有人可以解释为什么GetTickCount只使用一个结果时连续调用 3 次吗？
这不是组装的全部功能，只是具有 GetTickCount 多次的部分。我不认为这是编程中的人为错误谁会这样做？是某种优化技巧还是某些编译器搞砸了？
下面的输出来自 OllyDbg
00412EB0 /$ 81EC F4010000 SUB ESP,1F4
00412EB6 |. 53 PUSH EBX
00412EB7 |. 55 PUSH EBP
00412EB8 |. 56 PUSH ESI
00412EB9 |. 8B35 14B24D00 MOV ESI,DWORD PTR DS:[<&KERNEL32.GetTick>; kernel32.GetTickCount
00412EBF |. 57 PUSH EDI
00412EC0 |. FFD6 CALL ESI ; [GetTickCount
00412EC2 |. FFD6 CALL ESI ; [GetTickCount
00412EC4 |. FFD6 CALL ESI ; [GetTickCount
00412EC6 |. 8BC8 MOV ECX,EAX
00412EC8 |. B8 CDCCCCCC MOV EAX,CCCCCCCD
00412ECD |. F7E1 MUL ECX
00412ECF |. C1EA 03 SHR EDX,3
00412ED2 |. 895424 14 MOV DWORD PTR SS:[ESP+14],EDX
00412ED6 |. FFD6 CALL ESI ; [GetTickCount
00412ED8 |. FFD6 CALL ESI ; [GetTickCount
00412EDA |. FFD6 CALL ESI ; [GetTickCount
00412EDC |. 8BD0 MOV EDX,EAX
00412EDE |. B8 CDCCCCCC MOV EAX,CCCCCCCD
00412EE3 |. F7E2 MUL EDX
00412EE5 |. C1EA 03 SHR EDX,3
00412EE8 |. 895424 1C MOV DWORD PTR SS:[ESP+1C],EDX
00412EEC |. FFD6 CALL ESI ; [GetTickCount
00412EEE |. 8B0D A87C4300 MOV ECX,DWORD PTR DS:[437CA8]
00412EF4 |. 33C0 XOR EAX,EAX
00412EF6 |. 3BC8 CMP ECX,EAX
00412EF8 |. 894424 18 MOV DWORD PTR SS:[ESP+18],EAX
00412EFC |. A3 20594D00 MOV DWORD PTR DS:[4D5920],EAX
Hex-Ray 的反编译结果
DWORD GetTickCountDivideddBy10; // [sp+14h] [bp-1F0h]@1
unsigned int v41; // [sp+18h] [bp-1ECh]@1
DWORD GetTickCountDividedBy10; // [sp+1Ch] [bp-1E8h]@1
GetTickCount();
GetTickCount();
GetTickCountDivideddBy10 = GetTickCount() / 0xA;
GetTickCount();
GetTickCount();
GetTickCountDividedBy10 = GetTickCount() / 0xA;
GetTickCount();
v41 = 0;
dword_4D5920 = 0;