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Allocating static buffer in processing loop

Posted on 2003-11-21
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Last Modified: 2010-04-02
Are there any pitfalls to allocating memory in a processing loop. I am not a fan of this concept, but is there a problem doing this? Exactly what is occuring by doing this? What will be gained, or lost by allocating outside the loop and clearing the beffer before each use?

  for(ll->rewind(); ll->data() ; (*ll)++ )
  {
    char temp[80];
    sprintf(temp, (char *)"  %s", (char *)ll->data());
    msg_list.append(ll_strdup(temp));
  }
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Question by:rayskelton
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5 Comments
 
LVL 6

Expert Comment

by:GloomyFriar
ID: 9798161
The buffer is allocated in stack as usual.
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LVL 6

Expert Comment

by:GloomyFriar
ID: 9798221
void func1(void)
{
  char buff[80];

  for(int i=1; i<3; i++)
  {
    buff[i] = i;
  }
}

void func2(void)
{
  for(int i=1; i<3; i++)
  {
    char buff[80];
    buff[i] = i;
  }
}



484:  void func1(void)
485:  {
00402050 55                   push        ebp
00402051 8B EC                mov         ebp,esp
00402053 81 EC 94 00 00 00    sub         esp,94h
00402059 53                   push        ebx
0040205A 56                   push        esi
0040205B 57                   push        edi
0040205C 8D BD 6C FF FF FF    lea         edi,[ebp-94h]
00402062 B9 25 00 00 00       mov         ecx,25h
00402067 B8 CC CC CC CC       mov         eax,0CCCCCCCCh
0040206C F3 AB                rep stos    dword ptr [edi]
486:    char buff[80];
487:
488:    for(int i=1; i<3; i++)
0040206E C7 45 AC 01 00 00 00 mov         dword ptr [ebp-54h],1
00402075 EB 09                jmp         func1+30h (00402080)
00402077 8B 45 AC             mov         eax,dword ptr [ebp-54h]
0040207A 83 C0 01             add         eax,1
0040207D 89 45 AC             mov         dword ptr [ebp-54h],eax
00402080 83 7D AC 03          cmp         dword ptr [ebp-54h],3
00402084 7D 0C                jge         func1+42h (00402092)
489:    {
490:      buff[i] = i;
00402086 8B 4D AC             mov         ecx,dword ptr [ebp-54h]
00402089 8A 55 AC             mov         dl,byte ptr [ebp-54h]
0040208C 88 54 0D B0          mov         byte ptr [ebp+ecx-50h],dl
491:    }
00402090 EB E5                jmp         func1+27h (00402077)
492:  }


494:  void func2(void)
495:  {
00401520 55                   push        ebp
00401521 8B EC                mov         ebp,esp
00401523 81 EC 94 00 00 00    sub         esp,94h
00401529 53                   push        ebx
0040152A 56                   push        esi
0040152B 57                   push        edi
0040152C 8D BD 6C FF FF FF    lea         edi,[ebp-94h]
00401532 B9 25 00 00 00       mov         ecx,25h
00401537 B8 CC CC CC CC       mov         eax,0CCCCCCCCh
0040153C F3 AB                rep stos    dword ptr [edi]
496:    for(int i=1; i<3; i++)
0040153E C7 45 FC 01 00 00 00 mov         dword ptr [ebp-4],1
00401545 EB 09                jmp         func2+30h (00401550)
00401547 8B 45 FC             mov         eax,dword ptr [ebp-4]
0040154A 83 C0 01             add         eax,1
0040154D 89 45 FC             mov         dword ptr [ebp-4],eax
00401550 83 7D FC 03          cmp         dword ptr [ebp-4],3
00401554 7D 0C                jge         func2+42h (00401562)
497:    {
498:      char buff[80];
499:      buff[i] = i;
00401556 8B 4D FC             mov         ecx,dword ptr [ebp-4]
00401559 8A 55 FC             mov         dl,byte ptr [ebp-4]
0040155C 88 54 0D AC          mov         byte ptr buff[ecx],dl
500:    }
00401560 EB E5                jmp         func2+27h (00401547)
501:  }
0
 
LVL 6

Expert Comment

by:GloomyFriar
ID: 9798246
As you can see above the func1() and func2 are identical.
0
 

Author Comment

by:rayskelton
ID: 9798604
So there is no reallocation each iteration through the loop?
0
 
LVL 6

Accepted Solution

by:
GloomyFriar earned 500 total points
ID: 9799448
Yes. But its for MSVC compiler. I can't check it for other compilers.
But, imho, the other compilers will do the same.
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