X86 assembly - Handling the IDIV instruction

Question

I am currently writing a simple C compiler, that takes a .c file as input and generates assembly code (X86, AT&T syntax). Everyting is good, but when I try to execute a IDIVQ instruction, I get a floating-point exception. Here's my input:

int mymain(int x){   int d;   int e;   d = 3;   e = 6 / d;   return e; } 

And here is my generated code:

mymain: .LFB1:     .cfi_startproc     pushq   %rbp     .cfi_def_cfa_offset 16     movq    %rsp, %rbp     .cfi_offset 6, -16     .cfi_def_cfa_register 6     movq    %rdi, -40(%rbp)     movq    $3, -8(%rbp)     movq    $6, %rax     movq    -8(%rbp), %rdx     movq    %rdx, %rbx     idivq   %rbx     movq    %rax, -16(%rbp)     movq    -16(%rbp), %rax     leave     .cfi_def_cfa 7, 8     ret     .cfi_endproc .LFE1:     .size mymain, .-mymain 

According to http://www.cs.virginia.edu/~evans/cs216/guides/x86.html, idivq %rbx should produce 6/d (the quotient) in %rax. But I'm getting a floating-point exception, and I can't seem to find the problem.

Any help will be much appreciated!

Answer 1

The first part of Mysticials answer is correct, idiv does a 128/64 bit division, so the value of rdx, which holds the upper 64 bit from the dividend must not contain a random value. But a zero extension is the wrong way to go.

As you have signed variables, you need to sign extend rax to rdx:rax. There is a specific instruction for this, cqto (convert quad to oct) in AT&T and cqo in Intel syntax. AFAIK newer versions of gas accept both names.

movq    %rdx, %rbx cqto                  # sign extend rax to rdx:rax idivq   %rbx