指標訪問與數組訪問的效率分析

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作者:小馬

 

很多人都知道指標訪問比數組訪問效率高, 但很少人明白其中的原理, 這篇文章從彙編層級上分析一下原因.

我用一個"reverse"函數來作為例子, 分別用指標和數組的形式來實現這個函數.

//指標訪問void reverse(char *pStr,int nLen){ char *pStart=pStr;//指向字串開始 char *pEnd=pStr+nLen-1;//指向字串結尾 char chTmp;  while(pEnd > pStart) {   chTmp = *pStart;  *pStart = *pEnd;  *pEnd = chTmp;    pStart++;  pEnd--; }}

上面的函數在vc6下的反組譯碼代碼,我加了一些注釋方便理解

19:   void reverse(char *pStr,int nLen)20:   {00401070   push        ebp00401071   mov         ebp,esp00401073   sub         esp,4Ch00401076   push        ebx00401077   push        esi00401078   push        edi00401079   lea         edi,[ebp-4Ch]0040107C   mov         ecx,13h00401081   mov         eax,0CCCCCCCCh00401086   rep stos    dword ptr [edi]//調用子函數時進堆棧儲存資訊. 00401088   mov         eax,dword ptr [ebp+8]0040108B   mov         dword ptr [ebp-4],eax//char *pStart=pStr;0040108E   mov         ecx,dword ptr [ebp+0Ch]00401091   mov         edx,dword ptr [ebp+8]00401094   lea         eax,[edx+ecx-1]00401098   mov         dword ptr [ebp-8],eax//char *pEnd=pStr+nLen-1;   0040109B   mov         ecx,dword ptr [ebp-8]0040109E   cmp         ecx,dword ptr [ebp-4]//while(pEnd > pStart)004010A1   jbe         reverse+61h (004010d1)      004010A3   mov         edx,dword ptr [ebp-4]//chTmp = *pStart;004010A6   mov         al,byte ptr [edx]//取一個位元組004010A8   mov         byte ptr [ebp-0Ch],al    004010AB   mov         ecx,dword ptr [ebp-4]//*pStart = *pEnd;004010AE   mov         edx,dword ptr [ebp-8]004010B1   mov         al,byte ptr [edx]//從pEnd中取一個位元組004010B3   mov         byte ptr [ecx],al        004010B5   mov         ecx,dword ptr [ebp-8]// *pEnd = chTmp;004010B8   mov         dl,byte ptr [ebp-0Ch]004010BB   mov         byte ptr [ecx],dl         004010BD   mov         eax,dword ptr [ebp-4]//pStart++;004010C0   add         eax,1004010C3   mov         dword ptr [ebp-4],eax          004010C6   mov         ecx,dword ptr [ebp-8]//pEnd--;004010C9   sub         ecx,1004010CC   mov         dword ptr [ebp-8],ecx}

 

//數組訪問void reverse(char *pStr,int nLen){ int i = 0; int j = nLen-1; char chTmp;  while(j>i) {   chTmp=pStr[i];  pStr[i]=pStr[j];  pStr[j]=chTmp;    i++;  j--;     }}

 

這是上面代碼的反組譯碼

19:   void reverse(char *pStr,int nLen)20:   {004010D0   push        ebp004010D1   mov         ebp,esp004010D3   sub         esp,4Ch004010D6   push        ebx004010D7   push        esi004010D8   push        edi004010D9   lea         edi,[ebp-4Ch]004010DC   mov         ecx,13h004010E1   mov         eax,0CCCCCCCCh004010E6   rep stos    dword ptr [edi]004010E8   mov         dword ptr [ebp-4],0004010EF   mov         eax,dword ptr [ebp+0Ch]004010F2   sub         eax,1004010F5   mov         dword ptr [ebp-8],eax 004010F8   mov         ecx,dword ptr [ebp-8]004010FB   cmp         ecx,dword ptr [ebp-4]004010FE   jle         reverse+6Ah (0040113a)    00401100   mov         edx,dword ptr [ebp+8] //注意這裡00401103   add         edx,dword ptr [ebp-4] 00401106   mov         al,byte ptr [edx]00401108   mov         byte ptr [ebp-0Ch],al0040110B   mov         ecx,dword ptr [ebp+8] //注意這裡0040110E   add         ecx,dword ptr [ebp-4] 00401111   mov         edx,dword ptr [ebp+8] //注意這裡00401114   add         edx,dword ptr [ebp-8] 00401117   mov         al,byte ptr [edx]00401119   mov         byte ptr [ecx],al0040111B   mov         ecx,dword ptr [ebp+8] //注意這裡0040111E   add         ecx,dword ptr [ebp-8] 00401121   mov         dl,byte ptr [ebp-0Ch]00401124   mov         byte ptr [ecx],dl00401126   mov         eax,dword ptr [ebp-4]00401129   add         eax,10040112C   mov         dword ptr [ebp-4],eax0040112F   mov         ecx,dword ptr [ebp-8]00401132   sub         ecx,100401135   mov         dword ptr [ebp-8],ecx}

兩個函數反組譯碼後的代碼基本是一致的,但第二段彙編中多了幾個add語句, 這是因為在用數組下標的形式訪問字串時,總要對數組的首地址做加減操作才能定位到想要的值, 而指標的實現則一開始就copy了一個首地址，不需要再用加減來定位.顯然在這個例子中，用指標來實現reverse函數效率要高.