How to Solve C++ Error: member reference type is a pointer; did you mean to use ‘->’?

The error “member reference type is a pointer; did you mean to use ‘->'” occurs when using the dot . operator on a pointer to an object. We use the dot operator to access an object’s fields and methods.

You can solve the error by using the arrow operator -> if using a pointer to an object otherwise, apply the dot operator directly to the object.

This tutorial will go through the error in detail and how to solve it with code examples.

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Example

Let’s look at an example to reproduce the error.

#include <iostream>

// Declare Rectangle class
class Rectangle{

// Two data members of type int with private access 
    int width, height;

// Two member functions with public access declared

public:
    void set_values(int,int);
    int area() const {
        return width * height;
    }

};
// Define set_values function outside of the class

void Rectangle::set_values(int x, int y) {
    width = x;
    height = y;
}

int main() {
    // Create a pointer to object of Rectangle class

    auto *rec = new Rectangle();
    // Attempt to define width and height of rectangle

    rec.set_values(3, 4);
    // Attempt to calculate area of rectangle

    std::cout << "Area:  " << rec.area();
    return 0;

}

Let’s run the code to see what happens:

error: member reference type 'Rectangle *' is a pointer; did you mean to use '->'?
    rec.set_values(3, 4);
    ~~~^
       ->

The error occurs because we tried to apply the dot operator on a pointer to the Rectangle object rather than on the object itself. We can also encounter the error as “expression must have class type”.

Solution #1: Use ->

We can access members of the object through a pointer using the arrow -> operator instead of the dot operator. Let’s look at the revised code:

#include <iostream>

class Rectangle{
    int width, height;
public:
    void set_values(int,int);
    int area() const {
        return width * height;
    }

};

void Rectangle::set_values(int x, int y) {
    width = x;
    height = y;
}

int main() {
    // Create a pointer to object of Rectangle class
    auto *rec = new Rectangle();
    // Define width and height of rectangle using arrow operator
    rec->set_values(3, 4);
    // Attempt to calculate area of rectangle using arrow operator
    std::cout << "Area:  " << rec->area();
    return 0;

}

Let’s run the code to get the result:

Area:  12

We successfully calculated the area of the rectangle.

Solution #2: Access class object instead of pointer to object

We can also solve the error by applying the dot operator directly in the Rectangle object. Let’s look at the revised code:

#include <iostream>

class Rectangle{
    int width, height;
public:
    void set_values(int,int);
    int area() const {
        return width * height;
    }

};

void Rectangle::set_values(int x, int y) {
    width = x;
    height = y;
}

int main() {
    // Create an object of the Rectangle class
    Rectangle rec{};
    // Set width and height values using dot operator
    rec.set_values(3, 4);
    // Calculate area using dot operator
    std::cout << "Area:  " << rec.area();
    return 0;

}

Let’s run the code to get the result:

Area:  12

We successfully calculated the area of the rectangle.

Summary

Congratulations on reading to the end of this tutorial!

For further reading on C++, go to the articles:

• How to Convert Char Array to String in C++
• How to Reverse a String in C++

Have fun and happy researching!