def foo(x = [0]):
x[0] = x[0] + 1
return x[0]
def soo(x = None):
if x is None:
x = [0]
x[0] = x[0] + 1
return x[0]
foo()
1
foo() #See the behavior incremented by one
2
foo([1]) # but here based on given number
2
foo()
3
foo([1])
2
foo()
4
soo()
1
soo()
1
soo([1])
2
soo()
1
Why foo() is
quoting from docs:http://docs.python.org/reference/compound_stmts.html
*Default parameter values are evaluated when the function definition is
executed.* This means that the expression is evaluated once, when the
function is defined, and that that same “pre-computed” value is used for
each call.
This is addressed in the FAQ.
http://www.python.org/doc/faq/general/#why-are-default-values-shared-between-objects
jitendra gupta wrote:
def foo(x = [0]):
x[0] = x[0] + 1
return x[0]
def soo(x = None):
if x is None:
x = [0]
x[0] = x[0] + 1
return x[0]
foo()
1
foo() #See the behavior
