移位操做的疑问

以下一段代码：

#ifndef MAGICTYPE
#define MAGICTYPE(FIRST, SECOND) ((FIRST << 16) + SECOND)
#endif
long 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，不须要考虑溢出吗？

我第一反应是第二个问题不存在，由于c++对+操做符有一个规则，若是操做数类型长度大于int，则提高为操做数的类型进行+操做。

若是操做数类型长度小于+操做符，那么会被提高为int进行+操做。能够测试一下：

#include <stdio.h>

#ifndef MAGICTYPE
#define MAGICTYPE(FIRST, SECOND) ((FIRST << 16) + SECOND)
#endif
int 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;
}

结果以下：

signed short int length : 2
first   is : ffff
second  is : 2
msgtype is : 10001

关于第一个问题，我测试了几个场景：

第一种状况：

#include <stdio.h>
#include <iostream>

#ifndef MAGICTYPE
#define MAGICTYPE(FIRST, SECOND) ((FIRST << 16) + SECOND)
#endif
int 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)<<a)-1 : %lu\n", ((unsigned int)(1)<<a)-1);
        printf("((1)<<a) : %lx\n", ((1)<<a));
        return 0;
}

结果：

((unsigned int)(1)<<32)-1 : 4294967295
((unsigned int)(1)<<a)-1 : 0
((1)<<a) : 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，这涉及到移位操做！1<<32对于整形变量已经溢出了，因此这里的操做结果和机器强相关。

　　　　根据已经输出的结果能够推测，测试环境机器<<32以后溢出进行循环移位。操做结果又回到了1.最终操做结果就是0了

　　c、为了验证b操做中的推测，又写了c测试项来证实上述推测

第二种状况：

#include <stdio.h>
#include <iostream>

#ifndef MAGICTYPE
#define MAGICTYPE(FIRST, SECOND) ((FIRST << 16) + SECOND)
#endif
int 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 : 2
first   is : ffff
second  is : 2
msgtype is : ffffffffffff0000
msgtype is : ffffffffffff0002
first   is : 1
second  is : ffff
msgtype is : 1ffff

msgtpye为什么输出的不是ff000000呢，为什么多出来这么多个ff？求解惑