Does Python evaluate if's conditions lazily? [duplicate]

Answer 1

Python's laziness can be proved by the following code:

def foo():
    print('foo')
    return False

def bar():
    print('bar')
    return False

foo() and bar()         #Only 'foo' is printed

On the other hand,

foo() or bar()

would cause both 'foo' and 'bar' to be printed.

Answer 2

Yes, Python evaluates lazily, so foo2 will not be checked.

I use this all the time for grabbing items from dictionary-like objects if I don't know if the key exists:

if 'key' in mydict and mydict['key'] == 'heyyo!':
    do_stuff()

See @unutbu's answer for a fuller explanation.

Answer 3

It is really the or part that is short circuited:

>>> 1 or 1/0  #also 0 and 1/0
1
>>> 0 or 1/0  #also 1 and 1/0

Traceback (most recent call last):
  File "<pyshell#1240>", line 1, in <module>
    0 or 1/0
ZeroDivisionError: integer division or modulo by zero

Answer 4

and or is lazy

& | is not lazy

Answer 5

Yes, Python evaluates boolean conditions lazily.

The docs say,

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.

Answer 6

This isn't technically lazy evaluation, it's short-circuit boolean expressions.

Lazy evaluation has a somewhat different connotation. For example, true lazy evaluation would likely allow this

def foo(arg) :
    print "Couldn't care less"

foo([][0])

But Python doesn't.

Python is also nice in that it "echos" it's boolean arguments. For example, an or condition returns either it's first "truthy" argument or the last argument (if all arguments are "falsey"). An and condition does the inverse.

So "echo argument" booleans means

2 and [] and 1

evaluates to [], and

[] or 1 or 2

evaluates to 1