如下一段代码:
#ifndef MAGICTYPE#define MAGICTYPE(FIRST, SECOND) ((FIRST << 16) + SECOND)#endiflong getmagictype(unsigned short first, unsigned short second) { long msgtype = MAGICTYPE(first, second); return msgtype;}
a、(FIRST << 16),这里不会溢出吗?
b、(FIRST << 16) + SECOND 两个unsigned short相加,不需要考虑溢出吗?
我第一反应是第二个问题不存在,因为c++对+操作符有一个规则,如果操作数类型长度大于int,则提升为操作数的类型进行+操作。
如果操作数类型长度小于+操作符,那么会被提升为int进行+操作。可以测试一下:
#include#ifndef MAGICTYPE#define MAGICTYPE(FIRST, SECOND) ((FIRST << 16) + SECOND)#endifint main() { printf("signed short int length : %d\n", sizeof(signed short int)); unsigned short int first = ((1<<16) - 1); unsigned short int second = 2; long msgtype = 0; printf("first is : %x\n", first, first); printf("second is : %x\n", second, second); msgtype = first + second; printf("msgtype is : %lx\n", msgtype, msgtype); return 0;}
结果如下:
signed short int length : 2first is : ffffsecond is : 2msgtype is : 10001
关于第一个问题,我测试了几个场景:
第一种情况:
#include#include #ifndef MAGICTYPE#define MAGICTYPE(FIRST, SECOND) ((FIRST << 16) + SECOND)#endifint main() { unsigned int a=32; int k = 31; printf("((unsigned int)(1)<<32)-1 : %lu\n", ((unsigned int)(1)<<32)-1); printf("((unsigned int)(1)<
结果:
((unsigned int)(1)<<32)-1 : 4294967295((unsigned int)(1)<
有这样的推测:
a、((unsigned int)(1)<<32)-1 : 4294967295的结果是编译器把(1)<<32转换成unsigned int 0了,再进行减一。也就是(unsigned int)-1了,结果如上
b、((unsigned int)(1)<<a)-1 : 0中(1)<<a的结果是在运行时计算了,这涉及到移位操作!1<<32对于整形变量已经溢出了,所以这里的操作结果和机器强相关。
根据已经输出的结果可以推测,测试环境机器<<32之后溢出进行循环移位。操作结果又回到了1.最终操作结果就是0了
c、为了验证b操作中的推测,又写了c测试项来证明上述推测
第二种情况:
#include#include #ifndef MAGICTYPE#define MAGICTYPE(FIRST, SECOND) ((FIRST << 16) + SECOND)#endifint main() { printf("unsigned short int length : %d\n", sizeof(signed short int)); unsigned short int first = ((1<<16) - 1); unsigned short int second = 2; long msgtype = 0; printf("first is : %x\n", first, first); printf("second is : %x\n", second, second); msgtype = (first << 16); printf("msgtype is : %lx\n", msgtype, msgtype); msgtype = MAGICTYPE(first, second); printf("msgtype is : %lx\n", msgtype, msgtype); first = 1; second = ((1<<16) - 1); printf("first is : %x\n", first, first); printf("second is : %x\n", second, second); msgtype = MAGICTYPE(first, second); printf("msgtype is : %lx\n", msgtype, msgtype); return 0;}
测试结果:
unsigned short int length : 2first is : ffffsecond is : 2msgtype is : ffffffffffff0000msgtype is : ffffffffffff0002first is : 1second is : ffffmsgtype is : 1ffff
msgtpye为何输出的不是ff000000呢,为何多出来这么多个ff?求解惑