Thread: Question about pointer to structures

No, an array of structures is the same as an array of any plain old data. You're getting reference to the structure passed to the function. You don't need to make a pointer to the array. You seem to be confused about the use of pointers, though. They don't just "save memory" they're completely nessassary if you want to make any changes to the structure at all. If you pass by value and make a copy of your data, you can edit it all you want and not change the original, at all.

Originally Posted by KidMan

What if a want a function to edit an array of structures???

I thought I answered that. You pass it like an array of any other type. It works the same.

Code:

#include <stdio.h>

struct foo {
   int num1;
};

void setNum(struct foo obj1) {  // By value
     obj1.num1 = 10;
}

void setNum(struct foo *obj1) {  // By address
     obj1->num1 = 10;
}

void setNum(struct foo obj[], int size) {  // By array (which is also by address)
     int i;
     for (i = 0; i < size; i++)
        obj[i].num1 = i;
}

int main(void) {
   struct foo obj[2];
   obj[0].num1 = 5;
   obj[1].num1 = 7;
   
   printf("obj[0].num1 = %d\n", obj[0].num1);
   setNum(obj[0]);   // The first function
   printf("obj[0].num1 = %d\n", obj[0].num1);
   setNum(&obj[0]);  // The overloaded function for pointers
   printf("obj[0].num1 = %d\n", obj[0].num1);
   setNum(obj, 2);   // The overloaded function for arrays
   printf("obj[0].num1 = %d\nobj[1].num1 = %d\n", obj[0].num1, obj[1].num1);
   
   return 0;
}

/* Output:
obj[0].num1 = 5
obj[0].num1 = 5
obj[0].num1 = 10
obj[0].num1 = 0
obj[1].num1 = 1 */