if函数中的奇怪语句

Question

我从以下python函数得到了非常奇怪的输出：

s = "-8.8373347749999997e-08"
def roundNumb(numb, pos):
    print("roundNumb()", numb , pos)
    ret = ""
    if int(numb[pos]) < 5:
        ret = numb[:pos]
        print("if int(numb[pos]) < 5: ", int(numb[pos]), " ", ret)
    else:
        if int(numb[pos]) == 9:
            ret = roundNumb(numb , pos - 1)
            print("int(numb[pos]) == 9 ", int(numb[pos]), " ",ret)
        else:
            ret = numb[:(pos-1)] + str(int(numb[(pos-1)]) + 1)
            print("else:", ret)
    return ret


>>> roundNumb(s, 12)
roundNumb() -8.8373347749999997e-08 12
roundNumb() -8.8373347749999997e-08 11
if int(numb[pos]) < 5:  4   -8.83733477
int(numb[pos]) == 9  9   -8.83733477
'-8.83733477'

正如你所看到的，roundNumb函数被调用了两次，但我真的不明白为什么。它应该在第一个if语句结束：

if int(numb[pos]) < 5:

Answer 1

你调用函数 roundNumb(s, 12) // where numb[pos] = 9

然后函数转到else块 并执行if块

if int(numb[pos]) == 9:

如果您再次调用此函数，则这是此块执行之前的有趣部分

ret = roundNumb(numb , pos - 1) // and herer is function execution pause

print("int(numb[pos]) == 9 ", int(numb[pos]), " ",ret)

然后functoid执行函数的if块

if int(numb[pos]) < 5:

ret = numb[:pos]

print("if int(numb[pos]) < 5: ", int(numb[pos]), " ", ret)

然后执行此打印功能，然后执行暂停的功能恢复 然后执行这个打印函数

print("int(numb[pos]) == 9 ", int(numb[pos]), " ",ret)

有关递归的知识对理解这一点非常有帮助