Question

我的朋友最近向我提出了一项巨大挑战,即,如果不利用内在的(t)功能,就把一项投入转化为暗中......。基本上,他想要我来规范内在功能。迄今为止,我的法典只能处理消极和积极的数字,但精子会发出错误信息,我不想这样做,因为凭空功能能够通过四舍五入处理浮动。我的问题是,是否有办法加以规范? 我尝试利用这一轮功,因为你通常如何将浮游变成一种汇合,但我很快认识到,你会打下一个阵列。我也尝试读一读文件,但我也只字不提部分,因为至少我无法阅读文件。与此相关的Stack Overflow问题也没有提到如何与 de们合作。我的不完整守则如下:

num = input()
num = list(num)

multiplier = 1
new_num = 0
i = -1

while -1 * i != len(num) + 1:
    if num[i] == "1":
        new_num = 1 * multiplier + new_num
    elif num[i] == "2":
        new_num = 2 * multiplier + new_num
    elif num[i] == "3":
        new_num = 3 * multiplier + new_num
    elif num[i] == "4":
        new_num = 4 * multiplier + new_num
    elif num[i] == "5":
        new_num = 5 * multiplier + new_num
    elif num[i] == "6":
        new_num = 6 * multiplier + new_num
    elif num[i] == "7":
        new_num = 7 * multiplier + new_num
    elif num[i] == "8":
        new_num = 8 * multiplier + new_num
    elif num[i] == "9":
        new_num = 9 * multiplier + new_num
    elif num[i] == "0":
        pass
    elif num[i] == "-" and num[0] == "-":
        new_num = new_num * -1
    else:
        raise ValueError
    print(new_num)
    multiplier *=10
    i -=1

print(new_num)

Answer 1

这是我把扼杀变成暗中的方法。

numbers = "0123456789"

def string_to_int(s):
    num = 0
    start_index = 1 if s[0] == "-" else 0
    s = s.split(".")[0] # just get the part before "." if exist
    
    for i in range(start_index, len(s)):
        digit = numbers.find(s[i])
        if digit == -1:
            raise ValueError
            
        num = num * 10 + digit
        
    return num if start_index == 0 else -num 


s = "-123.4z5" 
print(string_to_int(s)) # -123

s2 = "123.45"
print(string_to_int(s2)) # 123

s3 = "123z.45" 
print(string_to_int(s3)) # raise ValueError

Answer 2

如果你只是试图复制确切的秘密功能,那么,如果你遇到不计数的char,你不得不以电文引发例外:

ValueError: invalid literal for int() with base 10: [the value]

除少数例外情况外,还不能忘记暗中支持白色空间和+象征。也许,你还必须管理第二个参数:基数。

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