Why does LLVM add two extra instructions for the same program?
I am compiling this C program and comparing the generated assembly code:
int main(){ return 0; }
GCC gives this main function (cc hello.c -S
):
_main:
LFB2:
pushq %rbp
LCFI0:
movq %rsp, %rbp
LCFI1:
movl $0, %eax
leave
ret
LLVM gives this main function (clang hello.c -S
):
_main:
Leh_func_begin0:
pushq %rbp
Ltmp0:
movq %rsp, %rbp
Ltmp1:
movl $0, %eax
movl $0, -4(%rbp)
popq %rbp
ret
Leh_func_end0:
What are movl $0, -4(%rbp)
and popq %rbp
ne开发者_如何学运维eded for? Moving something on the stack and popping it directly afterwards seems useless to me.
The movl $0, -4(%rbp)
instruction is dead, because this is unoptimized code. Try passing in -O
to both compilers to see what changes.
Actually, they're comparable. Leave is a high level instruction:
From the Intel manual:
16-bit: C9 LEAVE A Valid Valid Set SP to BP, then pop BP.
32-bit: C9 LEAVE A N.E. Valid Set ESP to EBP, then pop EBP.
64-bit: C9 LEAVE A Valid N.E. Set RSP to RBP, then pop RBP.
basically, leave is equivalent to
movq %rbp, %rsp
popq %rbp
It looks like LLVM is using a traditional function prolog/epilog, whereas GCC is taking advantage of the fact that the entry point doesn't need to clean up
精彩评论