const reference default-value

Question

Possible Duplicate:
how to initialize function arguments that are classes with default value

#include <string>

void foo1(const std::string& s = std::string());

void foo2(std::string& s = std::string());

void foo3(const std::string s = std::string());

void foo4(std::string s = std::string());

error at foo2(): default argument for ‘std::string& s’ has type ‘std::string {aka std::basic_string<char>}’

I understand the compiler's point, but I don't get how this does not apply to foo1() as well.

Solution 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.

Solution 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.

Comments

[aiao]

1. Can you elaborate (or direct me somewhere), why not, but ok to take for const reference 2. what is the life time of this temporary?

[David Schwartz]

Consider void myAbs(int &j) { if (j < 0) j = -j; } short q; ... myAbs(q);. This would (if it were allowed) create a temporary int, set it to its absolute value, then destroy it leaving q unchanged. Yuck. The const promises the compiler you're not going to do that.

[aiao]

what is the life time for this temporary in foo3() and foo4()?

[Johannes Schaub - litb]

@aiao the lifetime in all cases is that of the calling full expressions evaluation time. a reference function parameter does not lengthen the lifetime of an argument, nor does it shorten it.

[Remy Lebeau]

In C++11, you can haave a non-const reference to a temporary if you use an rvalue reference. foo2() in this example is using an lvalue reference instead.

[Kerrek SB]

@aiao: The lifetime of the temporary is the lifetime of the full-expression in which it occurs. It's actually the same for foo1 and foo3, but if foo3 the temporary is only used to copy-initialize the variable s.

[jaques-sam]

Does foo1 work when called from another dynamic library? I hear a colleague had trouble with this, but can't confirm.

[dragonxlwang]

and to be more precise, only local const reference can prolongs the lifetime of a temporary, with exceptions. Class initialization of const reference does not do so. stackoverflow.com/questions/2784262/…