C++ QueryPerformanceCounter返回负数

嘿，我正在计算函数执行所需的时间 我是这样做的： 定时器.cpp

long long int Timer :: clock1()
{
    QueryPerformanceCounter((LARGE_INTEGER*)&time1);
    return time1;
}
long long int Timer :: clock2()
{
    QueryPerformanceCounter((LARGE_INTEGER*)&time2);
    return time2;
}

#include "Timer.h"  //To allow the use of the timer class.

Timer query;

void print()
{
    query.clock1();
    //Loop through the elements in the array.
    for(int index = 0; index < num_elements; index++)
    {
        //Print out the array index and the arrays elements.
        cout <<"Index: " << index << "\tElement: " << m_array[index]<<endl;
    }
    //Prints out the number of elements and the size of the array.
    cout<< "\nNumber of elements: " << num_elements;
    cout<< "\nSize of the array: " << size << "\n";
    query.clock2();
    cout << "\nTime Taken : " << query.time1 - query.time2;
}

main.cpp

long long int Timer :: clock1()
{
    QueryPerformanceCounter((LARGE_INTEGER*)&time1);
    return time1;
}
long long int Timer :: clock2()
{
    QueryPerformanceCounter((LARGE_INTEGER*)&time2);
    return time2;
}

#include "Timer.h"  //To allow the use of the timer class.

Timer query;

void print()
{
    query.clock1();
    //Loop through the elements in the array.
    for(int index = 0; index < num_elements; index++)
    {
        //Print out the array index and the arrays elements.
        cout <<"Index: " << index << "\tElement: " << m_array[index]<<endl;
    }
    //Prints out the number of elements and the size of the array.
    cout<< "\nNumber of elements: " << num_elements;
    cout<< "\nSize of the array: " << size << "\n";
    query.clock2();
    cout << "\nTime Taken : " << query.time1 - query.time2;
}

#包括“Timer.h”//以允许使用Timer类。
定时器查询；
作废打印（）
{
query.clock1（）；
//循环遍历数组中的元素。
for（int index=0；indexcout您正在从开始时间减去结束时间

cout << "\nTime Taken : " << query.time1 - query.time2;

cout假设我从10秒开始，30秒结束。需要多长时间？20秒。要得到它，我们要做
30-10
；也就是说，第二次减去第一次

所以，也许你想要：

cout << "\nTime Taken : " << (query.time2 - query.time1);

cout
query.time1-query.time2
如果
time2
较大，则会为负值，对吗？正确，但我如何才能实现这两个变量的差异呢？
query.time2-query.time1