c++ - 接收参数的 lambda 函数类型

Question

我正在尝试接收 lambda 函数作为参数，但我遇到了它的类型问题。

下面是我调用函数的方式

profile(0, 1000, fps, [](ProfilerVariable<int> &fps){fps.counter++;}, [](ProfilerVariable<int> &fps){fps.counter=0;});

我是这样定义的:

template <typename T>
void profile(int index, int intervalInMilliseconds, ProfilerVariable<T>& profilerVariable,
                 std::function<void(ProfilerVariable<T>&)> const &increaseFunction, 
                 std::function<void(ProfilerVariable<T>&)> const &resetFunction)
    {
            increaseFunction(profilerVariable);
            //...

我看不出有什么不妥。返回类型为 void，它接受 ProfilerVariable& :

    /home/dev/orwell/cpp/common/ZLRTSPClient.cpp:38:26: error: no matching function for call to 'ZLRTSPClient::profile(int, int, ProfilerVariable<int>&, ZLRTSPClient::init()::<lambda(const toolkit::SockException&)>::<lambda(const Ptr&)>::<lambda(ProfilerVariable<int>&)>, ZLRTSPClient::init()::<lambda(const toolkit::SockException&)>::<lambda(const Ptr&)>::<lambda(ProfilerVariable<int>&)>)'
                         });
                          ^
In file included from /home/dev/orwell/cpp/common/RTSPClient.h:11:0,
                 from /home/dev/orwell/cpp/common/ZLRTSPClient.h:3,
                 from /home/dev/orwell/cpp/common/ZLRTSPClient.cpp:1:
/home/dev/orwell/cpp/common/Profiler.h:41:10: note: candidate: template<class T> void Profiler::profile(int, int, ProfilerVariable<T>&, const std::function<void(ProfilerVariable<T>&)>&, const std::function<void(ProfilerVariable<T>&)>&)
     void profile(int index, int intervalInMilliseconds, ProfilerVariable<T>& profilerVariable,
          ^~~~~~~
/home/dev/orwell/cpp/common/Profiler.h:41:10: note:   template argument deduction/substitution failed:
/home/dev/orwell/cpp/common/ZLRTSPClient.cpp:38:26: note:   'ZLRTSPClient::init()::<lambda(const toolkit::SockException&)>::<lambda(const Ptr&)>::<lambda(ProfilerVariable<int>&)>' is not derived from 'const std::function<void(ProfilerVariable<T>&)>'
                         });
                          ^

PS:const 是什么意思？在std::function<void(ProfilerVariable<T>&)> const &increaseFunction代表什么？

Answer 1

lambda 不是 std::function，因此无法推导 T。

你可以摆脱 std::function 并采用仿函数:

template <typename T, typename F1, typename >
void profile(int index,
             int intervalInMilliseconds,
             ProfilerVariable<T>& profilerVariable,
             F1 const &increaseFunction,
             F2 const &resetFunction) {/*..*/}

或者您可以使这些参数不可推导:

template <typename T>
void profile(int index,
             int intervalInMilliseconds,
             ProfilerVariable<T>& profilerVariable, // Deduce T only here
             std::type_identity_t<std::function<void(ProfilerVariable<T>&)>> const &increaseFunction, 
             std::type_identity_t<std::function<void(ProfilerVariable<T>&)>> const &resetFunction)

std::type_identity_t是 C++20，但可以为以前的版本实现。

否则你必须在调用站点中更改

• 明确:

  profile<int>(0, 1000, fps, [](ProfilerVariable<int> &fps){fps.counter++;}, [](ProfilerVariable<int> &fps){fps.counter=0;});
  

• 或使用“正确”的参数

  profile(0,
          1000,
          fps,
          std::function{[](ProfilerVariable<int> &fps){fps.counter++;}},
          std::function{[](ProfilerVariable<int> &fps){fps.counter=0;}});