c++ - 位运算的问题

Question

#include <iostream>

using namespace std;

int main()
{
    cout << (0b11 & (~0)) << endl;
    getchar();
    return 0;
}

为什么上面这个代码输出结果是3？就是想问为什么~0可以得到全是1的掩码，为什么不是只有一个1，像这样0000000000000000000001。

Answer 1

Assume your environment is 32-bit and int is 4 bytes. 0 is equivalent to 00000000 00000000 00000000 00000000, '~' bitwise inversion , and we get 11111111 11111111 11111111 11111111
. Also, what is 0b11? I don't think I've seen this way of writing before?

Answer 2

Because ~ is bitwise inversion. If all are 0, if you negate each one, they will all be 1

Answer 3

Since 0b11 is a binary number, the answer is obvious. The result of ～0 is 1, 0b11 is ANDed with 1, and the result remains unchanged, so the result is 3, the decimal result of 0b11.

Answer 4

0b11 is actually binary literal.
But this is a gcc extension. If you use vc++, it can only be used in 2015 Preview.