Why can braced-init-list not be used in ternary operator?

Question

My compiler is the latest VC++ 2013 RC.

int f(bool b) {     return {}; // OK     return b ?  1  : { }; // C2059: syntax error : '{'     return b ?  1  : {0}; // C2059: syntax error : '{'     return b ? {1} : {0}; // C2059: syntax error : '{' } 

Why can braced-init-list not be used in ternary operator?

Is this behavior defined as ill-formed by the C++ standard, or just a bug of the VC++ compiler?

Answer 1

Well, here's what the standard says about the braced-init-list (8.5.3.1):

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)

Since this doesn't mention the conditional operator, I guess your compiler is right. Also note that the conditional operator expects expressions on the both sides of : (5.16), and as far as I understand, a brace-initializer is not an expression.

Answer 2

It's a syntax error. A braced-init-list is not an expression, and it doesn't have a type or value category. braced-init-list is available at various locations in the C++ grammar, and the operands of conditional expression are not one of those locations. Therefore your code fails to even parse.

If you want to do this:

struct T { ... };  T f(bool b) {     return b ? {i,j,k} : {x,y}; } 

instead you could do this:

T f(bool b) {     return b ? T{i,j,k} : T{x,y}; } 

and I believe, although this requires a move constructor, it wouldn't use it and RVO would kick in.

Alternatively you could of course do this:

T f(bool b) {     if (b)         return {i,j,k};     else         return {x,y}; } 

to get all the advantages of list-initialization.