Question

作为一个初学者，我正在尝试编写类似的代码。当然，这是错误的方法，因为它根本不起作用。我想做的是：使用我在if循环内执行某操作时获得的结果，并在下次输入if时使用这些结果计算条件... 更确切地说，如果cond1 && cond2为TRUE，我想做同样的事情，但是即使情况发生（延迟1000）且值不低于前一个值，也不能尽早...

我不知道该如何编码，这是我第一次输入if和for其他... 你能帮我吗？？？非常感谢。

boolean cond_1 = a < 10;
boolean cond_2 = b < 15;
boolean cond_delay = (time_now - time_i_have_done_something) > 1000;
boolean cond_value = value_now > value_i_have_done_something;

if (cond_A1 == true && cond_A2 == true && cond_delay == true) {
 do something;
 long time_i_have_done_something = do_something.gettime();
 double value_i_have_done_something = do_something.getvalue();
}

我要实现的逻辑示例：

if (cond_A1 == true) { // firt occurence at 7 h 30 min 00 sec 000 ms
time = gettime();
value = getvalue();
IOrder order = engine.submitOrder("EURUSD", instrument, orderCommand, 1); 
}
if (cond_A1 == true) { // second occurence at 7 h 30 min 00 sec 190 ms
// do nothing because time between occurence#1 and occurence#2 < 1000 ms  
}
if (cond_A1 == true) { // firt occurence at 7 h 30 min 01 sec 050 ms
// execute this because time between occurence#1 and occurence#3 > 1000 ms 
time = gettime();
value = getvalue();
IOrder order = engine.submitOrder("EURUSD", instrument, orderCommand, 1); 
}

Answer 1

这将循环，并在调用操作doSomething()之前，如果if将检查条件是否为真，并且自从上次执行操作以来已经过了1秒。它还将等待第一动作延迟1秒钟，因为第一次呼叫不会被视为特殊情况。可以通过这样声明启动来解决此问题：

long start = System.nanoTime() - timeBetween;

我输入了一个值为5的重复变量。当然可以删除它。

我选择不sleep()线程，而选择yield()，以使处理器能够执行其他任务，同时还可以忙于循环程序：

import java.time.LocalDateTime;
import java.time.temporal.ChronoUnit;

class StackOverflowTest {

  public static void main(String [] args){
    StackOverflowTest test = new StackOverflowTest();
    test.notTooFast();
  }

  public void notTooFast(){
    boolean condition1  = true;
    long    timeBetween = 1_000_000_000;  // 1 second.
    long    start       = System.nanoTime();
    int     repeat      = 5;              // run the loop only 5 times. 

    while (repeat > 0) {  // or for an endless loop, use: while (true)

      // has timeBetween passed since last time?
      if (condition1 && (System.nanoTime() - start > timeBetween)) {

        start = System.nanoTime(); // reset the time
        doSomething(repeat--);     // don't forget to decrease the counter.

      } else {
        Thread.currentThread().yield();  // hint to yield this threads current use of a processor.
      }
    }
  }

  // just print something for the sake of the example:
  public void doSomething(int countdown){
    System.out.println(LocalDateTime
                         .now()
                         .truncatedTo(ChronoUnit.MILLIS)
                       + ": Doing something "
                       + countdown);
  }
}

它打印：

2020-07-15T13:30:34.409: Doing something 5
2020-07-15T13:30:35.350: Doing something 4
2020-07-15T13:30:36.350: Doing something 3
2020-07-15T13:30:37.349: Doing something 2
2020-07-15T13:30:38.350: Doing something 1

我想是那条线

doSomething(repeat--);

应该用这样的东西代替：

repeat--;
IOrder order = engine.submitOrder("EURUSD", instrument, orderCommand, 1);

您也可以使用{p>代替Thread.currentThread().yield();

try {        
  Thread.currentThread().sleep(1000); // 1 second.
} catch (InterruptedException ex) {
  // handle whichever way
  Thread.currentThread().interrupt();  // reset interrupt flag
}

这将使线程休眠1秒。

Java：if条件：使用if内的变量作为if的条件

1 个答案: