CODEWITHSUNDEEP
typesintpython-3.6
Dexter Siah Tze Ming
Dexter Siah Tze Ming

Reputation: 57

Checking if value from list is an integer

I'm trying to make a hexadecimal to decimal calculator and i'm trying to check if the input has integer or character.

How can I accomplish this?

def hex2dec(value):
    dict = {'A':10,'B':11,'C':12,'D':13,'E':14,'F':15}
    strVal = str(value)
    length = len(strVal)
    decimal = 0
    for i in range(0,length):
        try:
            int(strval[i:i+1])
            val = "int"
        except ValueError:
            val = "not int"

    return val

value = int(input("Enter Hexadecimal Value: "),16)
print("Value in Hexadecimal : ", hex2dec(value))

Whenever I run the codes and input any single digit or alphabet from A to F it returns int.

Anyone knows the solution to this?

Upvotes: 0

Views: 41

Answers (3)

Jongware
Jongware

Reputation: 22457

You always get "int" because the function is always called with an int. The variable value gets converted to an integer because of int(..) in the input line.

If you want to check first, feed the original string input into your function:

value = input("Enter Hexadecimal Value: ")

You don't need to check one character at a time, this will work as well:

try:
    int(strVal)
    val = "int"
except ValueError:
    val = "not int"

Note that there is no good reason to check in your function! Even if you seemingly enter an integer -- 123 --, you must still process this as a hexadecimal number. Checking if it's also an integer has no use at that point.

Upvotes: 0

Yuna
Yuna

Reputation: 798

Extending the previous answer, you could do:

if (strval[i:i+1] in dict.keys()) or (type(strval[i:i+1]) == int):
    val = 'int'
else
    'val = 'not int'

in order to check if the value is an int or in your dict (additionally you get some extensibility..)

Upvotes: 0

Andy G
Andy G

Reputation: 19367

With this line

value = int(input("Enter Hexadecimal Value: "),16)

you have already converted the provided value to an integer.

If you want to perform the conversion yourself, first accept what they have entered as a string (remove the use of int() in the above statement).

Also note that a submission of "12" is a valid hexadecimal value, with a decimal value of 18.

Upvotes: 1

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