Overloading a function that has default arguments

Question

In C++, if I overload a function like the following

void foo(int bar);
void foo(int bar, float baz = 0);

And then I call it

foo(1);

Which foo would be called?

Possible duplicate of [Conflicting overloaded methods with optional parameters](http://stackoverflow.com/questions/2674417/conflicting-overloaded-methods-with-optional-parameters) — user812786, Nov 12 '15 at 16:49
This is invalid. What is your reason for trying to do this? We can possibly find an alternative solution. — Jonathan Howard, Nov 12 '15 at 19:47

Yassine Imounachen · Accepted Answer · 2015-11-15T00:14:26.330

0

I tried, and it won't compile as David Haim said in his comment to my question.

edited Nov 15 '15 at 00:14

answered Nov 12 '15 at 19:31

score -2 · Answer 2 · edited May 23 '17 at 12:24

-2

Why would you even need the first? Seems redundant. The first one will be hit. This is a duplicate. Check here for answer.

edited May 23 '17 at 12:24

Community

answered Nov 12 '15 at 16:22

Always Learning

Why would I need the first, you mean? – Yassine Imounachen Nov 12 '15 at 17:00
Yes, I meant the first - edited for clarity. The second would cover both cases. But still, a dupe question. See my link for more explanation: "If two candidates are judged to be equally good, preference goes to a candidate that does not have optional parameters for which arguments were omitted in the call. This is a consequence of a general preference in overload resolution for candidates that have fewer parameters." – Always Learning Nov 12 '15 at 19:07
1

The "answer" you pointed to is about C# and OP asked about C++ which has different rules of overload resolution. – Paul Nov 12 '15 at 19:16
It is still considered ambiguous. From [learnCPP.com](http://www.learncpp.com/cpp-tutorial/77-default-parameters/) "However, it is important to note that default parameters do NOT count towards the parameters that make the function unique. Consequently, the following is not allowed: void printValues(int x); void printValues(int x, int y=20); If the caller were to call printValues(10), the compiler would not be able to disambiguate whether the user wanted printValues(int) or printValues(int, 20) with the default value." – Always Learning Nov 12 '15 at 19:26

2 Answers2