如何修改Python代码避免递归错误（macOS环境）

问题背景 #

我正在编写一个程序，用于对比每次成功生成两个相同随机数后，扩大随机数范围时的耗时，后续计划添加数据记录模块。目前程序运行报错：

ValueError: not allowed to raise maximum limit

报错出自这段代码：

resource.setrlimit(resource.RLIMIT_STACK, (8388608, -1))

程序因递归过深受限，我是编程新手，不知如何重构代码避免递归。注释掉sys.setrecursionlimit()后，程序能正常运行到范围约240时，因递归限制触发崩溃。我尝试提升递归限制和栈大小，但macOS限制此操作。

现有代码 #

import random
import time
import sys
import resource

#set the highest recursion limit
sys.setrecursionlimit(2147483617)

#change the stack size to unlimited, doesn't woork on macos
resource.setrlimit(resource.RLIMIT_STACK, (8388608, -1))

#These are just a counter for every attempt
ticker = 0

#This is the variable that determines the range that the random number can fall between
tick = 0

def random_number():

    #Don't know any other way to get the compare def to pull in num and num2 apart from using a global variable
    global num
    global num2

    #I commented out parts of the code that seem to break the program such as these next 3 comments
    #global tick
    #tickers()
    #tick = tick + 1
    num = random.randint (0, tick)
    num2 = random.randint (0, tick)
    return num and num2
    #del num, num2

def compare():
    global ticker
    global tick
    
    #Generate two random numbers
    random_number()

    #compare them
    while num != num2:
        ticker = ticker + 1
        #time.sleep(0.05)
        #print(ticker, 'attempts, the two guessed numbers were', num, '&', num2)
        #del num, num2
        compare()
    else:
        #time.sleep(0.05)
        tick = tick + 1
        print (ticker, 'attempts when guessing between',tick, 'numbers. Number found. It is', num)
        ticker = 0
        compare()

compare()

期望输出 #

程序能无限输出如下结果（后续将设置断点）：

2 attempts when guessing between 7 numbers. Number found. It is 6
14 attempts when guessing between 8 numbers. Number found. It is 4
6 attempts when guessing between 9 numbers. Number found. It is 3
23 attempts when guessing between 10 numbers. Number found. It is 3
28 attempts when guessing between 11 numbers. Number found. It is 7
2 attempts when guessing between 12 numbers. Number found. It is 4

解决方案：重构为非递归版本 #

核心思路是把递归调用改成循环结构，同时去掉全局变量，让代码更清晰：

改写后的代码 #

import random

def main():
    tick = 0  # 随机数范围上限，初始为0
    while True:  # 无限循环直到手动终止
        tick += 1
        ticker = 0
        while True:
            # 生成两个范围内的随机数
            num = random.randint(0, tick)
            num2 = random.randint(0, tick)
            ticker += 1
            if num == num2:
                print(f"{ticker} attempts when guessing between {tick} numbers. Number found. It is {num}")
                break  # 找到相同数后，进入下一个更大的范围

if __name__ == "__main__":
    main()

代码说明 #

1. 去掉全局变量：用函数内部变量替代，避免全局变量带来的逻辑混乱
2. 双层循环替代递归：
   • 外层循环负责逐步扩大随机数范围（对应原代码中tick递增的逻辑）
   • 内层循环负责不断生成随机数，直到两个数相等（对应原代码中递归调用compare()的重复尝试逻辑）
3. 简化逻辑：删除了不必要的random_number()函数，直接在循环内生成随机数，减少函数调用开销
4. 移除无效配置：去掉了sys.setrecursionlimit()和resource.setrlimit()相关代码，因为不再依赖递归

这样改写后，程序不会再出现递归过深的问题，可以持续运行并输出符合预期的结果。

内容的提问来源于stack exchange，提问作者jmac0014