Literal initialization for const references

Question

How does the following code work in C++? Is it logical?

const int &ref = 9;
const int &another_ref = ref + 6;

Why does C++ allow l

Answer 1

So you can write code like this:

void f( const string & s ) {
}

f( "foobar" );

Although strictly speaking what is actually happening here is not the literal being bound to a const reference - instead a temprary string object is created:

string( "foobar" );

and this nameless string is bound to the reference.

Note that it is actually quite unusual to create non-parameter reference variables as you are doing - the main purpose of references is to serve as function parameters and return values.