Really awkward (seemingly simple) bug with python integer comparisons

Question

I have the following piece of code which is not working the way I expect it to at all...

current_frame = 15 # just for showcasing purposes
g_ch = 7

if (current_frame != int(row[0])) and (int(row[1]) != g_ch):
                current_frame = int(row[0])
                print "curious================================="
                print current_frame
                print row
                print current_frame, " != ", int(row[0]), ", ", current_frame != int(row[0])
                print "========================================"

which prints for any specific case:

curious================================= 

15 

['15', '1', 'more data'] 15 != 15 , False

========================================

This should obviously never even enter the if statement, as the equality is showing false. Why is this happening?

edit: I have also tried this with != instead of 'is not', and gotten the same results.

Since you say this happens with != as will as is not, please omit current_frame = int(row[0]) and show print repr(current_frame) and print repr(row[0]) — unutbu
– unutbu, Commented Jul 19, 2010 at 13:49

Philipp · Accepted Answer · 2010-07-19 13:43:04Z

6

Value comparisons are done with the != operator, not with is not, which compares object identity.

Apart from that, I think it's an indentation problem.

answered Jul 19, 2010 at 13:43

Philipp

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1 Comment

ahhtwer Over a year ago

Should have mentioned that I had done this as well, and got the same result.

Mr_Pink · Accepted Answer · 2010-07-19 13:44:01Z

1

In short, you need to use == and !=, and not is. is compares object identity, not equality.

answered Jul 19, 2010 at 13:44

Mr_Pink

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Comments

starblue · Accepted Answer · 2010-07-19 13:51:47Z

0

You assign current_frame = int(row[0]) inside the if, which changes the value of the boolean expression.

answered Jul 19, 2010 at 13:51

starblue

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