函数指针数组的类模板参数推导适用于clang,但不适用于gcc

Question

以下代码:

#include <array>

template <int i>
auto f(){}

int main () {
    std::array{f<5>};
}

编译与clang 7.0,但与gcc 8.2一起失败

prog.cc: In function 'int main()':
prog.cc:7:20: error: class template argument deduction failed:
     std::array{f<5>};
                    ^
prog.cc:7:20: error: no matching function for call to 'array(<unresolved overloaded function type>)'
In file included from prog.cc:1:
/opt/wandbox/gcc-8.2.0/include/c++/8.2.0/array:244:5: note: candidate: 'template<class _Tp, class ... _Up> std::array(_Tp, _Up ...)-> std::array<typename std::enable_if<(is_same_v<_Tp, _Up> && ...), _Tp>::type, (1 + sizeof... (_Up))>'
     array(_Tp, _Up...)
     ^~~~~
/opt/wandbox/gcc-8.2.0/include/c++/8.2.0/array:244:5: note:   template argument deduction/substitution failed:
prog.cc:7:20: note:   couldn't deduce template parameter '_Tp'
     std::array{f<5>};
                    ^
In file included from prog.cc:1:
/opt/wandbox/gcc-8.2.0/include/c++/8.2.0/array:94:12: note: candidate: 'template<class _Tp, long unsigned int _Nm> array(std::array<_Tp, _Nm>)-> std::array<_Tp, _Nm>'
     struct array
            ^~~~~
/opt/wandbox/gcc-8.2.0/include/c++/8.2.0/array:94:12: note:   template argument deduction/substitution failed:
prog.cc:7:20: note:   couldn't deduce template parameter '_Tp'
     std::array{f<5>};
                ^

这段代码合法吗？如果没有,我该如何解决？

Answer 1

我相信这个计划是很好的。我自己最近也遇到了类似的问题。当 GCC 在推导数组参数的过程中需要推导占位符返回类型时，它似乎会遇到问题。void例如，显式指定返回类型将使 GCC 接受您的代码。

最终，解决方法是拆分声明。

auto *p = f<5>;
std::array{p};