我对Scott Meyer的书“ Effective Modern C ++”的第24条感到很兴奋。他提到可以编写C ++ 14 lambda来记录任意函数调用所花费的时间。
我仍处于学习C ++ 14功能的初期。我的尝试(Main.cpp)看起来像这样,用于测量成员函数调用的时间:
#include <chrono>
#include <iostream>
auto measure = [](auto&& function, auto&&... parameters) -> decltype(function)
{
const std::chrono::steady_clock::time_point startTimePoint =
std::chrono::steady_clock::now();
const auto returnValue = std::forward<decltype(function)>(function)(
std::forward<decltype(parameters)>(parameters)...);
const std::chrono::steady_clock::time_point stopTimePoint =
std::chrono::steady_clock::now();
const std::chrono::duration<double> timeSpan = std::chrono::duration_cast<
std::chrono::duration<double>>(stopTimePoint - startTimePoint);
std::cout << "Computation took " << timeSpan.count()
<< " seconds." << std::endl;
return returnValue;
};
class Test
{
public:
int computation(double dummy)
{
std::cout << "Received " << dummy << ". Computing..." << std::endl;
return 123;
}
};
int main(int, char**)
{
Test instance;
using Function = int (Test::*)(double);
Function function = instance.computation;
int result = measure(function, 1.0);
std::cout << "Result: " << result << std::endl;
return 0;
}
我收到以下编译错误:
..\src\Main.cpp: In function 'int main(int, char**)':
..\src\Main.cpp:43:36: error: cannot convert 'int (Test::*)(double)' to 'int' in initialization
int result = measure(function, 1.0);
^
..\src\Main.cpp: In instantiation of '<lambda(auto:1&&, auto:2&& ...)> [with auto:1 = int (Test::*&)(double); auto:2 = {double}; decltype (function) = int (Test::*&)(double)]':
..\src\Main.cpp:43:36: required from here
..\src\Main.cpp:9:69: error: must use '.*' or '->*' to call pointer-to-member function in 'std::forward<int (Test::*&)(double)>((* & function)) (...)', e.g. '(... ->* std::forward<int (Test::*&)(double)>((* & function))) (...)'
const auto returnValue = std::forward<decltype(function)>(function)(
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~^
std::forward<decltype(parameters)>(parameters)...);
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
显然我做错了,但我不知道该怎么做。有谁能够帮助我?非常感谢你!
答案 0 :(得分:2)
有两种方法可以完成此任务。
接受一个函数(或一个函数对象),返回修改后的函数,该函数执行与原始函数相同的操作,并计算时间。返回的对象类型不能与接受的参数类型相同。它必须是lambda(或自定义类类型,但lambda更简单)。当调用 returned 对象时,将执行实际测量。使用语法示例:
result = measure(foo)(param1, param2); // variant 1
auto measured_foo = measure(foo);
result = measured_foo(param1, param2); // variant 2
接受一个函数(或一个函数对象)及其参数,调用它并执行测量。返回类型是原始函数的类型。使用语法示例:
result = measure(foo, param1, param2);
您的measure
最接近第二个变体,唯一与此声明不符的是声明。这是正确的:
auto measure = [](auto&& function, auto&&... parameters) -> decltype(auto)
确切地说,这不是唯一的错误。如果被测函数返回参考,则返回类型将是错误的。要解决此问题,请替换
const auto returnValue = ...
使用
decltype(auto) returnValue = ...
在lambda体内
程序的另一处错误(但不是measure
本身)是您尝试使用成员函数的方式。
Function function = instance.computation;
这是行不通的。使用lambda或std::bind
创建绑定的成员函数。关于在stackoverflow上执行此操作的正确方法的疑问不计其数(很好的答案)。
Live demo(通过引用返回)。
如果您想要第一种创建测量函数的方法,请按以下步骤操作:
auto measure = [](auto&& function) -> decltype(auto)
{
return [=](auto&&... parameters) mutable -> decltype(auto) {
const std::chrono::steady_clock::time_point startTimePoint =
std::chrono::steady_clock::now();
decltype(auto) result = function(std::forward<decltype(parameters)>(parameters)...);
const std::chrono::steady_clock::time_point stopTimePoint =
std::chrono::steady_clock::now();
const std::chrono::duration<double> timeSpan = std::chrono::duration_cast<
std::chrono::duration<double>>(stopTimePoint - startTimePoint);
std::cout << "Computation took " << timeSpan.count()
<< " seconds." << std::endl;
return result;
};
};
Live demo(引用返回)。
特别注意decltype(auto)
的宗教用途。也是第二版的mutable
。
答案 1 :(得分:1)
不知道您正在尝试做什么,但是如果您要这样做,我想在这里我已经做了什么:
#include <chrono>
#include <iostream>
#include <functional>
auto measure = [](auto function, auto&&... parameters) -> decltype(function(parameters...))
{
const std::chrono::steady_clock::time_point startTimePoint =
std::chrono::steady_clock::now();
auto returnValue = function(parameters...);
const std::chrono::steady_clock::time_point stopTimePoint =
std::chrono::steady_clock::now();
const std::chrono::duration<double> timeSpan = std::chrono::duration_cast<
std::chrono::duration<double>>(stopTimePoint - startTimePoint);
std::cout << "Computation took " << timeSpan.count()
<< " seconds." << std::endl;
return returnValue;
};
class Test
{
public:
int computation(double dummy)
{
std::cout << "Received " << dummy << ". Computing..." << std::endl;
return 123;
}
};
int main(int, char**)
{
Test instance;
auto func = std::bind(&Test::computation, &instance, std::placeholders::_1);
int result = measure(func, 1.0);
std::cout << "Result: " << result << std::endl;
return 0;
}
答案 2 :(得分:0)
如果由于某种原因不喜欢采用成员函数的函数指针(请参见here),那么好的旧宏可以助您一臂之力。有些人建议尽量减少对宏的使用,但是在这种情况下,我发现它更加直观,易读和容易。可以使用以下宏对任意函数(包括调用返回类型的类的公共成员函数的调用)进行计时。
#define timefnrt(W, X, Z){\
time_t V = time(NULL);\
W = X;\
time_t Y = time(NULL);\
Z = difftime(Y, V);\
};
如果该函数返回一个void
,则可以这样计时:
#define timefnvoid(X, Z){\
time_t V = time(NULL);\
X;\
time_t Y = time(NULL);\
Z = difftime(Y, V);\
};