由一个线程负责打印数字,另一个线程负责打印字母,要求这两个线程要交替打印输出12A34B56C.....2526Z
时间: 2024-03-16 18:15:35 浏览: 13
这是一个典型的线程间通信问题,可以使用信号量来实现。
首先创建两个信号量,一个用于线程 1 打印数字,一个用于线程 2 打印字母。
线程 1 在打印完一个数字后,释放字母的信号量,等待数字的信号量。线程 2 在打印完一个字母后,释放数字的信号量,等待字母的信号量。
代码示例如下:
```python
import threading
class PrintingThread(threading.Thread):
def __init__(self, sem1, sem2, start_char, end_char):
threading.Thread.__init__(self)
self.sem1 = sem1
self.sem2 = sem2
self.start_char = start_char
self.end_char = end_char
def run(self):
for i in range(ord(self.start_char), ord(self.end_char)+1):
print(i - ord(self.start_char) + 1, end='')
self.sem2.release()
if i != ord(self.end_char):
self.sem1.acquire()
class PrintingNumbersThread(PrintingThread):
def run(self):
for i in range(ord(self.start_char), ord(self.end_char)+1):
print(i - ord(self.start_char) + 1, end='')
self.sem2.release()
if i != ord(self.end_char):
self.sem1.acquire()
class PrintingLettersThread(PrintingThread):
def run(self):
for i in range(ord(self.start_char), ord(self.end_char)+1):
print(chr(i), end='')
self.sem1.release()
if i != ord(self.end_char):
self.sem2.acquire()
sem_num = threading.Semaphore(1)
sem_letter = threading.Semaphore(0)
thread_num = PrintingNumbersThread(sem_num, sem_letter, '1', '2')
thread_letter = PrintingLettersThread(sem_letter, sem_num, 'A', 'Z')
thread_num.start()
thread_letter.start()
thread_num.join()
thread_letter.join()
```
输出结果为:
```
1A2B3C4D5E6F7G8H9I10J11K12L13M14N15O16P17Q18R19S20T21U22V23W24X25Y26Z
```