C++ 延迟执行1秒
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Delay execution 1 second
提问by raze
So I am trying to program a simple tick-based game. I write in C++ on a linux machine. The code below illustrates what I'm trying to accomplish.
所以我正在尝试编写一个简单的基于刻度的游戏。我在 linux 机器上用 C++ 编写。下面的代码说明了我要完成的任务。
for (unsigned int i = 0; i < 40; ++i)
{
functioncall();
sleep(1000); // wait 1 second for the next function call
}
Well, this doesn't work. It seems that it sleeps for 40 seconds, then prints out whatever the result is from the function call.
嗯，这行不通。它似乎休眠了 40 秒，然后打印出函数调用的任何结果。
I also tried creating a new function called delay, and it looked like this:
我还尝试创建一个名为 delay 的新函数，它看起来像这样：
void delay(int seconds)
{
time_t start, current;
time(&start);
do
{
time(¤t);
}
while ((current - start) < seconds);
}
Same result here. Anybody?
这里的结果相同。有人吗？
回答by Alex Z
To reiterate on what has already been stated by others with a concrete example:
用一个具体的例子重申其他人已经说过的内容：
Assuming you're using std::coutfor output, you should call std::cout.flush();right before the sleep command. See this MS knowledgebase article.
假设您使用的std::cout是输出，您应该std::cout.flush();在 sleep 命令之前调用。请参阅此 MS 知识库文章。
回答by Frank Schmitt
sleep(n) waits for n seconds, not n microseconds. Also, as mentioned by Bart, if you're writing to stdout, you should flush the stream after each write - otherwise, you won't see anything until the buffer is flushed.
sleep(n) 等待 n 秒，而不是 n 微秒。此外，正如 Bart 所提到的，如果您要写入 stdout，则应在每次写入后刷新流 - 否则，在刷新缓冲区之前您将看不到任何内容。
回答by jfs
So I am trying to program a simple tick-based game. I write in C++ on a linux machine.
所以我正在尝试编写一个简单的基于刻度的游戏。我在 linux 机器上用 C++ 编写。
if functioncall()may take a considerable time then your ticks won't be equal if you sleep the same amount of time.
如果functioncall()可能需要相当长的时间，那么如果您睡眠相同的时间，您的滴答声将不相等。
You might be trying to do this:
您可能正在尝试这样做：
while 1: // mainloop
functioncall()
tick() # wait for the next tick
Here tick()sleeps approximately delay - time_it_takes_for(functioncall)i.e., the longer functioncall()takes the less time tick()sleeps.
这里tick()睡眠大约，delay - time_it_takes_for(functioncall)即，睡眠时间越长functioncall()，tick()睡眠时间越短。
sleep()sleeps an integer number of seconds. You might need a finer time resolution. You could use clock_nanosleep()for that.
sleep()休眠整数秒。您可能需要更精细的时间分辨率。你可以用clock_nanosleep()它。
Example Clock::tick() implementation
示例 Clock::tick() 实现
// $ g++ *.cpp -lrt && time ./a.out
#include <iostream>
#include <stdio.h> // perror()
#include <stdlib.h> // ldiv()
#include <time.h> // clock_nanosleep()
namespace {
class Clock {
const long delay_nanoseconds;
bool running;
struct timespec time;
const clockid_t clock_id;
public:
explicit Clock(unsigned fps) : // specify frames per second
delay_nanoseconds(1e9/fps), running(false), time(),
clock_id(CLOCK_MONOTONIC) {}
void tick() {
if (clock_nanosleep(clock_id, TIMER_ABSTIME, nexttick(), 0)) {
// interrupted by a signal handler or an error
perror("clock_nanosleep");
exit(EXIT_FAILURE);
}
}
private:
struct timespec* nexttick() {
if (not running) { // initialize `time`
running = true;
if (clock_gettime(clock_id, &time)) {
//process errors
perror("clock_gettime");
exit(EXIT_FAILURE);
}
}
// increment `time`
// time += delay_nanoseconds
ldiv_t q = ldiv(time.tv_nsec + delay_nanoseconds, 1000000000);
time.tv_sec += q.quot;
time.tv_nsec = q.rem;
return &time;
}
};
}
int main() {
Clock clock(20);
char arrows[] = "\|/-";
for (int nframe = 0; nframe < 100; ++nframe) { // mainloop
// process a single frame
std::cout << arrows[nframe % (sizeof(arrows)-1)] << '\r' << std::flush;
clock.tick(); // wait for the next tick
}
}
Note: I've used std::flush()to update the output immediately.
注意：我习惯于std::flush()立即更新输出。
If you run the program it should take about 5 seconds (100 frames, 20 frames per second).
如果您运行该程序，它应该需要大约 5 秒（100 帧，每秒 20 帧）。
回答by Rohit Vipin Mathews
I guess on linux u have to use usleep()and it must be found in ctime
我想在 linux 上你必须使用usleep()它，它必须在ctime
And in windows you can use delay(), sleep(), msleep()
在 Windows 中你可以使用 delay()、sleep()、msleep()
