如何解决 a+b >c 和 a*b 和 a-b 的 integer overflow 问题

最新推荐文章于 2024-01-16 11:20:20 发布

guilanl

最新推荐文章于 2024-01-16 11:20:20 发布

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本文详细介绍如何在C/C++中检测整数加法、减法及乘法运算时可能出现的溢出情况，并提供具体的代码实现。对于加法和减法，通过比较操作数与INT_MAX或INT_MIN来判断是否会发生溢出；对于乘法，则需进一步考虑特殊值-1与INT_MIN相乘的情况。

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注： a 和 b 都是 int 型的。

To create a conforming program you need to test for overflowbefore generating said overflow. The method can be used with unsigned integers too

// for addition
#include <limits.h>
int a = <something>;
int x = <something>;
if ((x > 0) && (a > INT_MAX - x)) /* `a + x` would overflow */;
if ((x < 0) && (a < INT_MIN - x)) /* `a + x` would underflow */;

// for subtraction
#include <limits.h>
int a = <something>;
int x = <something>;
if ((x < 0) && (a > INT_MAX + x)) /* `a - x` would overflow */;
if ((x > 0) && (a < INT_MIN + x)) /* `a - x` would underflow */;

// for multiplication
#include <limits.h>
int a = <something>;
int x = <something>;
if (a > INT_MAX / x) /* `a * x` would overflow */;
if ((a < INT_MIN / x)) /* `a * x` would underflow */;
// there may be need to check for -1 for two's complement machines
if ((a == -1) && (x == INT_MIN)) /* `a * x` can overflow */
if ((x == -1) && (a == INT_MIN)) /* `a * x` (or `a / x`) can overflow */