Question

When I compile the following code to asm in GCC on cygwin:

int scheme_entry() {
  return 42;
}

using:

gcc -O3 --omit-frame-pointer -S test1.c

I get the following ‘ASM’ generated:

    .file   "test1.c"
    .text
    .p2align 4,,15
.globl _scheme_entry
    .def    _scheme_entry;  .scl    2;  .type   32; .endef
_scheme_entry:
    movl    $42, %eax
    ret

But the ‘MOVL’ command isn’t actually x86 ASM. From looking at the following lists:

http://ref.x86asm.net/geek.html#x0FA0

http://en.wikipedia.org/wiki/X86_instruction_listings

There is no MOVL command, but there is

CMOVL
CMOVLE
MOVLPS
MOVLPD
MOVLHPS

My question is – is gcc ASM “simplified ASM”? If so – how do I map it to ‘real ASM’?

Answer 1

As mentioned by ughoavgfhw, GCC outputs AT&T syntax by default, which is different to the Intel-style syntax you seem to be expecting. This behaviour, however, is configurable: you can request it to output Intel-style as follows:

gcc -masm=intel -O3 --omit-frame-pointer -S test1.c

with the key parameter being -masm=intel.

Using this command line, the assembly output I get (with a few unnecessary lines cut out for brevity) is as follows:

scheme_entry:
    mov eax, 42
    ret