c++ - const 引用默认值

Answer 1

You can't take a non-const reference to a temporary like foo2 does.

Notice that this isn't specifically default parameters. You get the same error for function variables: http://ideone.com/g7Tf7L

#include <string>
using std::string;

#include <iostream>
using std::cout; using std::endl;

int main()
{
    string s1        = string("s1"); // OK, copy it
    const string& s2 = string("s2"); // OK, const reference to it
    string& s3       = string("s3"); // ERROR! non-const reference not allowed!

    cout
            << s1 << ", "
            << s2 << ", "
            << s3 << endl;
    return 0;
}

When you take a const reference to a temporary, the lifetime of the temporary is extended to the lifetime of the reference (§12.2, quoted from my copy of C++11 draft n3337):

There are two contexts in which temporaries are destroyed at a different point than the end of the fullexpression.

...

The second context is when a reference is bound to a temporary. The temporary to which the reference is bound or the temporary that is the complete object of a subobject to which the reference is bound persists for the lifetime of the reference except:

• A temporary bound to a reference member in a constructor’s ctor-initializer (12.6.2) persists until the constructor exits.
• A temporary bound to a reference parameter in a function call (5.2.2) persists until the completion of the full-expression containing the call.
• The lifetime of a temporary bound to the returned value in a function return statement (6.6.3) is not extended; the temporary is destroyed at the end of the full-expression in the return statement.
• A temporary bound to a reference in a new-initializer (5.3.4) persists until the completion of the full-expression containing the new-initializer.

Answer 2

It may come as a surprise to you, but you can bind the value of a temporary expression to a constant reference, and the lifetime of the expression is extended to that of the reference. But you cannot do this with a non-constant (lvalue) reference.

Answer 3

The declarations of foo3 and foo4 are legal because the argument to those functions is not a reference.

The declaration of foo2 is illegal because you can't bind a non-const reference to a temporary.

So why is the declaration of foo1 legal? It's that very important const qualifier that makes this declaration legal.