移位操做的疑問

以下一段代碼：

#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)，這裏不會溢出嗎？

b、(FIRST << 16) + SECOND 兩個unsigned short相加，不須要考慮溢出嗎？

我第一反應是第二個問題不存在，由於c++對+操做符有一個規則，若是操做數類型長度大於int，則提高爲操做數的類型進行+操做。

若是操做數類型長度小於+操做符，那麼會被提高爲int進行+操做。能夠測試一下：

#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;
}

結果以下：

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

關於第一個問題，我測試了幾個場景：

第一種狀況：

#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;
}

結果：

((unsigned int)(1)<<32)-1 : 4294967295
((unsigned int)(1)<<a)-1 : 0
((1)<<a) : 1

有這樣的推測：

　　a、((unsigned int)(1)<<32)-1 : 4294967295的結果是編譯器把(1)<<32轉換成unsigned int 0了，再進行減一。也就是(unsigned int)-1了，結果如上

　　b、((unsigned int)(1)<<a)-1 : 0中(1)<<a的結果是在運行時計算了，這涉及到移位操做！1<<32對於整形變量已經溢出了，因此這裏的操做結果和機器強相關。

　　　　根據已經輸出的結果能夠推測，測試環境機器<<32以後溢出進行循環移位。操做結果又回到了1.最終操做結果就是0了

　　c、爲了驗證b操做中的推測，又寫了c測試項來證實上述推測

第二種狀況：

#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;
}

測試結果：

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爲什麼輸出的不是ff000000呢，爲什麼多出來這麼多個ff？求解惑