Getting segmentation fault while using malloc

问题

I asked my problem here The largest size of 2-D array in C and what I got answer there before declaration as duplicate by some people, I modified my code like this:

#include<stdio.h>
#include<stdlib.h>

int main()
{
    int i,j,n,m,p = 0 ;
    int sum[100000] = {0};

    scanf("%d%d%d",&n,&m,&p);

    /*for ( i = 0 ; i < n ; i++ )
    {
        for( j = 0 ; j < m ; j++ )
        {
            a[i][j] = j + 1 ;
        }
    }*/
    int **a = (int **) malloc(sizeof(int)*n);
    for(i=0; i<n; i++)
    {
        a[i] = (int *) malloc(sizeof(int)*m); 

       for(i=0; i<n; i++){
           for(j=0; j<m; j++){
            a[i][j] = j + 1 ;

           }
       }

    }
    return 0;
}

But still I am getting the same segmentation fault.Please help me to fix it and I hope this question will not be declared as duplicate. :P

回答1:

You're telling malloc() the wrong size.

This:

int **a = (int **) malloc(sizeof(int)*n);

should be:

int **a = malloc(n * sizeof *a);

Don't cast the result of malloc(), and use sizeof on the type the return pointer points at, i.e. int * which probably has a different size than int on your system.

Thus, thus:

a[i] = (int *) malloc(sizeof(int)*m);

should be:

a[i] = malloc(m * sizeof *a[i]);

Add code to make sure all malloc() succeed, i.e. never return NULL, before relying on the pointers being valid. Also make sure the initial scanf() returns 3, and print out the values of n, m and p. Also remove sum.

回答2:

int **a = (int **) malloc(sizeof(int)*n); is wrong. What you want here is an array of pointer, so you should do int **a = (int **) malloc(sizeof(int *)*n); instead.

来源：https://stackoverflow.com/questions/23474293/getting-segmentation-fault-while-using-malloc

标签

arrays

segmentation-fault

malloc