在C ++中分配给float变量时，为什么双引用值不会改变

Question

I am trying to understand the concept of assigning a float to const reference double and the value of double doesn't change if the float value updated.

float d = 2.0;
const double & f = d;
d = 3.0;
std::cout << d << " " << f << std::endl;

output:

 3 2

What is the reason behind this.

However this issue is not seen when we set the reference variable as the same type as the other variable.

Answer 1

However this issue is not seen when we set the reference variable as the same type as the other variable

That's the point; you can't bind reference to object with different type directly.

Given const double & f = d; , a temporary double will be constructed from d , then bound to the reference f . The modification on d has nothing to do with the temporary; they're two irrelevant objects. That's why you got different result when print out d and f .

BTW: Only the lvalue reference to const and rvalue reference could be bound to temporary, so const double & f = d; and double && f = d; work fine. Lvalue reference to non-const can't be bound to temporary, then double & f = d; won't work.