[Tutor] How default arg of function works

[Deepak Dixit]

I am learning python and working with function.

Here is my test program :-

program.py
------------------------------------
def test1(nums=[]):
        return nums


def test2(nums=[]):
        nums.append(len(nums));
        return nums

print "Calling test1"
print '=' * 40
print 'test1()', test1()
print 'test1([1,2])', test1([1,2])
print 'test1()', test1()
print 'test1([1,1,1])', test1([1,1,1])
print 'test1()', test1()

print "Calling test2"
print '=' * 40

print 'test2()', test2()
print 'test2([1,2,3])', test2([1,2,3])
print 'test2([1,2])', test2([1,2])
print 'test2()', test2()
print 'test2()', test2()

--------------------------------------------------

# python program.py


Calling test1
========================================
test1() [ ]
test1([1,2]) [1, 2]
test1() [ ]
test1([1,1,1]) [1, 1, 1]
test1() [ ]
Calling test2
========================================
test2() [0]
test2([1,2,3]) [1, 2, 3, 3]
test2([1,2]) [1, 2, 2]
test2() [0, 1]
test2() [0, 1, 2]


I am assuming that in test1() we are not doing any modification in the
passed list and because of that its working as per my expectation.
But when I am calling test2() with params and without params then both are
using different references.  Why ? Can you please help me to understand
this.

-- 
With Regards,
Deepak Kumar Dixit