I'm trying to create a 2D array (5 rows and 2 cols) dynamically. At the time of allocating the memory there's no sort of problem. However, while assigning actual values to the array I got a Segmentation Fault error.
While debugging I got to realize that the error pops up when i in the for loop has value 4
How can it be possible to get a segmentation fault error even after allocating with no problem the memory that will be used by the array?
#include <stdio.h>
#include <stdlib.h>
int main(int argc, char *argv[]){
int **ptr = (int **)malloc(sizeof(int)*5);
for(int i = 0; i<5; i++){
ptr[i] = (int *)malloc(sizeof(int)*2);
}
for(int i = 0; i<5; i++){
for(int j = 0; j<2; j++){
ptr[i][j] = i;
}
}
return 0;
}
How can it be possible to get a segmentation fault error even after allocating with no problem the memory that will be used by the array?
Code did not allocate correctly. sizeof(int)*5 is the wrong size in OP's code.
Avoid allocation errors. Size to the referenced object, not the type. It is easier to code right the first time, review and maintain. In OP's case sizeof ptr[0] is the correct size, the size of a pointer. sizeof(int) is incorrect as it is the size of a int, yet the size of a pointer is needed here.
Casting not needed.
Better code checks for allocation failures.
// int **ptr = (int **)malloc(sizeof(int)*5);
int **ptr = malloc(sizeof ptr[0] * 5);
if (ptr == NULL) Handle_OutOfMemory();