从 std::function 调用签名推导出模板参数 [英] Deduce template argument from std::function call signature

问题描述

考虑这个模板函数:

template<typename ReturnT>
ReturnT foo(const std::function<ReturnT ()>& fun)
{
    return fun();
}

为什么编译器不能从传递的调用签名中推导出ReturnT?

Why isn't it possible for the compiler to deduce ReturnT from the passed call signature?

bool bar() { /* ... */ }

foo<bool>(bar); // works
foo(bar); // error: no matching function call

推荐答案

std::function<bool()> bar;

foo(bar); // works just fine

C++ 无法从您的函数 bar 推导出返回类型，因为它必须先知道类型，然后才能找到所有使用您的函数指针的构造函数.

C++ can't deduce the return type from your function bar because it would have to know the type before it could find all the constructors that take your function pointer.

例如，谁能说 std::function 没有一个采用 bool (*)()?

For example, who's to say that std::function<std::string()> doesn't have a constructor taking a bool (*)()?

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