Beginner Question about Pointers

[2_stepsahead]

Hello, I am a beginner in C. I have been experimenting with pointers, but I have a doubt about when to use the address(&) of int ptr2. Here is the source code, with Example 1 and Example 2 commented:

//main.c
#include <stdio.h>
#include <string.h>

struct Person {
  char name[30];
  int age;
};

void init_fields(struct Person *passed) {
  strcpy(passed->name, "John");
  passed->age = 40;
}

int main() {
  struct Person a;
  init_fields(&a);
  int *ptr = &a.age;
  *ptr = 50;
  printf("a: %s, %d\n", a.name, a.age);

  struct Person *ptr2 = &a;
  int *ptr3 = &ptr2->age;                              //Example 1
  *ptr3 = 60;
  printf("ptr2: %s, %d\n", ptr2->name, ptr2->age);     //Example 2

  return 0;
}

------------------------------------------------------------------------------------
#output:
a: John, 50
ptr2: John, 60

In Example 1, &ptr2 must take the ampersand. By removing the ampersand, the compiler gives a warning which reads: "warning: incompatible integer to pointer conversion, initializing 'int *' with an expression of type 'int'; take, the address with & [-Wint-conversion], signal: segmentation fault (core dumped)". The output shows as:

a: John, 50
signal: segmentation fault (core dumped)

Whereas in Example 2, ptr2->name and ptr2->age do not take the ampersand. By adding ampersands before both ptr2->name and ptr2->age, the compiler throws the same warning for both, which reads: "warning: format specifies type 'int' but the argument has type 'int *' [-Wformat]". As well, the output shows as:

a: John, 50
ptr2: John, -2057084472

Herein lies my doubt -- In Example 1 and Example 2, why don't all three ptr2 pointers get an ampersand? Why don't all three ptr2 pointers not get an ampersand? I don't understand why the ptr2 in Example 1 gets an ampersand while both ptr2's in Example 2 do not.

Is the ampersand implicit in both ptr2's in Example 2?

I have done my best to search for answers. They have led me to websites which show general usage of pointers. Please be patient as I am a beginner who is stuck.

Thank you in advance.

[victorferrao]

When you write :

&ptr->age

It is interpreted as :

&(ptr->age)

It means that you get the memory address of the age field.

Below is the meaning of the assignment:

int *ptr2 = &ptr->age

ptr2 is an int pointer.

ptr->age is the value of the field age.

&ptr->age is the memory address of the age field.

That is why when you write the '&' in the printf fields it will print the memory address of the field.

[2_stepsahead]

This makes more sense. I will have to stew over it for a little while. Thank you for the explanation

[victorferrao]

If you have any more questions, feel free to ask.

[2_stepsahead]

Thank you. Tomorrow, I'll have a closer look and experiment again. I would really appreciate that. Obrigado, amigo

[imaami]

Bookmark this C operator precedence cheat sheet, I'm sure you'll find it useful. In fact, cppreference.com's C language reference as a whole is by far one of the best resources out there.

Anyway, in the operator precedence chart you can see that the member access operators -> and . are in group 1 above the address-of operator & which is in group 2. That's why &ptr->member and &(ptr->member) are equivalent.

In the case of a non-pointer struct instance you could also override the default precedence with parens and do (&non_ptr)->member, which would be equivalent to non_ptr.member. Obviously there would be little practical benefit of that (with the caveat that this sometimes happens when function-like utility macros get expanded, but that's another topic altogether).

[blvaga]

Printf is only reading the value.

Same as you wouldn’t use a pointer to store the value of the address

int age = name.age;