coding issue

[Lee]

I wirte a simple as following.
Could someone can tell me why the answer is different Buf1 and Buf2 ??
Thank for you help.

def fList_Test(pBuf, pX):
    pBuf.clear()
    _bufX = []
    if pX == 0:
        _bufX = ['a1', 'a2']
    elif pX == 1:
        _bufX = ['b1', 'b2']
    else:
        _bufX = ['c1', 'c2']
    pBuf.append(_bufX)

Buf1 = []
Buf2 = []
Buf_T = []
for x in range(3):
    _bufX = []
    fList_Test(Buf_T, x)
    Buf1.append(Buf_T)
    fList_Test(_bufX, x)
    Buf2.append(_bufX)

print (Buf1)
print (Buf2)

output:

Output:
[[['c1', 'c2']], [['c1', 'c2']], [['c1', 'c2']]]
[[['a1', 'a2']], [['b1', 'b2']], [['c1', 'c2']]]

[Ofnuts]

Because Buf1 contains three identical references to the same list, since Buf_T is only created once, while you get a new _bufX on every iteration on x. Try this at the end of your code:

for x in range(3):
    print ('Buf1[0] is Buf1[%d]: %s' % (x, Buf1[0] is Buf1[x]))
for x in range(3):
    print ('Buf2[0] is Buf2[%d]: %s' % (x, Buf2[0] is Buf2[x]))

Output:
Buf1[0] is Buf1[0]: True
Buf1[0] is Buf1[1]: True
Buf1[0] is Buf1[2]: True
Buf2[0] is Buf2[0]: True
Buf2[0] is Buf2[1]: False
Buf2[0] is Buf2[2]: False

Recommended: https://www.youtube.com/watch?v=_AEJHKGk9ns

[Lee]

Hi Ofnuts
Good answer and teaching web.
Thank you.