We can improve the efficiency of our code by buffering our input and output. We create an input buffer and read a whole sequence of bytes at one time. Then we fetch them one by one from the buffer.
We also create an output buffer. We store our output in it until it is full. At that time we ask the kernel to write the contents of the buffer to stdout.
The program ends when there is no more input. But we still need to ask the kernel to write the contents of our output buffer to stdout one last time, otherwise some of our output would make it to the output buffer, but never be sent out. Do not forget that, or you will be wondering why some of your output is missing.
%include 'system.inc' %define BUFSIZE 2048 section .data hex db '0123456789ABCDEF' section .bss ibuffer resb BUFSIZE obuffer resb BUFSIZE section .text global _start _start: sub eax, eax sub ebx, ebx sub ecx, ecx mov edi, obuffer .loop: ; read a byte from stdin call getchar ; convert it to hex mov dl, al shr al, 4 mov al, [hex+eax] call putchar mov al, dl and al, 0Fh mov al, [hex+eax] call putchar mov al, ' ' cmp dl, 0Ah jne .put mov al, dl .put: call putchar jmp short .loop align 4 getchar: or ebx, ebx jne .fetch call read .fetch: lodsb dec ebx ret read: push dword BUFSIZE mov esi, ibuffer push esi push dword stdin sys.read add esp, byte 12 mov ebx, eax or eax, eax je .done sub eax, eax ret align 4 .done: call write ; flush output buffer push dword 0 sys.exit align 4 putchar: stosb inc ecx cmp ecx, BUFSIZE je write ret align 4 write: sub edi, ecx ; start of buffer push ecx push edi push dword stdout sys.write add esp, byte 12 sub eax, eax sub ecx, ecx ; buffer is empty now ret
We now have a third section in the source code, named .bss. This section is not included in our executable file, and,
therefore, cannot be initialized. We use resb instead of
db. It simply reserves the requested size of uninitialized
memory for our use.
We take advantage of the fact that the system does not modify the registers: We
use registers for what, otherwise, would have to be global variables stored in the .data section. This is also why the UNIX® convention of passing parameters to system calls on
the stack is superior to the Microsoft convention of passing them in the registers: We
can keep the registers for our own use.
We use EDI and ESI as
pointers to the next byte to be read from or written to. We use EBX and ECX to keep count of the
number of bytes in the two buffers, so we know when to dump the output to, or read more
input from, the system.
Let us see how it works now:
% nasm -f elf hex.asm % ld -s -o hex hex.o % ./hex Hello, World! Here I come! 48 65 6C 6C 6F 2C 20 57 6F 72 6C 64 21 0A 48 65 72 65 20 49 20 63 6F 6D 65 21 0A ^D %
Not what you expected? The program did not print the output until we pressed ^D. That is easy to fix by inserting three lines of code to write
the output every time we have converted a new line to 0A. I
have marked the three lines with > (do not copy the > in your hex.asm).
%include 'system.inc' %define BUFSIZE 2048 section .data hex db '0123456789ABCDEF' section .bss ibuffer resb BUFSIZE obuffer resb BUFSIZE section .text global _start _start: sub eax, eax sub ebx, ebx sub ecx, ecx mov edi, obuffer .loop: ; read a byte from stdin call getchar ; convert it to hex mov dl, al shr al, 4 mov al, [hex+eax] call putchar mov al, dl and al, 0Fh mov al, [hex+eax] call putchar mov al, ' ' cmp dl, 0Ah jne .put mov al, dl .put: call putchar > cmp al, 0Ah > jne .loop > call write jmp short .loop align 4 getchar: or ebx, ebx jne .fetch call read .fetch: lodsb dec ebx ret read: push dword BUFSIZE mov esi, ibuffer push esi push dword stdin sys.read add esp, byte 12 mov ebx, eax or eax, eax je .done sub eax, eax ret align 4 .done: call write ; flush output buffer push dword 0 sys.exit align 4 putchar: stosb inc ecx cmp ecx, BUFSIZE je write ret align 4 write: sub edi, ecx ; start of buffer push ecx push edi push dword stdout sys.write add esp, byte 12 sub eax, eax sub ecx, ecx ; buffer is empty now ret
Now, let us see how it works:
% nasm -f elf hex.asm % ld -s -o hex hex.o % ./hex Hello, World! 48 65 6C 6C 6F 2C 20 57 6F 72 6C 64 21 0A Here I come! 48 65 72 65 20 49 20 63 6F 6D 65 21 0A ^D %
Not bad for a 644-byte executable, is it!
Note: This approach to buffered input/output still contains a hidden danger. I will discuss—and fix—it later, when I talk about the dark side of buffering.
Warning: This may be a somewhat advanced topic, mostly of interest to programmers familiar with the theory of compilers. If you wish, you may skip to the next section, and perhaps read this later.
While our sample program does not require it, more sophisticated filters often need to look ahead. In other words, they may need to see what the next character is (or even several characters). If the next character is of a certain value, it is part of the token currently being processed. Otherwise, it is not.
For example, you may be parsing the input stream for a textual string (e.g., when implementing a language compiler): If a character is followed by another character, or perhaps a digit, it is part of the token you are processing. If it is followed by white space, or some other value, then it is not part of the current token.
This presents an interesting problem: How to return the next character back to the input stream, so it can be read again later?
One possible solution is to store it in a character variable, then set a flag. We
can modify getchar to check the flag, and if it is set,
fetch the byte from that variable instead of the input buffer, and reset the flag. But,
of course, that slows us down.
The C language has an ungetc() function, just for
that purpose. Is there a quick way to implement it in our code? I would like you to
scroll back up and take a look at the getchar procedure and
see if you can find a nice and fast solution before reading the next paragraph. Then come
back here and see my own solution.
The key to returning a character back to the stream is in how we are getting the characters to start with:
First we check if the buffer is empty by testing the value of EBX. If it is zero, we call the read
procedure.
If we do have a character available, we use lodsb,
then decrease the value of EBX. The lodsb instruction is effectively identical to:
mov al, [esi] inc esi
The byte we have fetched remains in the buffer until the next time read is called. We do not know when that happens, but we do know
it will not happen until the next call to getchar. Hence,
to "return" the last-read byte back to the stream, all we have to do is decrease the
value of ESI and increase the value of EBX:
ungetc: dec esi inc ebx ret
But, be careful! We are perfectly safe doing this if our look-ahead is at most
one character at a time. If we are examining more than one upcoming character and call
ungetc several times in a row, it will work most of the
time, but not all the time (and will be tough to debug). Why?
Because as long as getchar does not have to call
read, all of the pre-read bytes are still in the buffer,
and our ungetc works without a glitch. But the moment getchar calls read, the contents of
the buffer change.
We can always rely on ungetc working properly on
the last character we have read with getchar, but not on
anything we have read before that.
If your program reads more than one byte ahead, you have at least two choices:
If possible, modify the program so it only reads one byte ahead. This is the simplest solution.
If that option is not available, first of all determine the maximum number of
characters your program needs to return to the input stream at one time. Increase that
number slightly, just to be sure, preferably to a multiple of 16—so it aligns
nicely. Then modify the .bss section of your code, and
create a small "spare" buffer right before your input buffer, something like this:
section .bss resb 16 ; or whatever the value you came up with ibuffer resb BUFSIZE obuffer resb BUFSIZE
You also need to modify your ungetc to pass the
value of the byte to unget in AL:
ungetc: dec esi inc ebx mov [esi], al ret
With this modification, you can call ungetc up to
17 times in a row safely (the first call will still be within the buffer, the remaining
16 may be either within the buffer or within the "spare").