Passing pointers to other function creating segmentaion fault

Question

 struct temp {
     int a;
 };
 int pass(struct temp **); //passing var
 int pass(struct temp ** var1) { //passing var
     (* var1)->a = 100;
     return 0;
 };
 int main() {
     struct temp * var2 = NULL;
     var2->a = 10;
     printf("\nbefore pass var2->a = %d", var2->a);
     pass(&var2);
     printf("\nafter pass var2->a = %d", var2->a); 
     return 0;
 }

bash$ !gc gcc sample2.c ; ./a.exe Segmentation fault (core dumped)

I am trying to update a's value inside the pass() function but keep getting seg fault. Not sure as to what's going wrong. Looking for the rectified code! TIA!

Support Ukraine · Answer 1 · 2018-11-15T07:34:47.240

Here

 struct temp * var2 = NULL;

you create a pointer to struct temp and initialize it to NULL. It is only a pointer - there is not any memory for a struct temp.

Then you do:

var2->a = 10;

which means you dereference a NULL pointer. That will fail.

Try this:

struct temp * var2 = malloc(sizeof *var2);  // Create pointer and allocate 
                                            // memory for a struct temp

instead.

BTW: You don't need to pass double-pointer to the function. Just int pass(struct temp *); will do.

So the complete code could be:

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

 struct temp {
     int a;
 };

 int pass(struct temp *); //passing var
 int pass(struct temp * var1) { //passing var
     var1->a = 100;
     return 0;
 };

 int main() {
     struct temp * var2 = malloc(sizeof *var2);
     var2->a = 10;
     printf("\nbefore pass var2->a = %d", var2->a);
     pass(var2);
     printf("\nafter pass var2->a = %d", var2->a); 
     free(var2);
     return 0;
 }

Passing a double pointer in code like this is only needed if you want to change the value of var2 in main. For instance if the function should allocate more memory (i.e. more struct temp elements).

It could be like:

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

 struct temp {
     int a;
 };

 int pass(struct temp **);
 int pass(struct temp ** var1) {
     free(*var1);                         // Free previous allocation
     *var1 = malloc(20 * sizeof **var1);  // Allocate 20 struct temp
     (*var1)[0].a = 100;
     (*var1)[1].a = 200;
     return 0;
 };

 int main() {
     struct temp * var2 = malloc(sizeof *var2);  // Allocate 1 struct temp
     var2->a = 10;
     printf("\npointer before pass var2 = %p", (void*)var2);
     printf("\nbefore pass var2[0].a = %d", var2[0].a);
     // printf("\nbefore pass var2[1].a = %d", var2[1].a);  // ILLEGAL HERE

     pass(&var2);

     printf("\npointer after pass var2 = %p", (void*)var2);
     printf("\nafter pass var2[0].a = %d", var2[0].a); 
     printf("\nafter pass var2[1].a = %d", var2[1].a);   // LEGAL HERE

     free(var2);
     return 0;
 }

Possible output:

pointer before pass var2 = 0x563d28afa010
before pass var2[0].a = 10
pointer after pass var2 = 0x563d28afb040
after pass var2[0].a = 100
after pass var2[1].a = 200

Notice how the value of the pointer changed.

Ok so the issue was missing heap memory allocation was. Thanks. But I am learning familiarize with passing using double pointers so please explain in that context. — km P, Nov 15 '18 at 07:18
Thanks. Just what I wanted. One last query. printf("\nbefore pass var2[0].a = %d", var2[0].a); <----- Are the pointer variables and arrays interrelated? Any specific reason why you used "[0]" or is it just an alternate form. I was able to run your code with "var[2]->a". — km P, Nov 15 '18 at 09:37
@kmP You ask: "Are the pointer variables and arrays interrelated?" Well, yes. Wrinting: `*somePointer` is the same as `somePointer[0]` and writing `*(somePointer + 5)` is the same as `somePointer[5]` Also see https://stackoverflow.com/questions/4622461/difference-between-pointer-index-and-pointer — Support Ukraine, Nov 15 '18 at 12:08

score 0 · Answer 2 · answered Nov 15 '18 at 07:01

In main you are doing var = NULL; and in the next line you do var2->a = 10; which means you are trying to assign a value to structure pointed by var, but var is not pointing to any structure as of yet.

Perhaps what you want to do is:

 struct temp {
     int a;
 };
 int pass(struct temp *); //passing var
 int pass(struct temp * var1) { //passing var
     var1->a = 100;
     return 0;
 };
 int main() {
     struct temp var2;
     var2.a = 10;
     printf("\nbefore pass var2->a = %d", var2.a);
     pass(&var2);
     printf("\nafter pass var2->a = %d", var2.a); 
     return 0;
 }

Passing pointers to other function creating segmentaion fault

2 Answers2