Python - logical evaluation order in "if" statement

Question

In Python we can do this:

if True or blah:
    print("it's ok") # will be executed

if blah or True: # will raise a NameError
    print("it's not ok")

class Blah:
    pass
blah = Blah()

if blah or blah.notexist:
    print("it's ok") # also will be executed

• Can somebody point me to documentation on this feature?
• Is it an implementation detail or feature of the language?
• Is it good coding style to exploit this feature?

Earthward

The blah or True one does not raise an exception for me, it prints. – Nedanedda

@TimS.: only if you define blah first. Note that blah is not yet defined at the top of the example, so a NameError is raised instead. – Pandybat

Answer 1

The or and and short circuit, see the Boolean operations documentation:

The expression x and y first evaluates x; if x is false, its value is returned; otherwise, y is evaluated and the resulting value is returned.

The expression x or y first evaluates x; if x is true, its value is returned; otherwise, y is evaluated and the resulting value is returned.

Note how, for and, y is only evaluated if x evaluates to a True value. Inversely, for or, y is only evaluated if x evaluated to a False value.

For the first expression True or blah, this means that blah is never evaluated, since the first part is already True.

Furthermore, your custom Blah class is considered True:

In the context of Boolean operations, and also when expressions are used by control flow statements, the following values are interpreted as false: False, None, numeric zero of all types, and empty strings and containers (including strings, tuples, lists, dictionaries, sets and frozensets). All other values are interpreted as true. (See the __nonzero__() special method for a way to change this.)

Since your class does not implement a __nonzero__() method (nor a __len__ method), it is considered True as far as boolean expressions are concerned.

In the expression blah or blah.notexist, blah is thus true, and blah.notexist is never evaluated.

This feature is used quite regularly and effectively by experienced developers, most often to specify defaults:

some_setting = user_supplied_value or 'default literal'
object_test = is_it_defined and is_it_defined.some_attribute

Do be wary of chaining these and use a conditional expression instead where applicable.

Pandybat

Answer 2

This is called short-circuiting and is a feature of the language:

http://docs.python.org/2/tutorial/datastructures.html#more-on-conditions

The Boolean operators and and or are so-called short-circuit operators: their arguments are evaluated from left to right, and evaluation stops as soon as the outcome is determined. For example, if A and C are true but B is false, A and B and C does not evaluate the expression C. When used as a general value and not as a Boolean, the return value of a short-circuit operator is the last evaluated argument.

Frieze

Answer 3

It's the way the operators logical operators, specifically or in python work: short circuit evaluation.

To better explain it, consider the following:

if True or False:
    print('True') # never reaches the evaluation of False, because it sees True first.

if False or True:
    print('True') # print's True, but it reaches the evaluation of True after False has been evaluated.

For more information see the following:

• The official python documentation on boolean operations
• A stack overflow question regarding short circuitry in python

Khania

Answer 4

With the or operator, values are evaluated from left to right. After one value evaluates to True, the entire statement evaluates to True (so no more values are evaluated).

• Official documentation
• It's a feature of the language
• There is nothing wrong with using its functionality

Chape