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?

• jimifiki: replace 'print x' with 'print x + 1', and you'll understand!
• Jim Garrison: @Nitesh: Welcome to SO. Please "accept" one of the correct answers by clicking the "check mark" at the left of the answer.

Answer 1

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
>>>

Answer 2

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

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

Answer 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?