Signed overflow:
void BreakFor() { for (int i = 1; i > 0; i++) { printf("%d \r\n", i); } }
The above procedure is not an infinite loop, when a signed integer number increased to the maximum and continue to carry plus a sign bit will be modified, thus becoming negative.
Increment, decrement
98: int nVarOne = argc; 012665EE 8B 45 08 mov eax,dword ptr [argc] 012665F1 89 45 F8 mov dword ptr [nVarOne],eax 99: int nVarTwo = argc; 012665F4 8B 45 08 mov eax,dword ptr [argc] 012665F7 89 45 EC mov dword ptr [nVarTwo],eax 100: nVarTwo = 5 + (nVarOne++); 012665FA 8B 45 the F8 MOV EAX, DWORD PTR [nVarOne] 012665FD 83 C0 05 the Add EAX, . 5 01.2666 million 89 45 EC MOV DWORD PTR [nVarTwo], EAX 01,266,603 8B 4D the F8 MOV ECX, DWORD PTR [nVarOne] 01,266,606 83 a C1 01 the Add ECX , 1 // postfix increment, plus one at the end of the statement
01,266,609 89 4D F8 mov DWORD ptr [nVarOne], ECX 101: = NVarTwo . 5 + (++ nVarOne) ; 0126660C 8B 45 the F8 MOV EAX, DWORD PTR [nVarOne] 0126660F 83 C0 01 the Add EAX, . 1 // prefix increment, the first increment 01,266,612 89 45 the F8 MOV DWORD PTR [nVarOne] , EAX 01,266,615 8B 4D the F8 MOV ECX, DWORD PTR [nVarOne] 01,266,618 83 a C1 05 the Add ECX, . 5 0126661B 89 4D EC MOV dword ptr [nVarTwo],ecx 102: 103: nVarOne = 5 + (nVarTwo--); 0126661E 8B 45 EC mov eax,dword ptr [nVarTwo] 01266621 83 C0 05 add eax,5 01266624 89 45 F8 mov dword ptr [nVarOne],eax 01266627 8B 4D EC mov ecx,dword ptr [nVarTwo] 0126662A 83 E9 01 sub ecx,1 0126662D 89 4D EC mov dword ptr [nVarTwo],ecx 104: nVarOne = 5 + (--nVarTwo); 01266630 8B 45 EC mov eax,dword ptr [nVarTwo] 01266633 83 E8 01 sub eax,1 01266636 89 45 EC mov dword ptr [nVarTwo],eax 01266639 8B 4D EC mov ecx,dword ptr [nVarTwo] 0126663C 83 C1 05 add ecx,5 0126663F 89 4D F8 mov dword ptr [nVarOne],ecx 105: }
Expressions truncated:
124 : (nNumber == 0 ) || (nNumber + = Accumulation (nNumber - . 1 )) ; 0126650E 83 7D 08 00 CMP DWORD PTR [nNumber], 0 01,266,512 74 15 JE Accumulation + 39h (01266529h) // preceding zero truncated expression jump directly 01,266,514 8B 45 08 MOV EAX, DWORD PTR [nNumber] 01,266,517 83 E8 01 Sub EAX, . 1 0126651A 50 Push EAX 0126651B E8 B4 the FE 8B the FF Call Accumulation (012519ABh) // 01266520 83 C4 04 add esp,4 01266523 03 45 08 add eax,dword ptr [nNumber] 01266526 89 45 08 mov dword ptr [nNumber],eax 125: return nNumber; 01266529 8B 45 08 mov eax,dword ptr [nNumber] 126: }
Conditional expression:
- Scheme 1: Expression 1 is a simple comparison, the difference between the expression 2 and expression 3 both equal to 1;
return argc == 5 ? 5 : 6;
00E44BB6 | 8B7D 08 | MOV EDI, DWORD PTR SS: [EBP + 0x8] | 00E44BB9 | 33C0 | XOR EAX, EAX | 00E44BBB | 83ff 05 | CMP EDI, 0x5 | 00E44BBE | 0F95C0 | setne Al | // difference is a , used setne, after obtaining the ZF value, negated, then into the AL, that is unequal to set AL 1 00E44BC1 | 83C0 05 | the Add eax, 0x5 |
- Scheme 2: Expression 1 is a simple comparison, the difference between the both 2 and 3 Expression Expression greater than 1;
return argc == 5 ? 4 : 10;
00E44BCF | BE 0A000000 | mov ESI, 0xA | 00E44BD4 | 83ff 05 | cmp EDI, 0x5 | 00E44BD7 | 8BC6 | mov eax, ESI | 00E44BD9 | BB 04000000 | mov EBX, 0x4 | 00E44BDE | 0F44C3 | cmove eax, EBX | // with equal time ebx - instead of eax> 4 -> 10
- Scheme 3: Expression 2 and Expression 3 both the difference is greater than 1;
return argc >= 8 ? 4 : 10;
00E44BEC | 83FF 08 | cmp edi,0x8 | 00E44BEF | 8BC6 | mov eax,esi | esi-->1000E44BF1 | 0F4DC3 | cmovge eax,ebx |//大于等于时，用ebx-->4代替eax-->10
- 方案4：表达式2和表达式3有一个为变量，于是无优化。
位运算：
141: unsigned int nVar = argc; 0126668E 8B 45 08 mov eax,dword ptr [argc] 01266691 89 45 F8 mov dword ptr [nVar],eax 142: nVar <<= 3; 01266694 8B 45 F8 mov eax,dword ptr [nVar] 01266697 C1 E0 03 shl eax,3 //无符号数<< 使用shl 逻辑左移 0126669A 89 45 F8 mov dword ptr [nVar],eax 143: nVar >>= 5; 0126669D 8B 45 F8 mov eax,dword ptr [nVar] 012666A0 C1 E8 05 shr eax,5 //无符号数>> 使用shr 逻辑右移 012666A3 89 45 F8 mov dword ptr [nVar],eax 144: 145: argc = argc << 3; 012666A6 8B 45 08 mov eax,dword ptr [argc] 144: 145: argc = argc << 3; 012666A9 C1 E0 03 shl eax,3 012666AC 89 45 08 mov dword ptr [argc],eax 146: argc = argc >> 5; 012666AF 8B 45 08 mov eax,dword ptr [argc] 012666B2 C1 F8 05 sar eax,5 //有符号数>> 使用sar 算术右移 012666B5 89 45 08 mov dword ptr [argc],eax 147: argc = argc | 0xFFFF0000; 012666B8 8B 45 08 mov eax,dword ptr [argc] 012666BB 0D 00 00 FF FF or eax,0FFFF0000h 012666C0 89 45 08 mov dword ptr [argc],eax 148: argc = argc & 0x0000FFFF; 012666C3 8B 45 08 mov eax,dword ptr [argc] 012666C6 25 FF FF 00 00 and eax,0FFFFh 012666CB 89 45 08 mov dword ptr [argc],eax 149: argc = argc ^ 0xFFFF0000; 012666CE 8B 45 08 mov eax,dword ptr [argc] 012666D1 35 00 00 FF FF xor eax,0FFFF0000h 012666D6 89 45 08 mov dword ptr [argc],eax 150: argc = ~argc; 012666D9 8B 45 08 mov eax,dword ptr [argc] 012666DC F7 D0 not eax 012666DE 89 45 08 mov dword ptr [argc],eax 151: 152: return argc; 012666E1 8B 45 08 mov eax,dword ptr [argc]