Why int(string) gives value error

Question

I was just wondering why when I take an integer of a string like int('string') why you recieve a value error in python 3.2 and not a type error. I see the definition of a type error is defined on the python site as the follows: Raised when an operation or function is applied to an object of inappropriate type. The associated value is a string giving details about the type mismatch.

isn't int the operator and the string is the inappropriate type? When I do this I receive a ValueError and do not understand the reason for this.

Here is the code So when i instantiate the class with a rank that has a string to try and purposely get an error I receive a valueerror and not a type as I would expect.

class Card:

#attributes
list_rank=   ["","Ace","2","3","4","5","6","7","8","9","10","Jack","Queen","King"]
list_suit={"d":"diamonds","c":"clubs","h":"hearts","s":"spades"}

#initialize rank and suit
def __init__(self,rank,suit):



self.rank=int(rank)

self.suit=suit




#return the rank of the card
def getRank(self):
    return(self.list_rank[self.rank])

#return the suit of the card
def getSuit(self):
    return(self.list_suit[self.suit])



#value of the cards
def bjValue(self):
    if(self.rank<10):
        return(self.rank)
    else:
        return(10)


#return the rank and suit of the card
def __str__(self):

    return (self.list_rank[self.rank]+" of "+self.list_suit[self.suit])

Answer 1

To understand the difference you must understand the difference between type and value. The type of "abc" is string, which you can check by running

type("abc")  # -> string

The value of the literal "abc" is simply the value itself. If you had an expression like "ab" + "c" the value would also be "abc".

That's why you get the ValueError and not a TypeError; int expects a string, which you gave it - so it's the correct type - but the value has to be something that can be interpreted as an integer and "abc" is not an integer obviously.

To try this out, see the difference between:

int('abc')
# throws:
# ValueError: invalid literal for int() with base 10: 'abc'

and

int(None)
# throws:
# TypeError: int() argument must be a string or a number, not 'NoneType'

The fact that "abc" doesn't immediately look like an integer doesn't stop us from interpreting it as one, however. In a different base it could still be interpreted as a number.

For example: binary numbers, as you probably know, only use 0s and 1s, and hexadecimal numbers are usually represented using the digits 0 - 9 and the letters a - f. The int function takes an optional argument that tells it which base to use, so if we try interpreting "abc" again but as a hexadecimal number, we get:

int('abc',  base=16)
# returns 2748

Answer 2

because int() accepts a string as a parameter for example int('1') will ouptut 1. so '1' is an appropriat value but 'a' is not