Suppose you are given the following code:

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class FooBar {
  public void foo() {
    for (int i = 0; i < n; i++) {
      print("foo");
    }
  }

  public void bar() {
    for (int i = 0; i < n; i++) {
      print("bar");
    }
  }
}

The same instance of FooBar will be passed to two different threads. Thread A will call foo() while thread B will call bar(). Modify the given program to output “foobar” n times.

Example:

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Input: n = 2
Output: "foobarfoobar"
Explanation: "foobar" is being output 2 times.

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import java.util.concurrent.Semaphore;

class FooBar {
    private int n;
    Semaphore bar = new Semaphore(0);
    Semaphore foo = new Semaphore(1);
    
    public FooBar(int n) {
        this.n = n;
    }

    public void foo(Runnable printFoo) throws InterruptedException {
        
        for (int i = 0; i < n; i++) {
            
        	// printFoo.run() outputs "foo". Do not change or remove this line.
            foo.acquire();
        	printFoo.run();
            bar.release();
        }
    }

    public void bar(Runnable printBar) throws InterruptedException {
        
        for (int i = 0; i < n; i++) {
            
            // printBar.run() outputs "bar". Do not change or remove this line.
            bar.acquire();
        	printBar.run();
            foo.release();
        }
    }
}

从 Semaphore bar = new Semaphore(0);和后面的bar.release();可以看出来，bar一开始初始化为0只是表明不能直接用，而不是说整个信道permit的数量都最多不超过0: bar.release()就创建了一个。只是表明一开始为0，不能直接用