CODEWITHSUNDEEP
python
INeedVerification
INeedVerification

Reputation: 39

How do I get return to print to console in this code?

def distance_from_zero(distance):
    if type(distance) == int or type(distance) == float:
        return abs(distance)
        print(distance)
    else:
        return "Nope"
        print distance_from_zero(distance)

distance_from_zero(18)

Upvotes: 0

Views: 228

Answers (5)

sabbahillel
sabbahillel

Reputation: 4425

This is the original code with the print statements put in the correct place. Note the infinite recursion error that still shows up.

def distance_from_zero(distance):
    if type(distance) == int or type(distance) == float:
        print(distance)
        return abs(distance)
    else:
        # This print forces infinite recursion because of distance
        # print distance_from_zero(distance)
        return "Nope"

This is a way of testing it

def distance_from_zero(distance):
    if type(distance) == int or type(distance) == float:
        print(distance)
        return abs(distance)
    else:
        try:
            distance = float(distance)
            print(distance)
            return abs(distance)
        except:
            print ("invalid distance specified", distance)
            return "Nope"

Of course, the initial if can be dropped since float(int) and float(float) still return a normal float.

def distance_from_zero(distance):
    try:
        distance = float(distance)
        print(distance)
        return abs(distance)
    except:
        print ("invalid distance specified", distance)
        return "Nope"

Upvotes: 0

linky00
linky00

Reputation: 65

I think that Python stops reading the definition after you've returned something, so you should swap the print and return commands around.

Upvotes: 0

mhawke
mhawke

Reputation: 87124

Firstly, why do you need to check the type of the argument before calling abs()?. Just call it. abs() knows best the types of operands that it can work with. If it sees one that it can't handle a TypeError exception will be raised and the caller can deal with it.

If you really want to handle the error in your function, use a try/except block instead of trying to guess the types that abs() supports, e.g. complex numbers are supported; abs(complex(1, 2)) is perfectly valid:

try:
    distance = abs(distance)
except TypeError as e:
    print('{!r} is not a valid distance'.format(distance))
    return None    # or some other error
else:
    print(distance)
    return distance

Finally, if you really do want to limit support to only int and float types, it is better to use isinstance() than type() because is supports inheritance:

if isinstance(distance, (int, float)):
    # it's a float or an int
    distance = abs(distance)

A couple of other things to note:

  • Call print() before calling return in the function. return terminates the function and returns execution to the calling code, so any statements after the return will not be executed.

  • You seem to want to print the value of distance after abs(), so you need to store the result in a variable (or perform the calculation twice):

    distance = abs(distance)
print(distance)
return distance

Upvotes: 0

Martín Muñoz del Río
Martín Muñoz del Río

Reputation: 1968

The code after return will never be executed.

You have to revert the order.

Something like:

print(distance)
return abs(distance)

Upvotes: 0

Atri
Atri

Reputation: 5831

You should print before returning. Once you return, your print statement is not reachable.

replace:

return abs(distance)
print(distance)

with: 

print(distance)
return abs(distance)

And this:

return "Nope"
print distance_from_zero(distance)

with: 

print distance_from_zero(distance)
return "Nope"

Upvotes: 2

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