將boost :: chrono計時機制包裝到類編譯錯誤中？

Question

我遇到的問題與在類似的類型t1_t0 = t1 - t0;相似類型之間聲明操作的結果類型有關t1_t0 = t1 - t0; 在a.cpp中 t1和t0分別是最后時間和初始時間，差值用於測量一次計算之間的時間跨度。 我嘗試在整個函數中使用auto代替變量boost::chrono::high_resolution_clock::time_point ， std::chrono::seconds boost::chrono::high_resolution_clock::time_point std::chrono::seconds和int但是在調用函數之前，在聲明變量時引起了奇怪的問題它。

它被稱為error: use of before deduction of 'auto'編譯錯誤error: use of before deduction of 'auto'因此我嘗試使用實際類型來編譯項目，現在遇到了麻煩。 如果在int main之上聲明了該類，則該項目的編譯效果很好（使用auto ）， 但是如果將該類分為頭文件和cpp文件，則該項目將無法正常工作 。

這是整個項目，請幫忙！

#include <iostream>
#include "a.h"

int main()
{
    a A;

    int timer = A.time();
    std::cout<<timer<<std::endl;

    return 0;
}

這是啊

#ifndef A_H
#define A_H

#include <iostream>
#include <boost/chrono/chrono_io.hpp>
#include <boost/date_time/posix_time/posix_time.hpp>
#include <boost/thread/thread.hpp>
#include <sstream>
#include <cassert>
#include <chrono>

class a
{
    private:

    public:

    a();
    ~a();
    int time();
};
#endif

這是a.cpp

#include "a.h"

a::a(){}
a::~a(){}
int a::time()//simple benchmark timer
{
    boost::chrono::high_resolution_clock::time_point t0, t1, t1_t0;
    t0 = boost::chrono::high_resolution_clock::now();

    //Run a process to measure time.

    t1 = boost::chrono::high_resolution_clock::now();

    t1_t0 = t1 - t0;//Not sure what the resulting type is here?

    std::chrono::seconds nsec = std::chrono::duration_cast<std::chrono::seconds>(t1_t0);
    //Not sure what the resulting type is here?

    return nsec.count();//Not sure what the return type is here?
}

這是編譯它的錯誤。

||=== Build: Debug in return_type_auto_from_class (compiler: GNU GCC Compiler) ===|
/home/Desktop/a.cpp||In member function ‘int a::time()’:|
/home/Desktop/a.cpp|14|error: no match for ‘operator=’ (operand types are ‘boost::chrono::steady_clock::time_point {aka boost::chrono::time_point<boost::chrono::steady_clock>}’ and ‘boost::common_type<boost::chrono::duration<long int, boost::ratio<1l, 1000000000l> >, boost::chrono::duration<long int, boost::ratio<1l, 1000000000l> > >::type {aka boost::chrono::duration<long int, boost::ratio<1l, 1000000000l> >}’)|
/usr/include/boost/chrono/time_point.hpp|156|note: candidate: constexpr boost::chrono::time_point<boost::chrono::steady_clock>& boost::chrono::time_point<boost::chrono::steady_clock>::operator=(const boost::chrono::time_point<boost::chrono::steady_clock>&)|
/usr/include/boost/chrono/time_point.hpp|156|note:   no known conversion for argument 1 from ‘boost::common_type<boost::chrono::duration<long int, boost::ratio<1l, 1000000000l> >, boost::chrono::duration<long int, boost::ratio<1l, 1000000000l> > >::type {aka boost::chrono::duration<long int, boost::ratio<1l, 1000000000l> >}’ to ‘const boost::chrono::time_point<boost::chrono::steady_clock>&’|
/usr/include/boost/chrono/time_point.hpp|156|note: candidate: constexpr boost::chrono::time_point<boost::chrono::steady_clock>& boost::chrono::time_point<boost::chrono::steady_clock>::operator=(boost::chrono::time_point<boost::chrono::steady_clock>&&)|
/usr/include/boost/chrono/time_point.hpp|156|note:   no known conversion for argument 1 from ‘boost::common_type<boost::chrono::duration<long int, boost::ratio<1l, 1000000000l> >, boost::chrono::duration<long int, boost::ratio<1l, 1000000000l> > >::type {aka boost::chrono::duration<long int, boost::ratio<1l, 1000000000l> >}’ to ‘boost::chrono::time_point<boost::chrono::steady_clock>&&’|
/home/Desktop/a.cpp|16|error: no matching function for call to ‘duration_cast(boost::chrono::steady_clock::time_point&)’|
/usr/include/c++/5/chrono|194|note: candidate: template<class _ToDur, class _Rep, class _Period> constexpr typename std::enable_if<std::chrono::__is_duration<_Tp>::value, _ToDur>::type std::chrono::duration_cast(const std::chrono::duration<_Rep, _Period>&)|
/usr/include/c++/5/chrono|194|note:   template argument deduction/substitution failed:|
/home/Desktop/a.cpp|16|note:   ‘boost::chrono::steady_clock::time_point {aka boost::chrono::time_point<boost::chrono::steady_clock>}’ is not derived from ‘const std::chrono::duration<_Rep, _Period>’|
||=== Build failed: 2 error(s), 0 warning(s) (0 minute(s), 2 second(s)) ===|

編輯后的代碼，它可以修復原始代碼，並且現在可以使用。

int a::time()
{
    boost::chrono::high_resolution_clock::time_point t0, t1;//, t1_t0;
    t0 = boost::chrono::high_resolution_clock::now();

    //Run a process to measure time.
    boost::this_thread::sleep_for( boost::chrono::milliseconds{ 2000});

    t1 = boost::chrono::high_resolution_clock::now();

    auto t1_t0 = t1 - t0;

    auto nsec = boost::chrono::duration_cast<boost::chrono::seconds>(t1_t0);

    return nsec.count();
}

Answer 1

1. 不要混合boost::chrono和std::chrono 。 使用一個或另一個。 我推薦std::chrono如果您的平台提供並支持它），否則請使用boost::chrono 。

2. 兩個time_point之間的差是duration ，而不是另一個time_point 。 例如，“明天”和“今天”可以被視為時間點（具有非常粗略的精度）。 tomorrow - today == 1 day 。 一天是持續時間。

3. 如果您希望最終結果以納秒為單位，則無需如此努力：

   nanoseconds nsec = t1 - t0;

4. 如何從a::time()返回nanoseconds而不是int呢？ 這樣，客戶端就具有類型安全的持續時間，而不是可能意味着任何含義的int 。

   return t1 - t0;