我正在尝试使用 C++ 和 OpenGL/GLUT 制作一个 Snake 克隆。然而，我一直在编程允许输入 Action 之间的短时间间隔时遇到问题。我已经尝试了一些计时方法，最后我为它创建了一个类(如下所示)。这似乎是对输入延迟进行编程的最佳方式(而不是 glutTimerFunc() 或 sleep())，因为计时器独立于游戏循环运行，而不是暂停整个程序。这很重要，因为我希望播放器能够随时暂停。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双倍时间限制(简单表示为双倍“限制”)。
为了测试该类，我设置了一个简单的循环控制台程序，该程序在计时器达到时间限制时显示来自用户的定向输入。它应该每 0.33 秒显示一次输入。相反，它以大约 0.8 秒的固定间隔显示输入，而不管时间限制的值是多少。 为什么它不在给定的时间间隔显示输入，为什么它有自己的时间限制？
这也恰好是我第一个没有教程的主要 C++/OpenGL 项目，因此欢迎对我的代码/方法提出任何意见或建议!
#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();
}
定时器.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;
};
定时器.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 1000/60 毫秒触发一次，您检查 Timer::checkTime，它调用 getElapsed。您的所有时间函数都以 double 定义，但您使用的是 time_t，它的分辨率仅为 1 秒。
因此，你会得到看起来像这样的东西(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开始时间相对于从 time() 使用的纪元开始的一秒的实际开始时间；
我怀疑如果您对其进行统计测量，平均时间接近 0.5 秒。
关于“决议”的回答问题:
返回当前时间的不同函数会返回不同级别的准确度。例如，现在，我屏幕右下角的时钟显示为“12:14 PM”。是正好 12:14 没有秒，还是 12:14 加 59 秒？我不能说，因为时钟显示的“分辨率”是一分钟。同样，我可能会说现在是 12 点一刻，如果我以“一刻钟”的分辨率报告时间，那么实际上是整点过 14 分钟。作为人类，我们一直都在不假思索地这样做。在软件中，您必须注意调用任何函数的这些细节。
如果您使用的是 Windows，则可以通过 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，分辨率在微秒范围内。
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;
Linux 上可能有类似的工具，但我不熟悉。
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();
}
定时器.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;
};
定时器.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;
}
关于C++ 蛇克隆 : timer function ignores given stop time and stops at it's own fixed time，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/24846663/