使用带有重载函数的std :: make_tuple时如何避免static_cast

Question

g ++说

错误:函数'constexpr std :: tuple的参数太多了

如果我在std::make_tuple通话中省略了static_cast

#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;
}

这是完整的错误消息:

$ 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.

如果我将主要功能更改为

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

程序编译得很好.有可能以某种方式忽略static_cast吗？似乎没必要提供两次类型func_t.

Answer 1

不要用std::make_tuple.使用braced-init-list作为:

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

现在编译器将选择适当的重载匹配func_t.

查看现场演示