Wrong math with Python?
Question:
Just starting out with Python, so this is probably my mistake, but…
I’m trying out Python. I like to use it as a calculator, and I’m slowly working through some tutorials.
I ran into something weird today. I wanted to find out 2013*2013, but I wrote the wrong thing and wrote 2013*013, and got this:
>>> 2013*013
22143
I checked with my calculator, and 22143 is the wrong answer! 2013 * 13 is supposed to be 26169.
Why is Python giving me a wrong answer? My old Casio calculator doesn’t do this…
Answers:
Because of octal arithmetic, 013 is actually the integer 11.
>>> 013
11
With a leading zero, 013 is interpreted as a base-8 number and 1*81 + 3*80 = 11.
Note: this behaviour was changed in python 3. Here is a particularly appropriate quote from PEP 3127
The default octal representation of integers is silently confusing to
people unfamiliar with C-like languages. It is extremely easy to
inadvertently create an integer object with the wrong value, because
‘013’ means ‘decimal 11’, not ‘decimal 13’, to the Python language
itself, which is not the meaning that most humans would assign to this
literal.
013 is an octal integer literal (equivalent to the decimal integer literal 11), due to the leading 0.
>>> 2013*013
22143
>>> 2013*11
22143
>>> 2013*13
26169
It is very common (certainly in most of the languages I’m familiar with) to have octal integer literals start with 0 and hexadecimal integer literals start with 0x. Due to the exact confusion you experienced, Python 3 raises a SyntaxError:
>>> 2013*013
File "<stdin>", line 1
2013*013
^
SyntaxError: invalid token
and requires either 0o or 0O instead:
>>> 2013*0o13
22143
>>> 2013*0O13
22143
This is mostly just expanding on @Wim’s answer a bit, but Python indicates the base of integer literals using certain prefixes. Without a prefix, integers are interpreted as being in base-10. With an “0x”, the integer will be interpreted as a hexadecimal int. The full grammar specification is here, though it’s a bit tricky to understand if you’re not familiar with formal grammars: http://docs.python.org/2/reference/lexical_analysis.html#integers
The table essentially says that if you want a long value (i.e. one that exceeds the capacity of a normal int), write the number followed by the letter “L” or “l”; if you want your number to be interpreted in decimal, write the number normally (with no leading 0); if you want it interpreted in octal, prefix it with “0”, “0o”, or “0O”; if you want it in hex, prefix it with “0x”; and if you want it in binary, prefix it with “0b” or “0B”.
Python’s ‘leading zero’ syntax for octal literals is a common gotcha:
Python 2.7.3
>>> 010
8
The syntax was changed in Python 3.x http://docs.python.org/3.0/whatsnew/3.0.html#integers
Just starting out with Python, so this is probably my mistake, but…
I’m trying out Python. I like to use it as a calculator, and I’m slowly working through some tutorials.
I ran into something weird today. I wanted to find out 2013*2013, but I wrote the wrong thing and wrote 2013*013, and got this:
>>> 2013*013
22143
I checked with my calculator, and 22143 is the wrong answer! 2013 * 13 is supposed to be 26169.
Why is Python giving me a wrong answer? My old Casio calculator doesn’t do this…
Because of octal arithmetic, 013 is actually the integer 11.
>>> 013
11
With a leading zero, 013 is interpreted as a base-8 number and 1*81 + 3*80 = 11.
Note: this behaviour was changed in python 3. Here is a particularly appropriate quote from PEP 3127
The default octal representation of integers is silently confusing to
people unfamiliar with C-like languages. It is extremely easy to
inadvertently create an integer object with the wrong value, because
‘013’ means ‘decimal 11’, not ‘decimal 13’, to the Python language
itself, which is not the meaning that most humans would assign to this
literal.
013 is an octal integer literal (equivalent to the decimal integer literal 11), due to the leading 0.
>>> 2013*013
22143
>>> 2013*11
22143
>>> 2013*13
26169
It is very common (certainly in most of the languages I’m familiar with) to have octal integer literals start with 0 and hexadecimal integer literals start with 0x. Due to the exact confusion you experienced, Python 3 raises a SyntaxError:
>>> 2013*013
File "<stdin>", line 1
2013*013
^
SyntaxError: invalid token
and requires either 0o or 0O instead:
>>> 2013*0o13
22143
>>> 2013*0O13
22143
This is mostly just expanding on @Wim’s answer a bit, but Python indicates the base of integer literals using certain prefixes. Without a prefix, integers are interpreted as being in base-10. With an “0x”, the integer will be interpreted as a hexadecimal int. The full grammar specification is here, though it’s a bit tricky to understand if you’re not familiar with formal grammars: http://docs.python.org/2/reference/lexical_analysis.html#integers
The table essentially says that if you want a long value (i.e. one that exceeds the capacity of a normal int), write the number followed by the letter “L” or “l”; if you want your number to be interpreted in decimal, write the number normally (with no leading 0); if you want it interpreted in octal, prefix it with “0”, “0o”, or “0O”; if you want it in hex, prefix it with “0x”; and if you want it in binary, prefix it with “0b” or “0B”.
Python’s ‘leading zero’ syntax for octal literals is a common gotcha:
Python 2.7.3
>>> 010
8
The syntax was changed in Python 3.x http://docs.python.org/3.0/whatsnew/3.0.html#integers