python 线程池顺序执行的方法实现

from concurrent.futures import ThreadPoolExecutor

def task(n):
    return f"Task {n} completed"

with ThreadPoolExecutor(max_workers=1) as executor:  # 单线程
    futures = [executor.submit(task, i) for i in range(1, 6)]
    for future in futures:
        print(future.result())

from concurrent.futures import ThreadPoolExecutor

def task(n):
    import time, random
    time.sleep(random.uniform(0, 1))  # 模拟随机耗时
    return f"Task {n} completed"

with ThreadPoolExecutor() as executor:
    # 提交任务并按顺序存储 Future 对象
    futures = [executor.submit(task, i) for i in range(1, 6)]
    # 按提交顺序逐个获取结果（会阻塞直到对应任务完成）
    for future in futures:
        print(future.result())  # 输出顺序始终是 1,2,3,4,5

from concurrent.futures import ThreadPoolExecutor
import threading

lock = threading.Lock()
current_step = 0

def sequential_task(n):
    global current_step
    while True:
        with lock:
            if n == current_step + 1:
                print(f"Executing Task {n}")
                current_step = n
                break

with ThreadPoolExecutor(max_workers=5) as executor:
    executor.map(sequential_task, range(1, 6))  # 严格按 1→2→3→4→5 执行

from concurrent.futures import ThreadPoolExecutor
import queue

def worker(q):
    while not q.empty():
        task_id = q.get()
        print(f"Processing Task {task_id}")
        q.task_done()

task_queue = queue.Queue()
for i in range(1, 6):
    task_queue.put(i)

with ThreadPoolExecutor() as executor:
    # 启动多个线程消费队列（实际按队列顺序执行）
    for _ in range(3):  # 3个工作线程
        executor.submit(worker, task_queue)
    task_queue.join()  # 等待所有任务完成