Ternary operator + C++11 constructor from initializer_list

Question

While developing an application, I had the following problem. I wanted to return an empty std::list<string> when a given function pointer was null, or the result of that function otherwise. This is a simplified version of my code:

typedef std::list<std::string> (*ParamGenerator)();  std::list<std::string> foo() {     /* ... */     ParamGenerator generator = ...;     if(generator)         return generator();     else         return {}; } 

However, I usually like to use the ternary (?:) operator in these cases, so I tried using it this way (as usual):

return generator ? generator() : {}; 

But got this error:

somefile.cpp:143:46: error: expected primary-expression before ‘{’ token somefile.cpp:143:46: error: expected ‘;’ before ‘{’ token 

Does this mean I can't use the ternary operator to return objects created using their constructor from an initializer_list? Is there any particular reason for that?

Answer 1

Standard writes in 8.5.4.1: List-initialization

Note: List-initialization can be used

• as the initializer in a variable definition (8.5)
• as the initializer in a new expression (5.3.4)
• in a return statement (6.6.3)
• as a function argument (5.2.2)
• as a subscript (5.2.1)
• as an argument to a constructor invocation (8.5, 5.2.3)
• as an initializer for a non-static data member (9.2)
• in a mem-initializer (12.6.2)
• on the right-hand side of an assignment (5.17)

Nothing of them is a ternary operator. The more minimalistic return 1?{}:{}; is invalid too, what you want is impossible.

Of course you can explicitly call the constructor std::list<std::string>{}, but I would recommend to write out the if-else-block as you already did.

Answer 2

When you do {} the compiler has no knowledge of the type you are expecting, so it's just a meaningless expression that the compiler doesn't know what to do with. Both sides of the : are evaluated separately, and only then will the compiler complain if the types don't match. I would just do this:

return generator ? generator() : std::list<std::string>();