如果我输入等于INTMax + 1的数字，计算机将不会返回-1 [英] Computer dosen't return -1 if I input a number equal to INTMax+1

问题描述

int类型是4字节长，我在Ubuntu下用C编写了一个小程序来打印我刚刚输入的数字.当我输入2147483648，即2 ^ 31时，它将打印2147483647而不是-1.当我输入大于2147483647的任何数字时，也会发生同样的事情.当我学习表单书时，为什么它不会溢出到-1，但似乎被截断为INT_Max，并且位级别发生了什么?

The type int is 4-byte long and I wrote a little procedure in C under Ubuntu to print the number I've just input. When I input 2147483648, i.e. 2^31, it prints 2147483647 rather than -1. The same thing happens when I input any number larger than 2147483647. Why doesn't it overflow to -1 as I learnt form book but seems like truncated to INT_Max and what happened in the bits level?

#include <stdio.h>
int main(){
  int x;
  scanf("%d",&x);
  printf("%d\n",x);
}

我弄错了.INT_Max + 1应该等于INT_Min.我修改了代码:

I made a mistake. INT_Max+1 should equal to INT_Min. I modified the code:

#include <stdio.h>
int main(){
  int x=2147483647;
  int y=x+1;
  printf("%d",y);
}

，输出为 -2147483648 现在我只是想知道当我调用函数scanf时会发生什么?我认为它会将大于2147483647的所有输入数字都截断为2147483647.

and the output is -2147483648 Now I'm just wondering what happened when I call the function scanf? I think it truncated all the input number larger than 2147483647 to 2147483647.

推荐答案

2147483647的表示形式，即INT_MAX是 0x7FFFFFFF .
如果添加1，您将得到未定义的行为.
这就是说，实际上，如果加1，您将得到 0x80000000 ，即-2147483648.
我不知道为什么您期望-1，因为其二进制编码为 0xFFFFFFFF .

The representation of 2147483647 ie INT_MAX is 0x7FFFFFFF.
If you add 1, you'll get an undefined behaviour.
This said, in practice, if you add 1, you'll get 0x80000000 ie -2147483648.
I don't know why you expect -1 as its binary encoding is 0xFFFFFFFF.

#include <stdio.h>
#include <errno.h>

int main(){
  int val1 = 2147483647;
  int val2 = val1+1;
  int x;

  printf("errno before : %d\n",errno);
  scanf("%d",&x);                        //enter 2147483648 or a larger value
  printf("errno after : %d\n\n",errno);

  printf("val1 = %d (0x%X)\n", val1, val1);
  printf("val2 = %d (0x%X)\n", val2, val2);
  printf("x    = %d (0x%X)\n", x, x);

  return 0;
}

输出:

errno before : 0
errno after : 34 //0 if the entered value is in the range of 4-bytes integer

val1 = 2147483647 (0x7FFFFFFF)
val2 = -2147483648 (0x80000000)
x    = 2147483647 (0x7FFFFFFF)

得到 x = 2147483647 的原因是 scanf 将值限制在可能的范围内.
如果在 scanf 调用后检查 errno ，您将看到它等于 ERANGE (代码34)

The reason why you get x=2147483647 is that scanf clamps the value to the possible range.
If you check errno after scanf call, you will see that it is equal to ERANGE (code 34)

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