# 1115.Print FooBar Alternately

**1115.Print FooBar Alternately**

难度:Medium

> Suppose you are given the following code:

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 1:

Input: n = 1
Output: "foobar"
Explanation: There are two threads being fired asynchronously. One of them calls foo(), while the other calls bar(). "foobar" is being output 1 time.
Example 2:

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

要交替输出foo和bar，在每个函数中，每次循环结束将当前线程锁起来，令下一个循环处于阻塞状态，待另一个函数的当前循环结束后，解锁，交替进行输出。

```
class FooBar {
private:
    int n;
    pthread_mutex_t t1=PTHREAD_MUTEX_INITIALIZER;
    pthread_mutex_t t2=PTHREAD_MUTEX_INITIALIZER;
public:
    FooBar(int n) {
        this->n = n;
        pthread_mutex_lock(&t1);

        
    }

    void foo(function<void()> printFoo) {
        
        for (int i = 0; i < n; i++) {
            pthread_mutex_lock(&t2);

        	// printFoo() outputs "foo". Do not change or remove this line.
        	printFoo();
            pthread_mutex_unlock(&t1);
        }
    }

    void bar(function<void()> printBar) {
        for (int i = 0; i < n; i++) {
            pthread_mutex_lock(&t1);

        	// printBar() outputs "bar". Do not change or remove this line.
        	printBar();
            pthread_mutex_unlock(&t2);

        }
    }
};
```

> 执行用时 :52 ms, 在所有 C++ 提交中击败了50.00%的用户\
> 内存消耗 :10.3 MB, 在所有 C++ 提交中击败了100.00%的用户