Why can't functions be overloaded by return type? [duplicate]

Answer 1

You can't overload on return types as it is not mandatory to use the return value of the functions in a function call expression.

For example, I can just say

GetVal();

What does the compiler do now?

Answer 2

The return type of functions is not a part of the mangled name which is generated by the compiler for uniquely identifying each function. Each of the following:

• Number of arguments
• Type of arguments
• Sequence of arguments

are the parameters which are used to generate the unique mangled name for each function. It is on the basis of these unique mangled names that compiler can understand which function to call even if the names are same(overloading).

Answer 3

There's only one function in C++ that can be overloaded by return type, the implicit conversion operator special function, named operator T() (with T varying).

Answer 4

No, you can't overload based on the return type.

From standard docs., Sec 13.1.2,

— Function declarations that differ only in the return type cannot be overloaded.

And regarding your library, each function might belong to a different namespace or else it ain't possible either.

Answer 5

You can try

1. template <typename T> T GetVal(T) {...} instead (using a dummy variable to perform resolution)
2. Use type conversions

Or just use the simplistic GetValBool etc.