–2 logarithm of st multiplied by st(1). That is why we placed 1 on st(1) before.

At this point, st contains the logarithm we have just calculated, st(1) contains the square of the actual f–number we saved for later.

frndint rounds the TOS to the nearest integer. fld1 pushes a 1. fscale shifts the 1 we have on the TOS by the value in st(1), effectively raising 2 to st(1).

Finally, fsqrt calculates the square root of the result, i.e., the nearest normalized f–number.

We now have the nearest normalized f–number on the TOS, the base–2 logarithm rounded to the nearest integer in st(1), and the square of the actual f–number in st(2). We are saving the value in st(2) for later.

But we do not need the contents of st(1) anymore. The last line, fstp st1, places the contents of st to st(1), and pops. As a result, what was st(1) is now st, what was st(2) is now st(1), etc. The new st contains the normalized f–number. The new st(1) contains the square of the actual f–number we have stored there for posterity.

At this point, we are ready to output the normalized f–number. Because it is normalized, we will not round it off to four significant digits, but will send it out in its full precision.

The normalized f-number is useful as long as it is reasonably small and can be found on our light meter. Otherwise we need a different method of determining proper exposure.

Earlier we have figured out the formula of calculating proper exposure at an arbitrary f–number from that measured at a different f–number.

Every light meter I have ever seen can determine proper exposure at f5.6. We will, therefore, calculate an "f5.6 multiplier," i.e., by how much we need to multiply the exposure measured at f5.6 to determine the proper exposure for our pinhole camera.

From the above formula we know this factor can be calculated by dividing our f–number (the actual one, not the normalized one) by 5.6, and squaring the result.

Mathematically, dividing the square of our f–number by the square of 5.6 will give us the same result.

Computationally, we do not want to square two numbers when we can only square one. So, the first solution seems better at first.


5.6 is a constant. We do not have to have our FPU waste precious cycles. We can just tell it to divide the square of the f–number by whatever 5.6 equals to. Or we can divide the f–number by 5.6, and then square the result. The two ways now seem equal.

But, they are not!

Having studied the principles of photography above, we remember that the 5.6 is actually square root of 2 raised to the fifth power. An irrational number. The square of this number is exactly 32.

Not only is 32 an integer, it is a power of 2. We do not need to divide the square of the f–number by 32. We only need to use fscale to shift it right by five positions. In the FPU lingo it means we will fscale it with st(1) equal to -5. That is much faster than a division.

So, now it has become clear why we have saved the square of the f–number on the top of the FPU stack. The calculation of the f5.6 multiplier is the easiest calculation of this entire program! We will output it rounded to four significant digits.

There is one more useful number we can calculate: The number of stops our f–number is from f5.6. This may help us if our f–number is just outside the range of our light meter, but we have a shutter which lets us set various speeds, and this shutter uses stops.

Say, our f–number is 5 stops from f5.6, and the light meter says we should use 1/1000 sec. Then we can set our shutter speed to 1/1000 first, then move the dial by 5 stops.

This calculation is quite easy as well. All we have to do is to calculate the base-2 logarithm of the f5.6 multiplier we had just calculated (though we need its value from before we rounded it off). We then output the result rounded to the nearest integer. We do not need to worry about having more than four significant digits in this one: The result is most likely to have only one or two digits anyway.

11.13.4 FPU Optimizations

In assembly language we can optimize the FPU code in ways impossible in high languages, including C.

Whenever a C function needs to calculate a floating–point value, it loads all necessary variables and constants into FPU registers. It then does whatever calculation is required to get the correct result. Good C compilers can optimize that part of the code really well.

It "returns" the value by leaving the result on the TOS. However, before it returns, it cleans up. Any variables and constants it used in its calculation are now gone from the FPU.

It cannot do what we just did above: We calculated the square of the f–number and kept it on the stack for later use by another function.

We knew we would need that value later on. We also knew we had enough room on the stack (which only has room for 8 numbers) to store it there.

A C compiler has no way of knowing that a value it has on the stack will be required again in the very near future.

Of course, the C programmer may know it. But the only recourse he has is to store the value in a memory variable.

That means, for one, the value will be changed from the 80-bit precision used internally by the FPU to a C double (64 bits) or even single (32 bits).

That also means that the value must be moved from the TOS into the memory, and then back again. Alas, of all FPU operations, the ones that access the computer memory are the slowest.

So, whenever programming the FPU in assembly language, look for the ways of keeping intermediate results on the FPU stack.

We can take that idea even further! In our program we are using a constant (the one we named PC).

It does not matter how many pinhole diameters we are calculating: 1, 10, 20, 1000, we are always using the same constant. Therefore, we can optimize our program by keeping the constant on the stack all the time.

Early on in our program, we are calculating the value of the above constant. We need to divide our input by 10 for every digit in the constant.

It is much faster to multiply than to divide. So, at the start of our program, we divide 10 into 1 to obtain 0.1, which we then keep on the stack: Instead of dividing the input by 10 for every digit, we multiply it by 0.1.

By the way, we do not input 0.1 directly, even though we could. We have a reason for that: While 0.1 can be expressed with just one decimal place, we do not know how many binary places it takes. We, therefore, let the FPU calculate its binary value to its own high precision.

We are using other constants: We multiply the pinhole diameter by 1000 to convert it from millimeters to microns. We compare numbers to 10000 when we are rounding them off to four significant digits. So, we keep both, 1000 and 10000, on the stack. And, of course, we reuse the 0.1 when rounding off numbers to four digits.

Last but not least, we keep -5 on the stack. We need it to scale the square of the f–number, instead of dividing it by 32. It is not by coincidence we load this constant last. That makes it the top of the stack when only the constants are on it. So, when the square of the f–number is being scaled, the -5 is at st(1), precisely where fscale expects it to be.

It is common to create certain constants from scratch instead of loading them from the memory. That is what we are doing with -5:

	fld1			; TOS =  1
	fadd	st0, st0	; TOS =  2
	fadd	st0, st0	; TOS =  4
	fld1			; TOS =  1
	faddp	st1, st0	; TOS =  5
	fchs			; TOS = -5

We can generalize all these optimizations into one rule: Keep repeat values on the stack!

Tip: PostScript® is a stack–oriented programming language. There are many more books available about PostScript than about the FPU assembly language: Mastering PostScript will help you master the FPU.

11.13.5 pinhole—The Code

;;;;;;; pinhole.asm ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
; Find various parameters of a pinhole camera construction and use
; Started:	 9-Jun-2001
; Updated:	10-Jun-2001
; Copyright (c) 2001 G. Adam Stanislav
; All rights reserved.

%include	'system.inc'

%define	BUFSIZE	2048

section	.data
align 4
ten	dd	10
thousand	dd	1000
tthou	dd	10000
fd.in	dd	stdin
fd.out	dd	stdout
envar	db	'PINHOLE='	; Exactly 8 bytes, or 2 dwords long
pinhole	db	'04,', 		; Bender's constant (0.04)
connors	db	'037', 0Ah	; Connors' constant
usg	db	'Usage: pinhole [-b] [-c] [-e] [-p <value>] [-o <outfile>] [-i <infile>]', 0Ah
usglen	equ	$-usg
iemsg	db	"pinhole: Can't open input file", 0Ah
iemlen	equ	$-iemsg
oemsg	db	"pinhole: Can't create output file", 0Ah
oemlen	equ	$-oemsg
pinmsg	db	"pinhole: The PINHOLE constant must not be 0", 0Ah
pinlen	equ	$-pinmsg
toobig	db	"pinhole: The PINHOLE constant may not exceed 18 decimal places", 0Ah
biglen	equ	$-toobig
huhmsg	db	9, '???'
separ	db	9, '???'
sep2	db	9, '???'
sep3	db	9, '???'
sep4	db	9, '???', 0Ah
huhlen	equ	$-huhmsg
header	db	'focal length in millimeters,pinhole diameter in microns,'
	db	'F-number,normalized F-number,F-5.6 multiplier,stops '
	db	'from F-5.6', 0Ah
headlen	equ	$-header

section .bss
ibuffer	resb	BUFSIZE
obuffer	resb	BUFSIZE
dbuffer	resb	20		; decimal input buffer
bbuffer	resb	10		; BCD buffer

section	.text
align 4
	call	write
	push	dword huhlen
	push	dword huhmsg
	push	dword [fd.out]
	add	esp, byte 12

align 4
	push	dword pinlen
	push	dword pinmsg
	push	dword stderr
	push	dword 4		; return failure

align 4
	push	dword biglen
	push	dword toobig
	push	dword stderr
	push	dword 5		; return failure

align 4
	push	dword iemlen
	push	dword iemsg
	push	dword stderr
	push	dword 1		; return failure

align 4
	push	dword oemlen
	push	dword oemsg
	push	dword stderr
	push	dword 2

align 4
	push	dword usglen
	push	dword usg
	push	dword stderr
	push	dword 3

align 4
global	_start
	add	esp, byte 8	; discard argc and argv[0]
	sub	esi, esi

	pop	ecx
	or	ecx, ecx
	je	near .getenv		; no more arguments

	; ECX contains the pointer to an argument
	cmp	byte [ecx], '-'
	jne	usage

	inc	ecx
	mov	ax, [ecx]
	inc	ecx

	cmp	al, 'o'
	jne	.i

	; Make sure we are not asked for the output file twice
	cmp	dword [fd.out], stdout
	jne	usage

	; Find the path to output file - it is either at [ECX+1],
	; i.e., -ofile --
	; or in the next argument,
	; i.e., -o file

	or	ah, ah
	jne	.openoutput
	pop	ecx
	jecxz	usage

	push	dword 420	; file mode (644 octal)
	push	dword 0200h | 0400h | 01h
	push	ecx
	jc	near oerr

	add	esp, byte 12
	mov	[fd.out], eax
	jmp	short .arg

	cmp	al, 'i'
	jne	.p

	; Make sure we are not asked twice
	cmp	dword [fd.in], stdin
	jne	near usage

	; Find the path to the input file
	or	ah, ah
	jne	.openinput
	pop	ecx
	or	ecx, ecx
	je near usage

	push	dword 0		; O_RDONLY
	push	ecx
	jc	near ierr		; open failed

	add	esp, byte 8
	mov	[fd.in], eax
	jmp	.arg

	cmp	al, 'p'
	jne	.c
	or	ah, ah
	jne	.pcheck

	pop	ecx
	or	ecx, ecx
	je	near usage

	mov	ah, [ecx]

	cmp	ah, '0'
	jl	near usage
	cmp	ah, '9'
	ja	near usage
	mov	esi, ecx
	jmp	.arg

	cmp	al, 'c'
	jne	.b
	or	ah, ah
	jne	near usage
	mov	esi, connors
	jmp	.arg

	cmp	al, 'b'
	jne	.e
	or	ah, ah
	jne	near usage
	mov	esi, pinhole
	jmp	.arg

	cmp	al, 'e'
	jne	near usage
	or	ah, ah
	jne	near usage
	mov	al, ','
	mov	[huhmsg], al
	mov	[separ], al
	mov	[sep2], al
	mov	[sep3], al
	mov	[sep4], al
	jmp	.arg

align 4
	; If ESI = 0, we did not have a -p argument,
	; and need to check the environment for "PINHOLE="
	or	esi, esi
	jne	.init

	sub	ecx, ecx

	pop	esi
	or	esi, esi
	je	.default	; no PINHOLE envar found

	; check if this envar starts with 'PINHOLE='
	mov	edi, envar
	mov	cl, 2		; 'PINHOLE=' is 2 dwords long
rep	cmpsd
	jne	.nextenv

	; Check if it is followed by a digit
	mov	al, [esi]
	cmp	al, '0'
	jl	.default
	cmp	al, '9'
	jbe	.init
	; fall through

align 4
	; We got here because we had no -p argument,
	; and did not find the PINHOLE envar.
	mov	esi, pinhole
	; fall through

align 4
	sub	eax, eax
	sub	ebx, ebx
	sub	ecx, ecx
	sub	edx, edx
	mov	edi, dbuffer+1
	mov	byte [dbuffer], '0'

	; Convert the pinhole constant to real
	cmp	al, '9'
	ja	.setconst
	cmp	al, '0'
	je	.processconst
	jb	.setconst

	inc	dl

	inc	cl
	cmp	cl, 18
	ja	near consttoobig
	jmp	short .constloop

align 4
	or	dl, dl
	je	near perr

	fild	dword [tthou]

	fild	dword [ten]
	fdivp	st1, st0

	fild	dword [thousand]
	mov	edi, obuffer

	mov	ebp, ecx
	call	bcdload

	fmul	st0, st2
	loop	.constdiv

	fadd	st0, st0
	fadd	st0, st0
	faddp	st1, st0

	; If we are creating a CSV file,
	; print header
	cmp	byte [separ], ','
	jne	.bigloop

	push	dword headlen
	push	dword header
	push	dword [fd.out]

	call	getchar
	jc	near done

	; Skip to the end of the line if you got '#'
	cmp	al, '#'
	jne	.num
	call	skiptoeol
	jmp	short .bigloop

	; See if you got a number
	cmp	al, '0'
	jl	.bigloop
	cmp	al, '9'
	ja	.bigloop

	; Yes, we have a number
	sub	ebp, ebp
	sub	edx, edx

	cmp	al, '0'
	je	.number0
	mov	dl, 1

	or	dl, dl		; Skip leading 0's
	je	.nextnumber
	push	eax
	call	putchar
	pop	eax
	inc	ebp
	cmp	ebp, 19
	jae	.nextnumber
	mov	[dbuffer+ebp], al

	call	getchar
	jc	.work
	cmp	al, '#'
	je	.ungetc
	cmp	al, '0'
	jl	.work
	cmp	al, '9'
	ja	.work
	jmp	short .number

	dec	esi
	inc	ebx

	; Now, do all the work
	or	dl, dl
	je	near .work0

	cmp	ebp, 19
	jae	near .toobig

	call	bcdload

	; Calculate pinhole diameter

	fld	st0	; save it
	fmul	st0, st3
	fld	st0
	fmul	st5
	sub	ebp, ebp

	; Round off to 4 significant digits
	fcom	st0, st7
	fstsw	ax
	jb	.printdiameter
	fmul	st0, st6
	inc	ebp
	jmp	short .diameter

	call	printnumber	; pinhole diameter

	; Calculate F-number

	fdivp	st1, st0
	fld	st0

	sub	ebp, ebp

	fcom	st0, st6
	fstsw	ax
	jb	.printfnumber
	fmul	st0, st5
	inc	ebp
	jmp	short .fnumber

	call	printnumber	; F number

	; Calculate normalized F-number
	fmul	st0, st0
	fld	st1
	fstp	st1

	sub	ebp, ebp
	call	printnumber

	; Calculate time multiplier from F-5.6

	fld	st0

	; Round off to 4 significant digits
	fcom	st0, st6
	fstsw	ax

	jb	.printfmul
	inc	ebp
	fmul	st0, st5
	jmp	short .fmul

	call	printnumber	; F multiplier

	; Calculate F-stops from 5.6

	fxch	st1

	sub	ebp, ebp
	call	printnumber

	mov	al, 0Ah
	call	putchar
	jmp	.bigloop

	mov	al, '0'
	call	putchar

align 4
	call	huh
	jmp	.bigloop

align 4
	call	write		; flush output buffer

	; close files
	push	dword [fd.in]

	push	dword [fd.out]


	; return success
	push	dword 0

align 4
	; Keep reading until you come to cr, lf, or eof
	call	getchar
	jc	done
	cmp	al, 0Ah
	jne	.cr

	cmp	al, 0Dh
	jne	skiptoeol

align 4
	or	ebx, ebx
	jne	.fetch

	call	read

	dec	ebx

	jecxz	.read
	call	write

	push	dword BUFSIZE
	mov	esi, ibuffer
	push	esi
	push	dword [fd.in]
	add	esp, byte 12
	mov	ebx, eax
	or	eax, eax
	je	.empty
	sub	eax, eax

align 4
	add	esp, byte 4

align 4
	inc	ecx
	cmp	ecx, BUFSIZE
	je	write

align 4
	jecxz	.ret	; nothing to write
	sub	edi, ecx	; start of buffer
	push	ecx
	push	edi
	push	dword [fd.out]
	add	esp, byte 12
	sub	eax, eax
	sub	ecx, ecx	; buffer is empty now

align 4
	; EBP contains the number of chars in dbuffer
	push	ecx
	push	esi
	push	edi

	lea	ecx, [ebp+1]
	lea	esi, [dbuffer+ebp-1]
	shr	ecx, 1


	mov	edi, bbuffer
	sub	eax, eax
	mov	[edi], eax
	mov	[edi+4], eax
	mov	[edi+2], ax

	sub	ax, 3030h
	shl	al, 4
	or	al, ah
	mov	[edi], al
	inc	edi
	loop	.loop

	fbld	[bbuffer]

	pop	edi
	pop	esi
	pop	ecx
	sub	eax, eax

align 4
	push	ebp
	mov	al, [separ]
	call	putchar

	; Print the integer at the TOS
	mov	ebp, bbuffer+9
	fbstp	[bbuffer]

	; Check the sign
	mov	al, [ebp]
	dec	ebp
	or	al, al
	jns	.leading

	; We got a negative number (should never happen)
	mov	al, '-'
	call	putchar

	; Skip leading zeros
	mov	al, [ebp]
	dec	ebp
	or	al, al
	jne	.first
	cmp	ebp, bbuffer
	jae	.leading

	; We are here because the result was 0.
	; Print '0' and return
	mov	al, '0'
	jmp	putchar

	; We have found the first non-zero.
	; But it is still packed
	test	al, 0F0h
	jz	.second
	push	eax
	shr	al, 4
	add	al, '0'
	call	putchar
	pop	eax
	and	al, 0Fh

	add	al, '0'
	call	putchar

	cmp	ebp, bbuffer
	jb	.done

	mov	al, [ebp]
	push	eax
	shr	al, 4
	add	al, '0'
	call	putchar
	pop	eax
	and	al, 0Fh
	add	al, '0'
	call	putchar

	dec	ebp
	jmp	short .next

	pop	ebp
	or	ebp, ebp
	je	.ret

	mov	al, '0'
	call	putchar
	dec	ebp
	jne	.zeros


The code follows the same format as all the other filters we have seen before, with one subtle exception:

We are no longer assuming that the end of input implies the end of things to do, something we took for granted in the character–oriented filters.

This filter does not process characters. It processes a language (albeit a very simple one, consisting only of numbers).

When we have no more input, it can mean one of two things:

  • We are done and can quit. This is the same as before.

  • The last character we have read was a digit. We have stored it at the end of our ASCII–to–float conversion buffer. We now need to convert the contents of that buffer into a number and write the last line of our output.

For that reason, we have modified our getchar and our read routines to return with the carry flag clear whenever we are fetching another character from the input, or the carry flag set whenever there is no more input.

Of course, we are still using assembly language magic to do that! Take a good look at getchar. It always returns with the carry flag clear.

Yet, our main code relies on the carry flag to tell it when to quit—and it works.

The magic is in read. Whenever it receives more input from the system, it just returns to getchar, which fetches a character from the input buffer, clears the carry flag and returns.

But when read receives no more input from the system, it does not return to getchar at all. Instead, the add esp, byte 4 op code adds 4 to ESP, sets the carry flag, and returns.

So, where does it return to? Whenever a program uses the call op code, the microprocessor pushes the return address, i.e., it stores it on the top of the stack (not the FPU stack, the system stack, which is in the memory). When a program uses the ret op code, the microprocessor pops the return value from the stack, and jumps to the address that was stored there.

But since we added 4 to ESP (which is the stack pointer register), we have effectively given the microprocessor a minor case of amnesia: It no longer remembers it was getchar that called read.

And since getchar never pushed anything before calling read, the top of the stack now contains the return address to whatever or whoever called getchar. As far as that caller is concerned, he called getchar, which returned with the carry flag set!

Other than that, the bcdload routine is caught up in the middle of a Lilliputian conflict between the Big–Endians and the Little–Endians.

It is converting the text representation of a number into that number: The text is stored in the big–endian order, but the packed decimal is little–endian.

To solve the conflict, we use the std op code early on. We cancel it with cld later on: It is quite important we do not call anything that may depend on the default setting of the direction flag while std is active.

Everything else in this code should be quite clear, providing you have read the entire chapter that precedes it.

It is a classical example of the adage that programming requires a lot of thought and only a little coding. Once we have thought through every tiny detail, the code almost writes itself.

11.13.6 Using pinhole

Because we have decided to make the program ignore any input except for numbers (and even those inside a comment), we can actually perform textual queries. We do not have to, but we can.

In my humble opinion, forming a textual query, instead of having to follow a very strict syntax, makes software much more user friendly.

Suppose we want to build a pinhole camera to use the 4x5 inch film. The standard focal length for that film is about 150mm. We want to fine–tune our focal length so the pinhole diameter is as round a number as possible. Let us also suppose we are quite comfortable with cameras but somewhat intimidated by computers. Rather than just have to type in a bunch of numbers, we want to ask a couple of questions.

Our session might look like this:

% pinhole


What size pinhole do I need for the focal length of 150?
150	490	306	362	2930	12
Hmmm... How about 160?
160	506	316	362	3125	12
Let's make it 155, please.
155	498	311	362	3027	12
Ah, let's try 157...
157	501	313	362	3066	12
156	500	312	362	3047	12
That's it! Perfect! Thank you very much!

We have found that while for the focal length of 150, our pinhole diameter should be 490 microns, or 0.49 mm, if we go with the almost identical focal length of 156 mm, we can get away with a pinhole diameter of exactly one half of a millimeter.

11.13.7 Scripting

Because we have chosen the # character to denote the start of a comment, we can treat our pinhole software as a scripting language.

You have probably seen shell scripts that start with:

#! /bin/sh



...because the blank space after the #! is optional.

Whenever UNIX is asked to run an executable file which starts with the #!, it assumes the file is a script. It adds the command to the rest of the first line of the script, and tries to execute that.

Suppose now that we have installed pinhole in /usr/local/bin/, we can now write a script to calculate various pinhole diameters suitable for various focal lengths commonly used with the 120 film.

The script might look something like this:

#! /usr/local/bin/pinhole -b -i
# Find the best pinhole diameter
# for the 120 film

### Standard

### Wide angle
30, 40, 50, 60, 70

### Telephoto
100, 120, 140

Because 120 is a medium size film, we may name this file medium.

We can set its permissions to execute, and run it as if it were a program:

% chmod 755 medium
% ./medium

UNIX will interpret that last command as:

% /usr/local/bin/pinhole -b -i ./medium

It will run that command and display:

80	358	224	256	1562	11
30	219	137	128	586	9
40	253	158	181	781	10
50	283	177	181	977	10
60	310	194	181	1172	10
70	335	209	181	1367	10
100	400	250	256	1953	11
120	438	274	256	2344	11
140	473	296	256	2734	11

Now, let us enter:

% ./medium -c

UNIX will treat that as:

% /usr/local/bin/pinhole -b -i ./medium -c

That gives it two conflicting options: -b and -c (Use Bender's constant and use Connors' constant). We have programmed it so later options override early ones—our program will calculate everything using Connors' constant:

80	331	242	256	1826	11
30	203	148	128	685	9
40	234	171	181	913	10
50	262	191	181	1141	10
60	287	209	181	1370	10
70	310	226	256	1598	11
100	370	270	256	2283	11
120	405	296	256	2739	11
140	438	320	362	3196	12

We decide we want to go with Bender's constant after all. We want to save its values as a comma–separated file:

% ./medium -b -e > bender
% cat bender
focal length in millimeters,pinhole diameter in microns,F-number,normalized F-number,F-5.6 multiplier,stops from F-5.6