c++ - [[nodiscard]] 在 std::function 返回类型定义中？

Question

我想知道是否有办法拥有这样的东西:

using CallbackType = std::function<[[nodiscard]]bool(void)>; 

(我知道上面的代码不会被编译并提示 nodiscard 不能应用于类型!)
我的目标是强制回调的调用者检查它的返回值!

Answer 1

你可以做这样的事情

#include <iostream>
#include <utility>

template <typename R, typename... Args>
struct Function {
  using Fn = R (*)(Args...);

  Fn fn;

  explicit Function(Fn fn) : fn{fn} {}

  [[nodiscard]] R operator()(Args... args) {
    return (*fn)(std::forward<Args>(args)...);
  }
};

template <typename R, typename... Args>
Function(R (*)(Args...)) -> Function<R, Args...>;


bool bar(const int& x) { return x % 2 == 0; }

int main() {
  Function{&bar}(10);
}

警告:

Compiler stderr

<source>: In function 'int main()':

<source>:24:17: warning: ignoring return value of 'R Function<R, Args>::operator()(Args ...) [with R = bool; Args = {const int&}]', declared with attribute 'nodiscard' [-Wunused-result]

   24 |   Function{&bar}(10);

      |   ~~~~~~~~~~~~~~^~~~

<source>:12:19: note: declared here

   12 |   [[nodiscard]] R operator()(Args... args) {

      |                   ^~~~~~~~

编辑:
扩展到成员函数 (+const) + lambda(带推导指南)

template <typename T, typename = std::void_t<>>
struct Function;

template <typename R, typename... Args>
struct Function<R (*)(Args...)> {
  using Fn = R (*)(Args...);

  Fn fn;

  explicit Function(Fn fn) : fn{fn} {}

  [[nodiscard]] R operator()(Args... args) {
    return fn(std::forward<Args>(args)...);
  }
};

template <typename T, typename R, typename... Args>
struct Function<R (T::*)(Args...)> {
  using MFn = R (T::*)(Args...);

  T* t;
  MFn mfn;

  Function(T* t, MFn mfn) : t{t}, mfn{mfn} {}

  [[nodiscard]] R operator()(Args... args) {
    return (t->*mfn)(std::forward<Args>(args)...);
  }
};

template <typename T, typename R, typename... Args>
struct Function<R (T::*)(Args...) const> {
  using MFn = R (T::*)(Args...) const;

  const T* t;
  MFn mfn;

  Function(const T* t, MFn mfn) : t{t}, mfn{mfn} {}

  [[nodiscard]] R operator()(Args... args) {
    return (t->*mfn)(std::forward<Args>(args)...);
  }
};

template <typename T>
struct Function<T, std::void_t<decltype(&T::operator())>> final
    : Function<decltype(&T::operator())> {
  explicit Function(const T& t)
      : Function<decltype(&T::operator())>(&t, &T::operator()) {}
};

template<typename T>
Function(T) -> Function<T>;

template<typename T, typename R, typename ...Args>
Function(T*, R(T::*)(Args...)) -> Function<R(T::*)(Args...)>;

template<typename T, typename R, typename ...Args>
Function(const T*, R(T::*)(Args...) const) -> Function<R(T::*)(Args...) const>;

void foo() {}

struct Foo
{
    void foo() const {}
};

int main() {
  Function{&foo}();
  Function{[]{}}();
  Foo foo{};
  Function{&foo, &Foo::foo}();
}