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Why does a negative value give segmentation fault for the program which is to find pairs of value from the array which has the given sum?

开发者 https://www.devze.com 2023-02-17 01:03 出处:网络
#include <stdio.h> void findpairs(int arr[], int arr_size, int sum) { int i, temp; int hash[100] = {0};
#include <stdio.h>

void findpairs(int arr[], int arr_size, int sum)
{
  int i, temp;
  int hash[100] = {0}; 

  for(i = 0; i < arr_size; i++)
  {
    temp = sum - arr[i];
    if(hash[temp] == 1)
    {
      printf("Pair with given sum %d is (%d, %d) \n", sum, arr[i], temp);
    }
    hash[arr[i]] = 1;
  }
}

int main()
{
    int A[] = {4,-4,9,2,1,6,5,11};
    int sum =7;
    int arr_size = 8;
    findpairs(A, arr_size, sum);
    开发者_如何转开发return 0;
}

link to the same program


hash[arr[i]] = 1;
...
hash[-4] = 1;

Your hash array has indices from 0 to 99, so you're accessing out-of-bounds data.

Your next out-of-bounds access will be for the value 9,

temp = 7 - 9;
if (hash[temp] == 1)


Because trying to access a negative index is bad.

temp = sum - arr[i];
if(hash[temp] == 1)

or

hash[arr[i]] = 1;


temp = sum - arr[i]

What if sum < arr [i]? hash[temp] == 1 would segfault on most implementations.

Similar behaviour in case of hash[arr[i]] = 1; if arr[i] is negative.

Technically speaking accessing out of bound array element invokes Undefined Behaviour.

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