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Asked: 2026-05-26T11:09:14+00:00

I have created this code: #include <stdio.h> typedef unsigned int uint; uint in[2]={1,2},out[2]={3,4}; int

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I have created this code:

#include <stdio.h>

typedef unsigned int uint;
uint in[2]={1,2},out[2]={3,4};

int main() {

    in[0]=out[0]/10;
}

and compiled it with GCC (v4.4.5,no optimizations) on Linux, the resulting assembly is:

0000000000400474 <main>:
400474:       55                      push   rbp
400475:       48 89 e5                mov    rbp,rsp
400478:       8b 05 ae 03 20 00       mov    eax,DWORD PTR [rip+0x2003ae]        # 0082c <out>
40047e:       89 45 fc                mov    DWORD PTR [rbp-0x4],eax
400481:       ba cd cc cc cc          mov    edx,0xcccccccd
400486:       8b 45 fc                mov    eax,DWORD PTR [rbp-0x4]
400489:       f7 e2                   mul    edx
40048b:       89 d0                   mov    eax,edx
40048d:       c1 e8 03                shr    eax,0x3
400490:       89 05 8e 03 20 00       mov    DWORD PTR [rip+0x20038e],eax        # 600824 <in>
400496:       c9                      leave
400497:       c3                      ret
400498:       90                      nop
400499:       90                      nop
40049a:       90                      nop
40049b:       90                      nop
40049c:       90                      nop
40049d:       90                      nop
40049e:       90                      nop
40049f:       90                      nop

Now, the question is: what is this code doing on line #5 ?

40047e:       89 45 fc                mov    DWORD PTR [rbp-0x4],eax

isn’t it storing the value it got from out[0] again in some place in memory? Why so? I didn’t tell it to read and write immediatly to some location.

Now, this temporal variable appears again at the address 400486 on line #7:

400486:       8b 45 fc                mov    eax,DWORD PTR [rbp-0x4]

It looks like in this example GCC is producing very inefficient code, and it will evict the cache line because of these temporal storages. Please confirm, maybe there is something I am not getting.

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1 Answer

  1. Editorial Team

    GCC makes very inefficient code when compiling on -O0 – what you’re seeing is basically a raw translation of its internal representation of the program. This internal representation includes a number of temporary variables, and your load/store pair here is a value passing through such a temporary. On higher optimization levels these kinds of useless loads/stores will mostly be eliminated; however on -O0 even the simplest of analysis is disabled.

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