Operator «arrow» pointer to member structure

Before reading this article, make sure you are familiar with structures and pointers

In English, this operator is called member selection operator

Let's start with some background:

Let's say we have a Website

Let this structure have three members name, year, rating

And we have a Website object called site1.

We can create a pointer to this object. Remember that:

The type of pointer should correspond to the type of object it points to.

So the syntax would be this:

struct Website * ptr;

So we set the right kind of pointer.

Now you can point to the desired object.

ptr = &heihei.ru

As usual. We've created a pointer and it points to the correct place.

But the structure has members: name, year, rating, so the question arises - how to address them ?

This problem is solved by ->.

We can write ptr->name, ptr->year, ptr->rating and everything is done.

#include <stdio.h> struct Website { char name[80]; int year; float rating; }; int main() { struct Website site1={"heihei.ru", 2018, 30.5}; struct Website *ptr; By analogy with pointers such as char or int, we create // pointer of type struct Website ptr = &site1; printf("%s %d %.1f\n", ptr->name, ptr->year, ptr->rating); return 0; }

gcc -o mso_example mso_example.c
./mso_example

heihei.ru 2018 30.5

Exercise

Replace the line

printf("%s %d %.1f\n", ptr->name, ptr->year, ptr->rating);

With

printf("%s %d %.1f\n", *ptr.name, *ptr.year, *ptr.rating);

Examine the error the compiler will produce and try to fix it.

The solution is just below

Solution

My compiler. GCC produces the following error

mso_example.c: In function ‘main’: mso_example.c:19:32: error: ‘ptr’ is a pointer; did you mean to use ‘->’? printf("%s %d %.1f\n", *ptr.name, *ptr.year, *ptr.rating); ^ -> mso_example.c:19:43: error: ‘ptr’ is a pointer; did you mean to use ‘->’? printf("%s %d %.1f\n", *ptr.name, *ptr.year, *ptr.rating); ^ -> mso_example.c:19:52: error: ‘ptr’ is a pointer; did you mean to use ‘->’? printf("%s %d %.1f\n", *ptr.name, *ptr.year, *ptr.rating); ^ ->

It tries to parse the dot between ptr and name and suggests using an arrow instead of the dot.

It doesn't see the * before ptr because the dot operator has higher priority than *

Enclose all *ptr in parentheses (*ptr)

printf("%s %d %.1f\n", (*ptr).name, (*ptr).year, (*ptr).rating);

Article author: Andrei Olegovich