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

Question

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;

Possibly a duplicate of this: https://stackoverflow.com/questions/27463785/cant-pass-temporary-object-as-reference `"The idea is that a function taking a non-const reference parameter is stating that it wants to modify the parameter and allowing it to go back to the caller. Doing so with a temporary is meaningless and most likely an error."` — selbie, Oct 13 '19 at 04:53
@selbie Thanks. And I want to know what if I use a move copy construction to push_back the temporary into vector? — Zhen Yang, Oct 13 '19 at 05:12

yumetodo · Answer 1 · 2019-10-13T05:56:01.077

0

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);
}

edited Oct 13 '19 at 05:56

answered Oct 13 '19 at 04:55

yumetodo

I want to use the resize function to reallocate memory of a object. So how to overload the function? – Zhen Yang Oct 13 '19 at 05:25
1

What he's saying is to have two versions of `resize`. One declared as `void resize(size_t n)` and another as `void resize(size_t n, std::string& s)` – selbie Oct 13 '19 at 05:31
OK, I get it. Thanks. – Zhen Yang Oct 13 '19 at 07:18

1 Answers1