解决python递归函数及递归次数受到限制的问题_python

def sum(n):
    if n > 0:              
        return n+sum(n-1)
    else:
        return 0           
new_sum = sum(10)
print(new_sum)
#output
55

def factorial(n):
    if n == 1:
        return 1
    else:
        return n*factorial(n-1)
new_sum = factorial(5)
print(new_sum)
#output
120

def sum(n):
    if n > 0:
        return 1+sum(n-1)  
    else:
        return 0
new_sum = sum(1000)
print(new_sum)

import sys
sys.setrecursionlimit(1500)#可递归次数修改为1500
def sum(n):
    if n > 0:
        return 1+sum(n-1)  
    else:
        return 0
new_sum = sum(1000)
print(new_sum)
#output
1000

function recursionTest()
    local times = 0
    if times < 10 then
        times = times + 1
        recursionTest()
    end
    if times == 0 then
        print("outer round")
    end
end

local recurTimes = 0
function recursionTest2()
    if recurTimes < 10 then
        recurTimes = recurTimes + 1
        recursionTest2()
    end
end

function recursion(str, t)
    str = str or "first time "
    t = t or 0
    t = t + 1
    print(str, t)
    if t < 10 then
        recursion("times:", t)
    end
    if t == 1 then
        print("outer round")
    end
end