How to concatenate `Object` with a string?

Question

How to concatenate Object with a string (primitive) without overloading and explicit type cast (str())?

class Foo:
    def __init__(self, text):
        self.text = text

    def __str__(self):
        return self.text


_string = Foo('text') + 'string'

Output:

Traceback (most recent call last):
  File "test.py", line 10, in <module>
      _string = Foo('text') + 'string'

TypeError: unsupported operand type(s) for +: 'type' and 'str'

operator + must be overloaded? Is there other ways (just wondering)?

PS: I know about overloading operators and type casting (like str(Foo('text')))

Why don't you want to use str to force the object into a string? — Marcin
– Marcin, Commented Feb 16, 2012 at 16:09
Python doesn't have casting. When you call str, you create a new string object by the return value of MyType.__str__. Casting is taking the same data in memory and telling the compiler/interpreter that it is another object type. — Daenyth
– Daenyth, Commented Feb 16, 2012 at 16:17
__str__ does return the string. But it isn't called when + is used, because python doesn't know what + means in this case. You have to tell it explicitly. — senderle
– senderle, Commented Feb 16, 2012 at 16:21

Sven Marnach · Accepted Answer · 2012-02-16 16:08:48Z

26

Just define the __add__() and __radd__() methods:

class Foo:
    def __init__(self, text):
        self.text = text
    def __str__(self):
        return self.text
    def __add__(self, other):
        return str(self) + other
    def __radd__(self, other):
        return other + str(self)

They will be called depending on whether you do Foo("b") + "a" (calls __add__()) or "a" + Foo("b") (calls __radd__()).

answered Feb 16, 2012 at 16:08

Sven Marnach

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5 Comments

Daenyth Over a year ago

IMO this is not good practice. Calling str explicitly is the correct pattern. Anyway, you're still calling str here, you're just hiding it.

senderle Over a year ago

@Daenyth, I think it depends on the circumstances. This isn't appropriate for any old object; but for something meant to emulate a string, it's perfectly reasonable.

Sven Marnach Over a year ago

@Daenyth: I agree with you, but there might be exceptions.

Sven Marnach Over a year ago

@senderle: The first thing I'd try to emulate a string is to derive from str. Only if this is no option for some reason, I'd go with the approach above.

senderle Over a year ago

@SvenMarnach, yes, that's what I meant by "emulate." Perhaps I was being unclear; in my mind, a type that inherits from str is doing something more than "emulating" a string.

Constantinius · Accepted Answer · 2012-02-16 16:38:39Z

7

_string = Foo('text') + 'string'

The problem with this line is that Python thinks you want to add a string to an object of type Foo, not the other way around.

It would work though if you'd write:

_string = "%s%s" % (Foo('text'), 'string')

EDIT

You could try it with

_string = 'string' + Foo('text')

In this case your Foo object should be automatically casted to a string.

edited Feb 16, 2012 at 16:38

answered Feb 16, 2012 at 16:06

Constantinius

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4 Comments

Sven Marnach Over a year ago

Are you sure about your edit? I'm pretty sure Python won't cast to string automatically.

tomas Over a year ago

@Constantinius, Thanx for the anwer, but I know about type casting. I want to understand how to return the real string from __str__

JBernardo Over a year ago

There's no such thing as Python casting objects to strings. Either the __add__ method will do it explicitly or no one will do.

Gandaro Over a year ago

I thought of '{}{}'.format(Foo('text'), 'string') first. Maybe this is also an option...

jcollado · Accepted Answer · 2012-02-16 16:08:32Z

1

If that makes sense for your Foo object, you can overload the __add__ method as follows:

class Foo:
    def __init__(self, text):
        self.text = text

    def __str__(self):
        return self.text

    def __add__(self, other):
        return str(self) + other

_string = Foo('text') + 'string'
print _string

Example output:

textstring

answered Feb 16, 2012 at 16:08

jcollado

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