Question

我遇到的问题是在a.cpp中写为t1_t0 = t1 - t0;的类似类型之间的操作类型声明。 t1和t0分别是最终和初始时间，差异用于衡量一次计算之间的时间跨度。我尝试在整个函数中使用auto代替变量boost::chrono::high_resolution_clock::time_point，std::chrono::seconds和int，但是在调用函数之后声明变量会导致奇怪的问题在它之前。

它被称为error: use of before deduction of ‘auto’编译错误，因此我尝试使用真实类型编译项目，现在我遇到了这个问题。如果该类在auto 之上声明，则该项目编译正常（使用int main）但如果该类被分隔为头文件和cpp文件则无法正常工作。

这是整个项目，请帮忙！

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

int main()
{
    a A;

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

    return 0;
}

这是a.h

#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

这是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?
}

以下是编译它的错误。

||=== 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)) ===|

编辑过的代码，这确实修复了原始代码，现在可以正常工作。

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

请勿混用boost::chrono和std::chrono。使用其中一个。如果您的平台提供并支持std::chrono，我建议您使用boost::chrono。
两个time_point之间的差异是duration，而不是另一个time_point。例如，“明天”和“今天”可以被认为是时间点（具有非常粗略的精度）。 tomorrow - today == 1 day。一天是持续时间。
如果您希望最终结果为纳秒，则无需如此努力：

nanoseconds nsec = t1 - t0;
如何从nanoseconds返回int而不是a::time()？这样，客户端具有类型安全的持续时间，而不是int，这可能意味着什么。

return t1 - t0;

将boost :: chrono计时机制包装成类编译错误？

1 个答案: