c语言冒泡算法循环次数,C语言冒泡算法汇编解读

3: void _sort(int arr[], int n ) {

004010B0 push ebp

004010B1 mov ebp,esp

004010B3 sub esp,50h

004010B6 push ebx

004010B7 push esi

004010B8 push edi

004010B9 lea edi,[ebp-50h]

004010BC mov ecx,14h

004010C1 mov eax,0CCCCCCCCh

004010C6 rep stos dword ptr [edi]

4: int len = n,

004010C8 mov eax,dword ptr [ebp+0Ch]

004010CB mov dword ptr [ebp-4],eax

5: _i = 0,

004010CE mov dword ptr [ebp-8],0

6: _j = 0,

004010D5 mov dword ptr [ebp-0Ch],0

7: _t = 0;

004010DC mov dword ptr [ebp-10h],0

8: // len = sizeof(arr) / sizeof(int);

9: for(_i ; _i < len; _i++) {

004010E3 jmp _sort+3Eh (004010ee)

004010E5 mov ecx,dword ptr [ebp-8]

004010E8 add ecx,1

004010EB mov dword ptr [ebp-8],ecx

004010EE mov edx,dword ptr [ebp-8]

004010F1 cmp edx,dword ptr [ebp-4]

004010F4 jge _sort+0A1h (00401151)

10: for(_j = _i; _j < len; _j++) {

004010F6 mov eax,dword ptr [ebp-8]

004010F9 mov dword ptr [ebp-0Ch],eax

004010FC jmp _sort+57h (00401107)

004010FE mov ecx,dword ptr [ebp-0Ch]

00401101 add ecx,1

00401104 mov dword ptr [ebp-0Ch],ecx

00401107 mov edx,dword ptr [ebp-0Ch]

0040110A cmp edx,dword ptr [ebp-4]

0040110D jge _sort+9Fh (0040114f)

11: if (arr[_i] > arr[_j]) {

0040110F mov eax,dword ptr [ebp-8]

00401112 mov ecx,dword ptr [ebp+8]

00401115 mov edx,dword ptr [ebp-0Ch]

00401118 mov esi,dword ptr [ebp+8]

0040111B mov eax,dword ptr [ecx+eax*4]

0040111E cmp eax,dword ptr [esi+edx*4]

00401121 jle _sort+9Dh (0040114d)

12: _t = arr[_j];

00401123 mov ecx,dword ptr [ebp-0Ch]

00401126 mov edx,dword ptr [ebp+8]

00401129 mov eax,dword ptr [edx+ecx*4]

0040112C mov dword ptr [ebp-10h],eax

13: arr[_j] = arr[_i];

0040112F mov ecx,dword ptr [ebp-0Ch]

00401132 mov edx,dword ptr [ebp+8]

00401135 mov eax,dword ptr [ebp-8]

00401138 mov esi,dword ptr [ebp+8]

0040113B mov eax,dword ptr [esi+eax*4]

0040113E mov dword ptr [edx+ecx*4],eax

14: arr[_i] = _t;

00401141 mov ecx,dword ptr [ebp-8]

00401144 mov edx,dword ptr [ebp+8]

00401147 mov eax,dword ptr [ebp-10h]

0040114A mov dword ptr [edx+ecx*4],eax

15: }

16: }

0040114D jmp _sort+4Eh (004010fe)

17: }

0040114F jmp _sort+35h (004010e5)

18: //return ;

19: }

请各位大礼告知如何解读汇编代码。