c++ - C++ 中的右移给出不寻常的结果（无符号 64 位）

Question

我正处于可怕的位移世界中。我有以下代码：

我正在转移这个号码：140638023551944 >> 5。

根据http://www.binaryhexconverter.com/decimal-to-binary-converter的 140638023551944 的二进制表示是

1000011000011111011101000111

右移 5，我预计：0000010000110000111110111010

但相反，我得到 4394938235998，即 111111111101000110101110110111110001011110。

在我看来，这个数字与原来的数字几乎没有任何关系。我看不到另一个存在的模式。这很奇怪。

代码大致如下：

uint64_t n, index, tag;
uint64_t one = 1;
uint64_t address = 140638023551944;
/*left shift to get index into the last index.length() number of slots*/               
cout << "original address is " << address << " " << "\n";
n = (address >> 5);
cout << "after right shifting away offset bits " << n << "\n";

“地址”填充了正确的整数 140638023551944。我已经验证过了。

这是什么奇怪的行为？它与这个模拟器一致： http: //www.miniwebtool.com/bitwise-calculator/bit-shift/ ?data_type=10&number=140638023551944&place=5&operator=Shift+Right ！但我很确定右移不应该那样工作！

Answer 1

// EVERYTHING WORKS CORRECTLY!

#include <cassert>   // assert()
#include <iostream>  // cout
#include <cstdint>   // UINT64_MAX

using namespace std;

int main() {
    uint64_t n, index, tag;
    uint64_t one = 1;
    uint64_t address = 140638023551944;
    /*left shift to get index into the last index.length() number of slots*/
    cout << "original address is " << address << " " << "\n";
    n = (address >> 5);
    cout << "after right shifting away offset bits " << n << "\n";

    {   // Everything works correctly!
        assert( 140638023551944>>5 == 140638023551944/32 );
        assert( 140638023551944>>5 == 4394938235998 );

        assert( 140638023551944/32 == 4394938235998 );
        assert( 140638023551944 < UINT64_MAX );
    }
}