+- Python Forum (https://python-forum.io)
+-- Forum: Python Coding (https://python-forum.io/forum-7.html)
+--- Forum: General Coding Help (https://python-forum.io/forum-8.html)
+--- Thread: coding issue (/thread-2003.html)
coding issue - Lee - Feb-10-2017
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']]]RE: coding issue - Ofnuts - Feb-10-2017
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]: FalseRE: coding issue - Lee - Feb-13-2017
Hi Ofnuts
Good answer and teaching web.
Thank you.