Why I need to define a default reference parameter as const so that I can assign a lvalue to it? [duplicate]

Question

This question already has an answer here:

• Can't pass temporary object as reference 4 answers

I have defined a function with default parameter:

void resize(size_t n, std::string &s = std::string()); // error: initial value to non-const must be lvalue 

right way:

void resize(size_t n, const std::string &s = std::string());

I want to know why the const makes a difference?

I know that a const variable can be assigned an lvalue. Are they the same question?

const int a = 42;

Answer 1

Add overload. That is a most simple, easy way to resolve.

void resize(size_t n, std::string &s); 
//add overload
void resize(size_t n)
{
    std::string s;
    resize(n, s);
}