为什么位移一个int向上产生负数？ [英] Why does bit-shifting an int upwards produce a negative number?

问题描述

我是新来的位操作技巧和我写了一个简单的code看到在一个号码即做单位移位的输出。 2

I am new to bit manipulations tricks and I wrote a simple code to see the output of doing single bit shifts on a single number viz. 2

#include <iostream>
int main(int argc, char *argv[])
{

  int num=2;

 do
   {
     std::cout<<num<<std::endl;
     num=num<<1;//Left shift by 1 bit.

   } while (num!=0);


  return 0;
}

此的输出如下。

The output of this is the following.

2
4
8
16
32
64
128
256
512
1024
2048
4096
8192
16384
32768
65536
131072
262144
524288
1048576
2097152
4194304
8388608
16777216
33554432
67108864
134217728
268435456
536870912
1073741824
-2147483648

显然，不断位转移到1位离开，会导致零它上面做，但为什么电脑输出负数的在最后的终止循环之前（因为NUM转身零）??

Obviously, continuously bit shifting to the left by 1 bit, will result in zero as it has done above, but why does the computer output a negative number at the very end before terminating the loop (since num turned zero)??

然而，当我把 INT NUM = 2 按无符号整数NUM = 2 然后我得到的输出不相同
这最后一个号码是这段时间显示为阳性，即 2147483648 而不是 -2147483648

However when I replace int num=2 by unsigned int num=2 then I get the same output except that the last number is this time displayed as positive i.e. 2147483648 instead of -2147483648

我使用的是 GCC 编译器在Ubuntu Linux

I am using the gcc compiler on Ubuntu Linux

推荐答案

这是因为 INT 是一个符号整数。在二进制补码再presentation ，整数的符号被确定最上面位。

That's because int is a signed integer. In the two's-complement representation, the sign of the integer is determined by the upper-most bit.

一旦你已经转移了1到最高（符号）位，它翻转为负。

Once you have shifted the 1 into the highest (sign) bit, it flips negative.

当您使用无符号，没有符号位。

When you use unsigned, there's no sign bit.

0x80000000 = -2147483648 for a signed 32-bit integer.
0x80000000 =  2147483648 for an unsigned 32-bit integer.

编辑：

注意，严格来说，签署整数溢出在C / C未定义行为++。海湾合作委员会在这方面的行为是不完全一致：

Note that strictly speaking, signed integer overflow is undefined behavior in C/C++. The behavior of GCC in this aspect is not completely consistent:

• NUM = NUM​​＆LT;＆LT; 1; 或 NUM＆LT;＆LT; = 1; 通常作为上述行为


• NUM + = NUM​​; 或 NUM * = 2; 实际上可能的go插入GCC 一个无限循环。

• num = num << 1; or num <<= 1; usually behaves as described above.
• num += num; or num *= 2; may actually go into an infinite loop on GCC.

这篇关于为什么位移一个int向上产生负数？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！