C语言的隐式类型转换


看了一篇博文,该文章中有如下描述

 1 #include   
 2   
 3 int main()  
 4 {  
 5     unsigned short a = 1;  
 6     unsigned short b = 0;  
 7    
 8     if (a < (b-1)) //a和b-1的结果-1都被转换为int类型进行比较,1<-1,显然false  
 9     {  
10         printf("in if\n");  
11     }  
12   
13     return 0;  
14       
15 }  

预期结果时打印“in if”,但实际运行时却发现啥都没有输出。改为如下代码就可以正常输出“in if”,此时与预期相符

 1 #include   
 2   
 3 int main()  
 4 {  
 5     unsigned short a = 1;  
 6     unsigned int b = 0;  
 7    
 8     if (a < (b-1)) //b-1的结果-1,被转换为0xffffffff,1<0xffffffff,显然true啊  
 9     {  
10         printf("in if\n");  
11     }  
12   
13     return 0;  
14       
15 }  

使用汇编,直接对比两者结果(公司安全不让发送外网图片..),发现主要差异在如下5行,前四行中对short进行了扩展,由2字节扩展为4字节,高位填充0。导致运行结果与与预期不符合的原因就在第5行:jge 8048461 

 1 08048425 
: 2 8048425: 8d 4c 24 04 lea 0x4(%esp),%ecx 3 8048429: 83 e4 f0 and $0xfffffff0,%esp 4 804842c: ff 71 fc pushl -0x4(%ecx) 5 804842f: 55 push %ebp 6 8048430: 89 e5 mov %esp,%ebp 7 8048432: 51 push %ecx 8 8048433: 83 ec 14 sub $0x14,%esp 9 8048436: 66 c7 45 f6 01 00 movw $0x1,-0xa(%ebp) 10 804843c: 66 c7 45 f4 00 00 movw $0x0,-0xc(%ebp) 11 8048442: 0f b7 45 f6 movzwl -0xa(%ebp),%eax 12 8048446: 0f b7 55 f4 movzwl -0xc(%ebp),%edx 13 804844a: 83 ea 01 sub $0x1,%edx 14 804844d: 39 d0 cmp %edx,%eax 15 804844f: 7d 10 jge 8048461 0x3c> 16 8048451: 83 ec 0c sub $0xc,%esp 17 8048454: 68 00 85 04 08 push $0x8048500 18 8048459: e8 a2 fe ff ff call 8048300 19 804845e: 83 c4 10 add $0x10,%esp 20 8048461: b8 00 00 00 00 mov $0x0,%eax 21 8048466: 8b 4d fc mov -0x4(%ebp),%ecx 22 8048469: c9 leave 23 804846a: 8d 61 fc lea -0x4(%ecx),%esp 24 804846d: c3 ret 25 804846e: 66 90 xchg %ax,%ax

汇编中的数值是没有符号之分的,但指令是有符号分别的,上述使用的jge即是有符号比较的方式,导致比较结果为:if (1<-1),为false,不会打印字符串

指令 含义 运算符号
jbe unsigned below or equal (lower or same) <=
jae unsigned above or equal (higher or same) >=
jb unsigned below (lower) <
ja unsigned above (higher) >
jle signed less or equal <=
jge signed greater or equal >=
jl signed less than <
jg signed greater than >