Conditional operator in member-initialization list
c++
Solution
The result of a conditional operator is always a fixed type determined at compile time by finding a common type that both options can be converted to. (The exact rules are a little involved, but in common use it usually 'does the right thing'.)
In your example the simplest thing to do is to let that type be a temporary `foo` and then use the copy constructor to intialize `_f` in bar.
You can do this as follows.
_f( condition ? foo("string") : foo() )
Problem
Suppose I have this class: ``` class foo { public: foo() { } foo(const std::string& s) : _s(s) { } private: std::string _s; }; ``` Which is a member of another class: ``` class bar { public: bar(bool condition) : _f(condition ? "go to string constructor" : **go to empty ctor**) { } private: foo _f; }; ``` When initializing `_f` in `bar`'s member initialization list I would like to choose which constructor of `foo` to invoke based on `condition`. What can I put instead of `go to empty ctor` to make this work? I thought of putting `foo()`, is there another way?