小编use*_*167的帖子

请参阅下面的简单代码:

#include <iostream>
#include <stdlib.h>

using namespace std;


int main(void)
{
    unsigned long currentTrafficTypeValueDec;
    long input;
    input=63;
    currentTrafficTypeValueDec = (unsigned long) 1LL << input; 
    cout << currentTrafficTypeValueDec << endl;
    printf("%u \n", currentTrafficTypeValueDec);
    printf("%ld \n", currentTrafficTypeValueDec);
    return 0;
}

为什么printf()显示带负值的currentTrafficTypeValueDec(unsigned long)？

输出是:

9223372036854775808
0
-9223372036854775808 

基于以下简单程序,按位左移运算符仅适用于32位.这是真的吗？

#include <iostream>
#include <stdlib.h>

using namespace std;


    int main(void)
    {
        long long currentTrafficTypeValueDec;
        int input;
        cout << "Enter input:" << endl;
        cin >> input;
        currentTrafficTypeValueDec = 1 << (input - 1); 
        cout << currentTrafficTypeValueDec << endl;
        cout << (1 << (input - 1)) << endl;

        return 0;

    }

该计划的输出:

Enter input:
30
536870912
536870912

Enter input:
62
536870912
536870912

我怎么能生产64位掩码？

有没有办法将长度= 128*2 + 1的字母数字字符串压缩为最短的唯一表示？