如何在Processing中每10秒向变量添加+1? [英] How to add +1 to variable every 10 seconds in Processing?

问题描述

请原谅我的无知，但我遇到了一个问题，尽管这个想法很简单，但对我目前在使用 Processing 编程方面的知识来说是个挑战.你看，我需要每 10 秒向变量添加 1 个单位.这是代码:

Excuse my ignorance, but I ran into a problem that turned out to be challenging for my current knowledge in programming with Processing, even though the idea is quite simple. You see, I need to add 1 unit to a variable every 10 seconds. This is the code:

int i = 0;

void setup()
{
  frameRate(60);
}

void draw()
{

  int time = (millis() % 10000) / 1000;

  if (time == 9)
  {
    i++;
  } else {}

  System.out.println("-------------------------------\n" +
                     "Timer: " + time + "\n"
                   + "Adding 1 every 10 seconds: : " + i + "\n"
                   + "-------------------------------");
}

问题是因为 draw() 每秒循环 60 次，一旦 time 达到 9 秒，它就会使 if 语句执行 60 次，它结束每 10 秒向 i 添加 60，我只需要添加 1.

The problem is that because draw() loops 60 times per second, as soon as time reaches 9 the second it last makes the if statement to be executed 60 times and it ends adding 60 to i every 10 seconds and I just need to be adding 1.

我尝试应用某种算法来减去不必要的数字，就像这样:

I tried to apply some kind of algorithm that subtracts the unnecessary numbers as they increase like so:

int i = 1;
int toSubstract = 0; //Variable for algorithm

void setup()
{
  frameRate(60);
}

void draw()
{

  int time = (millis() % 10000) / 1000;

  if (time == 9)
  {
    i++;
    algToSubstract();
  } else {}



  System.out.println("-------------------------------\n" +
                     "Timer: " + time + "\n"
                   + "Adding 1 every 10 seconds: : " + i + "\n"
                   + "-------------------------------");
}

void algToSubstract() //<--- This is the algorithm
{
  i = i - toSubstract;
  toSubstract++;

  if (toSubstract > 59)
  {
    toSubstract = 0;
  } else {}
}

...但我无法让它工作.这个想法是这样的:

...but I couldn't make it work. The idea was something like this:

time 达到 9，if 语句执行，i = 1 and toSubstract = 0.

time reaches 9, if statement executes, i = 1 and toSubstract = 0.

i 增加 1 所以 i = 2.

i increases 1 so i = 2.

i = i - toSusbract (i = 2 - 0 所以 i = 2).

i = i - toSusbract (i = 2 - 0 so i = 2).

toSusbract 增加 1，所以 toSusbract = 1.

toSusbract increases 1 so toSusbract = 1.

i 增加 1 所以 i = 3.

i increases 1 so i = 3.

i = i - toSusbract (i = 3 - 1 所以 i = 2).

i = i - toSusbract (i = 3 - 1 so i = 2).

toSusbract 增加 1 所以 toSusbract = 2.

toSusbract increases 1 so toSusbract = 2.

...过程继续...

toSubstract 大于 59，因此重新启动为 0.

toSubstract gets bigger than 59 so it is restarted to 0.

time 不再是 9.

推荐答案

Ringo 有一个非常好的解决方案.

Ringo has a solution that's perfectly fine.

另一种可以轻松做到这一点的方法是:

Another way you can do this easily is:

bool addOnce = false;
void draw()
{
  int time = (millis() % 10000) / 1000;
  if (time == 9)
  {
      if(!addOnce) {
          addOnce = true;
          i++;
      }
  } else { addOnce = false; }
}

只要time == 9，我们只会通过一次if(!addOnce).

As long as time == 9, we'll only get through if(!addOnce) one time.

更改后，我们重置标志.

After it changes, we reset the flag.

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