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Home/ Questions/Q 7866781
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Editorial Team
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Editorial Team
Asked: June 3, 20262026-06-03T00:30:23+00:00 2026-06-03T00:30:23+00:00

I have the following lines of code: struct c_obj_thing *object = NULL; c_obj_initalizer(object); //

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I have the following lines of code:

struct c_obj_thing *object = NULL;
c_obj_initalizer(object);
// at this point, (object == NULL) is 'true'
printf("Value: %d\n", object->value); // this segfaults

Here is the definition for c_obj_initalizer:

int c_obj_initalizer(struct c_obj_thing *objParam) {
  objParam = malloc(sizeof(struct c_obj_thing));
  objParam->pointerThing = NULL;
  objParam->value = 0;
  return 0;
}

Why does the malloc in the c_obj_initalizer method not stay attached to the pointer passed as a parameter when the method returns? It seems like the call to the initalizer doesn’t do anything. I realize that passing an actual c_obj_thing not as a pointer would mean that the changes made in the initalizer did not return, but I thought that dynamic memory perpetuated throughout the entire program.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-03T00:30:24+00:00Added an answer on June 3, 2026 at 12:30 am

    Because when you call a function it is sending a copy of the pointer, when you change it in the function you don’t change in the calling method. You need to malloc it before initializer.

    For example:

    struct c_obj_thing *object = malloc(sizeof(struct c_obj_thing));
    c_obj_initalizer(object);
    printf("Value: %d\n", object->value); // this segfaults
    
    
    int c_obj_initalizer(struct c_obj_thing *objParam) {  
      objParam->pointerThing = NULL;
      objParam->value = 0;
      return 0;
    }
    
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