Avoiding static_cast when using std :: make_tuple with overloaded functions

Question

g ++ says

error: too many arguments for function 'constexpr std :: tuple

if I left static_cast in a call to std::make_tuple

 #include <tuple> typedef int (*func_t)(); int number() { return 2; } double number(bool a) { return 1.2; } int main() { // With a static_cast it compiles without any error // std::tuple<func_t> tup = std::make_tuple(static_cast<func_t>(number)); std::tuple<func_t> tup = std::make_tuple(number); return 0; } 

Here is the complete error message:

 $ g++ -std=c++11 test.cc test.cc: In function 'int main()': test.cc:31:54: error: too many arguments to function 'constexpr std::tuple<typename std::__decay_and_strip<_Elements>::__type ...> std::make_tuple(_Elements&& ...) [with _Elements = {}; typename std::__decay_and_strip<_Elements>::__type = <type error>]' In file included from test.cc:1:0: /usr/include/c++/4.7/tuple:844:5: note: declared here test.cc:31:54: error: conversion from 'std::tuple<>' to non-scalar type 'std::tuple<int (*)()>' requested $ g++ --version g++ (Ubuntu/Linaro 4.7.2-2ubuntu1) 4.7.2 Copyright (C) 2012 Free Software Foundation, Inc. This is free software; see the source for copying conditions. There is NO warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. 

If you change the main function to

 int main() { std::tuple<func_t> tup = std::make_tuple(static_cast<func_t>(number)); return 0; } 

The program compiles just fine. Is there any way to leave static_cast? It seems inappropriate to provide the type func_t twice.

Answer 1

Do not use std::make_tuple . Use braced-init-list like:

 std::tuple<func_t> tup { number }; 

Now the compiler will choose the appropriate func_t overload func_t .

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