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• Decoding equivalent assembly code of C code...

  - by puffadder

  Hi All, Wanting to see the output of the compiler (in assembly) for some C code, I wrote a simple program in C and generated its assembly file using gcc. The code is this: #include <stdio.h> int main() { int i = 0; if ( i == 0 ) { printf("testing\n"); } return 0; } The generated assembly for it is here (only the main function): _main: pushl %ebpz movl %esp, %ebp subl $24, %esp andl $-16, %esp movl $0, %eax addl $15, %eax addl $15, %eax shrl $4, %eax sall $4, %eax movl %eax, -8(%ebp) movl -8(%ebp), %eax call __alloca call ___main movl $0, -4(%ebp) cmpl $0, -4(%ebp) jne L2 movl $LC0, (%esp) call _printf L2: movl $0, %eax leave ret I am at an absolute loss to correlate the C code and assembly code. All that the code has to do is store 0 in a register and compare it with a constant 0 and take suitable action. But what is going on in the assembly ? Thanks in advance.

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• Switch case assembly level code

  - by puffadder

  Hi All, I am programming C on cygwin windows. After having done a bit of C programming and getting comfortable with the language, I wanted to look under the hood and see what the compiler is doing for the code that I write. So I wrote down a code block containing switch case statements and converted them into assembly using: gcc -S foo.c Here is the C source: switch(i) { case 1: { printf("Case 1\n"); break; } case 2: { printf("Case 2\n"); break; } case 3: { printf("Case 3\n"); break; } case 4: { printf("Case 4\n"); break; } case 5: { printf("Case 5\n"); break; } case 6: { printf("Case 6\n"); break; } case 7: { printf("Case 7\n"); break; } case 8: { printf("Case 8\n"); break; } case 9: { printf("Case 9\n"); break; } case 10: { printf("Case 10\n"); break; } default: { printf("Nothing\n"); break; } } Now the resultant assembly for the same is: movl $5, -4(%ebp) cmpl $10, -4(%ebp) ja L13 movl -4(%ebp), %eax sall $2, %eax movl L14(%eax), %eax jmp *%eax .section .rdata,"dr" .align 4 L14: .long L13 .long L3 .long L4 .long L5 .long L6 .long L7 .long L8 .long L9 .long L10 .long L11 .long L12 .text L3: movl $LC0, (%esp) call _printf jmp L2 L4: movl $LC1, (%esp) call _printf jmp L2 L5: movl $LC2, (%esp) call _printf jmp L2 L6: movl $LC3, (%esp) call _printf jmp L2 L7: movl $LC4, (%esp) call _printf jmp L2 L8: movl $LC5, (%esp) call _printf jmp L2 L9: movl $LC6, (%esp) call _printf jmp L2 L10: movl $LC7, (%esp) call _printf jmp L2 L11: movl $LC8, (%esp) call _printf jmp L2 L12: movl $LC9, (%esp) call _printf jmp L2 L13: movl $LC10, (%esp) call _printf L2: Now, in the assembly, the code is first checking the last case (i.e. case 10) first. This is very strange. And then it is copying 'i' into 'eax' and doing things that are beyond me. I have heard that the compiler implements some jump table for switch..case. Is it what this code is doing? Or what is it doing and why? Because in case of less number of cases, the code is pretty similar to that generated for if...else ladder, but when number of cases increases, this unusual-looking implementation is seen. Thanks in advance.

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• Need for prefixing a function with (void)...

  - by puffadder

  Hi All, I recently came across a rather unusual coding convention wherein the call for a function returning "void" is prefixed with (void). e.g. (void) MyFunction(); Is it any different from the function call like: MyFunction(); Has it got any advantage or is it yet another needless but there coding convention of some sort?

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