为什么以下程序会发生溢出？ [英] Why may an overflow occur in the following program?

问题描述

void main () {
  int i;
  if (i < 0) { i = -i; };
}

任何人都可以帮助我理解为什么在上述程序中可能出现溢出？

Can anyone help me to understand why an overflow may occur in the above program?

推荐答案

可能会发生溢出，因为二进制补码中的整数表示范围不对称：最小负数的大小可以表示的是可以表示的最高正数的大小加一。例如，在32位系统上，值为 -2,147,483,648 和 2,147,483,647 。这就是为什么否定 -2,147,483,648 会导致溢出：否定的结果，正值 2,147,483,648 ，无法表示在相同大小的 int 中。

An overflow may occur because the range of integer representation in two's complement is not symmetric: the magnitude of the smallest negative number that can be represented is the magnitude of the highest positive number that can be represented, plus one. For example, on a 32-bit system the values are -2,147,483,648 and 2,147,483,647. That's why negating -2,147,483,648 would result in an overflow: the result of negation, a positive value 2,147,483,648, cannot be represented in an int of the same size.

注意这个问题的反面不是这样的：否定正面数字不会导致溢出：

Note that the inverse of this problem is not true: negating a positive number would not result in an overflow:

if (i > 0) { i = -i; } // No overflow here

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