If statement fault in python

Question

I executed the following code:

import sys

x=sys.argv[1]

print x;

if x > 1:
    print "It's greater than 1"

and here is the output:

C:\Python27>python abc.py 0
0
It's greater than 1

How the hell it's greater than 1? in fact the if condition should fail, is there any fault with my code?

replace 'print x' with 'print x + 1', and you'll understand! — jimifiki
– jimifiki, Commented Feb 4, 2012 at 7:37
@Nitesh: Welcome to SO. Please "accept" one of the correct answers by clicking the "check mark" at the left of the answer. — Jim Garrison
– Jim Garrison, Commented Feb 4, 2012 at 7:44

RanRag · Accepted Answer · 2012-02-04 07:33:11Z

5

Because type of x in x=sys.argv[1] is str.

import sys
x = sys.argv[1]
print type(x)

Output =<type 'str'>

So in python,

>>> '0'>1
True

Therefore you need

>>> int('0')>1
False
>>>

answered Feb 4, 2012 at 7:33

RanRag

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Comments

Gabe · Accepted Answer · 2012-02-04 07:30:29Z

3

x is a string but 1 is an integer, so the comparison is of mismatched types. You need something like if int(x) > 1:.

answered Feb 4, 2012 at 7:30

Gabe

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1 Comment

jaysun Over a year ago

oh awesome, thanks. I'm a beginner just started to learn python

Community · Accepted Answer · 2017-05-23 12:20:02Z

3

This is testing x (a string). try using:

if int(x) > 1: 
     print "It's greater than 1"

I got curious about this and found:

How does Python compare string and int?

edited May 23, 2017 at 12:20

CommunityBot

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answered Feb 4, 2012 at 7:33

Niall Byrne

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