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, Feb 04 '12 at 07:37
@Nitesh: Welcome to SO. Please "accept" one of the correct answers by clicking the "check mark" at the left of the answer. — Jim Garrison, Feb 04 '12 at 07:44

score 5 · Accepted Answer · answered Feb 04 '12 at 07:33

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 04 '12 at 07:33

RanRag

score 3 · Answer 2 · answered Feb 04 '12 at 07:30

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 04 '12 at 07:30

Gabe

oh awesome, thanks. I'm a beginner just started to learn python – jaysun Feb 04 '12 at 07:33

score 3 · Answer 3 · edited May 23 '17 at 12:20

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:

edited May 23 '17 at 12:20

Community

answered Feb 04 '12 at 07:33

Niall Byrne

3 Answers3