Python：十六进制转换代码函数中的语义错误

Question

I have written a function for my 4 hexit hexadecimal to decimal converter, and while it works, there is a semantic error I can't figure out. Could you please tell me what the error is?

def convert16_10(hexa):
    den = 0
    for i in range(4):
        if hexa[i] == "A":
            den += (16 ** 3-i) * 10
        elif hexa[i] == "B":
            den += (16 ** 3-i) *11
        elif hexa[i] == "C":
            den += (16 ** 3-i) * 12
        elif hexa[i] == "D":
            den += (16 ** 3-i) * 13
        elif hexa[i] == "E":
            den += (16 ** 3-i) * 14
        elif hexa[i] == "F":
            den += (16 ** 3-i) * 15
        elif hexa[i] not in "ABCDEF":
            den += (16 ** 3-i) * i
    return den

Answer 1

您可以使用以下代码简单地将其转换：

i = int(hexa, 16)

Answer 2

• You should use parentheses in the power in (16 ** 3-i) . Should be (16 ** (3-i)) .
• You should multiply the digit after converting it to an integer instead of the index in * i . Should be * int(hexa[i]) .

Solution:

def convert16_10(hexa):
    den = 0
    for i in range(4):
        if hexa[i] == "A":
            den += (16 ** (3-i)) * 10
        elif hexa[i] == "B":
            den += (16 ** (3-i)) *11
        elif hexa[i] == "C":
            den += (16 ** (3-i)) * 12
        elif hexa[i] == "D":
            den += (16 ** (3-i)) * 13
        elif hexa[i] == "E":
            den += (16 ** (3-i)) * 14
        elif hexa[i] == "F":
            den += (16 ** (3-i)) * 15
        elif hexa[i] not in "ABCDEF":
            den += (16 ** (3-i)) * int(hexa[i])
    return den

print(convert16_10("007F")) # => 127