为什么左移无符号数32次不会产生ZERO？

问题描述

#include <iostream>

using namespace std;

int main()
{
   cout << "sizeof(unsigned int): " << sizeof(unsigned int) << endl;
   unsigned a = 1;
   int i = 0;
   while (a) {
        cout << "a[" << i++ << "]: " << a << endl;
        a <<= 1;
   }

   cout << "a[" << i << "]: " << a << endl;

   unsigned b = 1;
   unsigned c = (b << 31);
   unsigned d = (b << 32);

   cout << "c: " << c << endl;
   cout << "d: " << d << endl;

   return 0;
}

/ *输出 * /

/* Output http://www.compileonline.com/compile_cpp_online.php */

Compiling the source code....
$g++ main.cpp -o demo -lm -pthread -lgmpxx -lgmp -lreadline 2>&1
main.cpp: In function 'int main()':
main.cpp:19:23: warning: left shift count >= width of type [enabled by default]

Executing the program....
$demo

sizeof(unsigned int): 4
a[0]: 1
a[1]: 2
a[2]: 4
a[3]: 8
a[4]: 16
a[5]: 32
a[6]: 64
a[7]: 128
a[8]: 256
a[9]: 512
a[10]: 1024
a[11]: 2048
a[12]: 4096
a[13]: 8192
a[14]: 16384
a[15]: 32768
a[16]: 65536
a[17]: 131072
a[18]: 262144
a[19]: 524288
a[20]: 1048576
a[21]: 2097152
a[22]: 4194304
a[23]: 8388608
a[24]: 16777216
a[25]: 33554432
a[26]: 67108864
a[27]: 134217728
a[28]: 268435456
a[29]: 536870912
a[30]: 1073741824
a[31]: 2147483648
a[32]: 0
c: 2147483648
d: 1

问题>正如你可以看到 a [32 等于0.现在为什么 d is not 0 but 1 ？

Question> As you can see a[32 is equal to 0. Now why d is NOT 0 but 1?

推荐答案

这是C / C ++中未定义的行为。

This is undefined behavior in C/C++.

标准特别留下这个未定义，因为不同的CPU将做不同的事情，转变。具体来说，在32位Intel上，我相信只有移位量的低5位被CPU使用，其余位被忽略。如果我记得正确，PowerPC和64位Intel都使用低6位，并忽略其余。

The standard specifically leaves this undefined because different CPUs will do different things when presented with such a shift. Specifically, on 32-bit Intel, I believe only the low 5 bits of the shift amount are used by the CPU, and the rest of the bits are just ignored. If I remember correctly, PowerPC and 64-bit Intel both use the low 6 bits and ignore the rest.

更高级的语言可能会尝试通过更正结果在逻辑上更一致，但是像C / C ++这样的低级语言被设计为接近金属，并且将希望为< 运算符。

A higher level language might attempt to smooth this out by correcting the results to be more logically consistent, but a low-level language like C/C++ is designed to be "close to the metal" and will want to generate a single bit-shift instruction for the << operator.

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