如何不等待线程在python中完成

From the documentation ,请 target

你不能把你的功能传递给 ,您正在传递函数的返回值,因此函数在传递给 threading.Thread t.start()

你应该使用 args

只需更改此行:

 t = threading.Thread(target=whatever(i))

t = threading.Thread(target=whatever, args=(i,))

你似乎没有意识到这行的作用: t = threading.Thread(target=whatever(i)) 开始

target t = threading.Thread(target=whatever(1)) ,的 whatever 你已经在原来的线程中运行了,你需要 threading

from time import sleep
import threading

def whatever(i):
  sleep(5)
  print("Hey! It's me number " + str(i))

for i in range(3):
  t = threading.Thread(target=whatever, args=(i,))
  t.start()

from time import sleep
import threading

def whatever(i):
  print("Hey! It's me number " + str(i))

for i in range(3):
  t = threading.Thread(target=whatever, args=(i,))
  t.start()

但是你必须考虑一件事。

把它包装成一个lambda。

from time import sleep
import threading

def whatever(i):
  sleep(5)
  print("Hey! It's me number " + str(i))

for i in range(3):
  t = threading.Thread(target=lambda: whatever(i))
  t.start()