CODEWITHSUNDEEP
csegmentation-fault
Assasins
Assasins

Reputation: 1633

segmentation fault when trying to get the pointer to an int

I get a segmentation fault when trying to run the code below

#include <stdio.h>


int main(){

    int* tes1;
    int* tes2;

    *tes1=55;

    *tes2=88;



    printf("int1 %p int2 %p \n",tes1,tes2);

    return 0;
}

why is that?

Upvotes: 0

Views: 71

Answers (2)

ApproachingDarknessFish
ApproachingDarknessFish

Reputation: 14323

You need to allocate the pointers, otherwise they point to garbage memory:

int* tes1; //random initial value
int* tes2; //random initial value

In order to make them point to assignable memory, use the malloc function:

int* tes1 = malloc(sizeof(int)); //amount of memory to allocate (in bytes)
int* tes2 = malloc(sizeof(int));

Then you safely use the pointers:

*tes1=55;
*tes2=88;

But when you're done, you should free the memory using the free function:

free(tes1);
free(tes2);

This will release the memory back to the system and prevent a memory leak.

Upvotes: 3

karthikr
karthikr

Reputation: 99670

You are declaring pointers, and then trying to define values of pointer of the pointers

#include <stdio.h>

int main(){

    int* tes1;
    int* tes2;
    tes1=55; //remove the * here
    tes2=88;
    printf("int1 %p int2 %p \n",tes1,tes2);

    return 0;
}

Upvotes: 1

Related Questions