Integer check not working

Question

This is quite possible the most perplexing piece of code I've ever seen, and i can't seem to figure out what's wrong with it.

a = 5
if a == 0 or 1 : 
    print "hi"

When run, it prints "hi" which by all means confuses me. Please provide some insight onto how i can make it not return hi and behave normally. (it should be returning nothing)

You'll have to check each value separately: `if a == 0 or a == 1:` or *else* you'll be *short-circuiting* the first condition with an integer. — Moses Koledoye, Oct 04 '16 at 15:38
`1` evalutes to `True` in Python - so you get `if False or True:` which evaluates to `True` — LocalVolatility, Oct 04 '16 at 15:39
You should study operator precedence. Also, if you want to check if a variable is one of some set of values, use `if a in (0, 1):` — Kyle Strand, Oct 04 '16 at 15:40
"This is quite possible the most perplexing piece of code I've ever seen". You must not have been programming very long :) — John Coleman, Oct 04 '16 at 15:41

score 0 · Answer 1 · answered Oct 04 '16 at 15:39

0

a = 5
if (a == 0) or (a == 1): #changed from if a == 0 or 1
    print ("hi")
#does not print 'hi'

Check a against 0 and 1 independently.

answered Oct 04 '16 at 15:39

Rafael

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It's not that you're wrong; it's just that this answer is already covered in a duplicate. – Makoto Oct 04 '16 at 15:43
Good catch. I didn't check. – Rafael Oct 04 '16 at 15:44

score 0 · Answer 2 · answered Oct 04 '16 at 16:01

0

It should work in this way:

  while not a = 5:
          if a in [0, 1]:
              print "Hi"

answered Oct 04 '16 at 16:01

Kian

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Integer check not working

2 Answers2