Can I pass a default value to a reference for a std::string ?

Question

void doStuff( std::string const & s1, std::string const & s2="");

I was wondering if this code is legal in C++, for the s2 string. I want to have a default argument, but passing a reference and having an empty string as default. Will a temporary be created, and the reference will point to that temporary, or is it illegal C++?

billz · Accepted Answer · 2013-11-06T10:02:11.307

15

Yes this is legal. const will ensure temporary be lasting till function doStuff finishes.

§ 12.2.5

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.

edited Nov 06 '13 at 10:02

answered Nov 06 '13 at 09:53

billz

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score 4 · Answer 2 · answered Sep 29 '15 at 18:43

4

Better will be

void doStuff( std::string const & s1, std::string const & s2 = std::string());

To avoid extra temporary const char *. (Your variant has 2 temporaries: a const char * and empty std::string).

Or, using user-defined literals (C++14):

void doStuff( std::string const & s1, std::string const & s2 = ""s);

answered Sep 29 '15 at 18:43

vladon

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Just for reference: One could also use `std::string const & s2 = {}` as short c++11-compatible version – Meuchelfix77 Oct 16 '19 at 20:51

score 2 · Answer 3 · answered Nov 06 '13 at 10:00

That to understand the semantic it is better to split the original statemenet .

void doStuff( std::string const & s1, std::string const & s2="");

into two statements

void doStuff( std::string const & s1, std::string const & s2);
doStuff( SomeString, "" );

In the call of the function the second argument is implicitly converted to an object of type std::string:

s2 = std::string( "" );

So in fact in the body of the function you will have

std::string const &s2 = std::string( "" );

That is constant reference s2 will refer temporary object std::string( "" ).

Can I pass a default value to a reference for a std::string ?

3 Answers3