I'm working on a program that takes hexadecimal numbers with Python and turns them into decimal numbers
def hexaToDecimal(hexaString):
decimalNum = 0
length = len(hexaString)
for num in range(0,length):
x=hexaString[num]
if 'a'<=x<='f' or 'A'<=x<='F' or '0'<x<='9':
decimalNum = decimalNum + chToHexa(x) * (16**(length-1))
length = length -1
(Optimized)
def chToHexa(x):
if x=='A' or 'a':
return 10
elif x=='B' or 'b':
return 11
elif x=='C' or 'c':
return 12
elif x=='D' or 'd':
return 13
elif x=='E' or "e":
return 14
elif x=='F' or 'f':
return 15
else:
return x
```
ChToHexa is a code to correct the hexadecimal number if there is an alphabet
No matter what number you put in the alphabet, only the value of 10 is returned in chToHexa
I have fixed the code, but I wonder why the number was returned to 100,000
(Unlike the preview, the text below appears in the column with the code written on it.)(T)
def chToHexa(x):
if x=='A' or 'a':
return 10
elif x=='B' or 'b':
return 11
elif x=='C' or 'c':
return 12
elif x=='D' or 'd':
return 13
elif x=='E' or "e":
return 14
elif x=='F' or 'f':
return 15
else:
return x
In if x=='A' or 'a': instead of if x=='A' or 'a': you must give a condition like if x=='A' or 'a':
The or 'a' after is always true.
Younghoon explained it well.
In addition, you can use x in ('A', 'a') or x.lower()=='a' in the condition of the if statement.
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