Printing variable value using Reference to const gives different result

Question

I am learning about references in C++. So i am trying out different examples to better understand the concept. One example that i could not understand is given below:

double val = 4.55;
const int &ref = val;
    
std::cout << ref <<std::endl; //prints 4 instead of 4.55

I want to know what is the problem and how can i solve it?

Answer 1

The problem is that the reference ref is bound to a temporary object of type int that has the value 4 . This is explained in more detail below.

When you wrote:

const int &ref = val;

a temporary of type int with value 4 is created and then the reference ref is bound to this temporary int object instead of binding to the variable val directly. This happens because the type of the variable val on the right hand side is double while on the left hand side you have a reference to int . But for binding a reference to a variable the types should match .

For solving the problem you should write:

const double &ref = val; //note int changed to double on the left hand side

The above statement means ref is a reference to const double . This means we cannot change the variable val using ref .

If you want to be able to change val using ref then you could simply write:

double &ref = val;

Answer 2

const int &ref = val;

Here you initialize ref as int type so it holds value of int type so,Implicit Type Conversion takes place.Read more about type conversion here