Python的互斥鎖與信號(hào)量詳解

import threading
# 聲明互斥鎖
lock=threading.Rlock();
def handle(sid):# 功能實(shí)現(xiàn)代碼
lock.acquire() #加鎖
# writer codeing
lock.relase() #釋放鎖

# 聲明信號(hào)量：
sema=threading.Semaphore(0); #無上限檢查
sema=threading.BuderedSeamphore(1) #有上限檢查設(shè)置
5
apple=1
def consumner():
seam.acquire(); # ‐1
9
if apple==1:
pass
else: sema2.release();#+ 1
def product():
seam.relarse(); # +1
if apple==1:
pass
else:
print("消費(fèi)：",apple);

# -*- coding: utf-8 -*-
"""
Created on Mon Sep 9 21:49:30 2019

@author: DGW-PC
"""
# 信號(hào)量解決生產(chǎn)者消費(fèi)者問題
import random;
import threading;
import time;

# 聲明信號(hào)量
sema=threading.Semaphore(0);# 必須寫參數(shù) 0 表示可以使用數(shù)
sema2=threading.BoundedSemaphore(1);

apple=1;

def product():#生產(chǎn)者
  global apple;
  apple=random.randint(1,100);
  time.sleep(3);
  print("生成蘋果:",apple);
  #sema2.release(); # +1
  if apple==1:
     pass
  else: sema2.release();#+ 1 
    
def consumer():
  print("等待");
  sema2.acquire();# -1
  if apple==1:
    pass
  else:    
    print("消費(fèi)：",apple);
threads=[];

for i in range(1,3):
  t1=threading.Thread(target=consumer);
  t2=threading.Thread(target=product);
  t1.start();
  t2.start();
  threads.append(t1);
  threads.append(t2);
for x in threads:
  x.join();