计算机这么想的返回-1，如果我输入一个数目等于INTMax +1 [英] Computer dosen't return -1 if I input a number equal to INTMax+1

问题描述

int类型为4个字节长，我在C写在Ubuntu下一个小步骤打印我有只输入数字。当我输入2147483648，即2 ^ 31，它打印2147483647，而不是-1。同样的事情发生时，我比投入较大2147483647任意数量为什么没有溢出-1作为我学习形式的书，但似乎截断INT_MAX，并在位水平发生了什么？

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);
}

我犯了一个错误。 INT_MAX + 1应该等于INT_MIN。我修改了code：

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);
}

和输出 -2147483648
现在，我只是想知道发生了什么事，当我调用该函数scanf函数？我想这一切截断比2147483647较大输入的号码为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即INT_MAX的重新presentation是为0x7FFFFFFF 。结果
如果你加1，你会得到一个未定义行为。结果
这就是说，在实践中，如果添加1，您将获得为0x80000000 即-2147483648。结果
我不知道为什么你期望-1作为二进制编码为 0xFFFFFFFF的。

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)

为什么你的原因 X = 2147483647 是 scanf函数钳制值的可能范围。

如果检查错误号在 scanf函数电话，你会看到，它等于 ERANGE （code 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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