检查整数溢出
原文:https://www.geeksforgeeks.org/check-for-integer-overflow/
写一个“C”函数，int addOvf(int* result，int a，int b)如果没有溢出，该函数将结果= sum a+b 放入“result”中并返回 0。否则返回-1。不允许使用长整型和加法检测溢出的解决方案。
方法 1 只有两个数的符号相同，且和的符号与数的符号相反，才能有溢出。
1) Calculate sum
2) If both numbers are positive and sum is negative then return -1
Else
If both numbers are negative and sum is positive then return -1
Else return 0
C++
#include <bits/stdc++.h>
using namespace std;
/* Takes pointer to result and two numbers as
arguments. If there is no overflow, the function
places the resultant = sum a+b in “result” and
returns 0, otherwise it returns -1 */
int addOvf(int* result, int a, int b)
{
*result = a + b;
if(a > 0 && b > 0 && *result < 0)
return -1;
if(a < 0 && b < 0 && *result > 0)
return -1;
return 0;
}
// Driver code
int main()
{
int *res = new int[(sizeof(int))];
int x = 2147483640;
int y = 10;
cout<<addOvf(res, x, y);
cout<<"\n"<<*res;
return 0;
}
// This code is contributed by rathbhupendra
C
#include<stdio.h>
#include<stdlib.h>
/* Takes pointer to result and two numbers as
arguments. If there is no overflow, the function
places the resultant = sum a+b in “result” and
returns 0, otherwise it returns -1 */
int addOvf(int* result, int a, int b)
{
*result = a + b;
if(a > 0 && b > 0 && *result < 0)
return -1;
if(a < 0 && b < 0 && *result > 0)
return -1;
return 0;
}
int main()
{
int *res = (int *)malloc(sizeof(int));
int x = 2147483640;
int y = 10;
printf("%d", addOvf(res, x, y));
printf("\n %d", *res);
getchar();
return 0;
}
Output:
-1
-2147483646
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