我正在尝试使用C和OpenGL / GLUT制作Snake克隆.但是,我一直无法编程允许在运动之间输入的短时间间隔.我已经尝试了几种计时方法,最后我为它做了一个课程(你将在下面看到).这似乎是编程输入延迟的最佳方式(而不是glutTimerFunc()或sleep()),因为计时器与游戏循环分开运行,而不是将整个程序置于保持状态.这很重要,因为我希望玩家能够随时暂停.不幸的是,我现在也遇到了这种方法的问题.我的计时器类似乎忽略了我给它的时间限制的双倍(简单地表示为双“限制”).
为了测试该类,我设置了一个简单的循环控制台程序,在计时器达到时间限制时显示用户的方向输入.它应该每0.33秒显示一次输入.相反,它以固定的间隔显示输入,看起来相隔约0.8秒,无论时间限制给出了什么值.为什么它不会在给定的时间间隔显示输入,为什么它会使它自己的时间限制？
这也恰好是我没有教程的第一个主要的C / OpenGL项目,因此对我的代码/方法的任何评论或建议都表示赞赏！
#include <iostream>
#include "timer.h"
// Include all files necessary for OpenGL/GLUT here.
using namespace std;
Timer timer;
// Insert necessary OpenGL/GLUT code for display/looping here.
void update(int value)
{
if (timer.checkTime())
{
if (GetAsyncKeyState(VK_LEFT))
cout << "You pressed LEFT!" << endl;
else if (GetAsyncKeyState(VK_RIGHT))
cout << "You pressed RIGHT!" << endl;
else if (GetAsyncKeyState(VK_UP))
cout << "You pressed UP!" << endl;
else if (GetAsyncKeyState(VK_DOWN))
cout << "You pressed DOWN!" << endl;
}
glutTimerFunc(1000/60, update, 0);
glutPostRedisplay();
}
timer.h
#pragma once
#include <time.h>
class Timer
{
public:
Timer();
bool checkTime(double limit = 0.33);
private:
double getElapsed();
time_t start;
time_t now;
double elapsed;
bool running;
};
timer.cpp
#include "timer.h"
Timer::Timer()
{
running = false;
}
bool Timer::checkTime(double limit)
{
elapsed = getElapsed();
if (elapsed < limit)
{
return false;
}
else if (elapsed >= limit)
{
running = false;
return true;
}
}
double Timer::getElapsed()
{
if (! running)
{
time(&start);
running = true;
return 0.00;
}
else
{
time(&now);
return difftime(now, start);
}
}
最佳答案 glutTimer每触发1000/60毫秒,你就会检查调用getElapsed的Timer :: checkTime.你有你所有的时间函数定义为double,但你使用time_t,其分辨率只有1秒.
因此,你得到的东西看起来像这样(模拟数字)
start time: 1234.5 seconds
glutTimer: 1234.516 seconds (getElapsed = 0) // checkTime returns false
glutTimer: 1234.532 seconds (getElapsed = 0)
...
glutTimer: 1234.596 seconds (getElapsed = 0)
glutTimer: 1235.012 seconds (getElapsed = 1) // condition finally returns true
...
因此,实际延迟取决于您何时设置相对于time()使用的Epoch的实际开始时间的开始时间;
我怀疑如果统计测量它平均接近0.5秒.
回答有关“决议”的问题：
返回当前时间的不同功能返回不同的准确度.例如,现在,我屏幕右下角的时钟显示为“12:14 PM”.这恰好是12:14而没有秒,或12:14和59秒？我无法分辨,因为时钟显示的“分辨率”是一分钟.同样地,我可能会说,如果我以“一刻钟”的分辨率报告时间,那么它是在12小时后的一个季度,当时真的是14分钟.作为人类,我们一直这样做而不考虑它.在软件中,您必须了解所调用的任何功能的这些细节.
如果您使用的是Windows,则可通过QueryPerformanceCounter API获得高分辨率计时器.在大多数平台上,性能计数器是基于硬件的,具有微秒级的分辨率.
LARGE_INTEGER StartingTime, EndingTime, ElapsedMicroseconds;
LARGE_INTEGER Frequency;
QueryPerformanceFrequency(&Frequency); // get number of ticks per second
QueryPerformanceCounter(&StartingTime); // get starting # of ticks
// Activity to be timed
QueryPerformanceCounter(&EndingTime); // get ending # of ticks
ElapsedMicroseconds.QuadPart = EndingTime.QuadPart - StartingTime.QuadPart;
//
// We now have the elapsed number of ticks, along with the
// number of ticks-per-second. We use these values
// to convert to the number of elapsed microseconds.
// To guard against loss-of-precision, we convert
// to microseconds *before* dividing by ticks-per-second.
//
ElapsedMicroseconds.QuadPart *= 1000000;
ElapsedMicroseconds.QuadPart /= Frequency.QuadPart;
在Linux上可能有类似的工具,但我不熟悉它.
试试这个：
void update(int value)
{
if (timer.hasTicked())
{
if (GetAsyncKeyState(VK_LEFT))
cout << "You pressed LEFT!" << endl;
else if (GetAsyncKeyState(VK_RIGHT))
cout << "You pressed RIGHT!" << endl;
else if (GetAsyncKeyState(VK_UP))
cout << "You pressed UP!" << endl;
else if (GetAsyncKeyState(VK_DOWN))
cout << "You pressed DOWN!" << endl;
}
else if (!timer.isRunning())
{
timer.start();
}
glutTimerFunc(1000/60, update, 0);
glutPostRedisplay();
}
timer.h
// this class provides a timer that can be polled and will allow the user to tell if a period has elapsed.
// note that this timer does NOT throw any events for timeout.
class PollTimer
{
public:
PollTimer();
// assuming time limit is a number of msec that and fits within a normal integer rather than the 64 bit
// variant (that would be a LONG LONG delay).
void setTimeout(int msDelay);
// Timers generally expose start/stop and it’s not generally a good idea to make a single function
// that overloads complex combinations of behaviors as here both the start & elapsed operations.
// admit this is a simple case, but generally it’s a bad design pattern that leads to “evil”.
void start();
void stop();
bool isRunning();
// Has the timer expired since the last poll
bool hasTicked();
private:
LARGE_INTEGER startTime;
LARGE_INTEGER frequency; // per second
int delay; // in milliseconds
bool running;
};
timer.cpp
#include "timer.h"
PollTimer::PollTimer()
{
// omitting error check for hardware that doesn’t support this.
QueryPerformanceFrequency(& frequency); // get number of ticks per second
running = false;
}
void PollTimer::setTimeout(int msDelay)
{
delay = msDelay;
}
void PollTimer::start()
{
QueryPerformanceCounter(&startTime);
running = true;
}
void PollTimer::stop()
{
running = false;
}
bool PollTimer::isRunning()
{
return running;
}
bool PollTimer::hasTicked()
{
if (!running)
return false;
LARGE_INTEGER now;
QueryPerformanceCounter(&now);
LARGE_INTEGER ElapsedMilliseconds;
ElapsedMilliseconds.QuadPart = now.QuadPart - startTime.QuadPart;
ElapsedMilliseconds.QuadPart *= 1000000;
ElapsedMilliseconds.QuadPart /= frequency.QuadPart; // now microseconds
ElapsedMilliseconds.QuadPart /= 1000; // milliseconds
bool fExpired = ( ElapsedMilliseconds.HighPart > 0 || ElapsedMilliseconds.LowPart >= delay ) ;
if (fExpired)
{
// reset start time
start(); // don’t copy/paste code you can call.
}
return fExpired;
}
