The shortest code to tell if a number is even or odd

Question

One of my colleagues proposed us a challenge: to write the shortest C/C++ program to determine if a number is even or odd.

These are the requirements:

Get a number from keyboard, at run-time (not from command line)
Output the string pari if the number is even
Output the string dispari if the number is odd
We can assume that the input is legal
We can assume that the machine has a i386 architecture
Code preprocessing is not allowed (e.g.: use #defines)
[edit after comments] The code can be considered valid if it compiles with gcc 4.x

This is the best attempt we had so far (file test.c, 50B):

main(i){scanf("%d",&i);puts("dispari"+3*(~i&1));}

Do you think we can go further?

_{(*): "dispari" is the italian for odd and "pari" is the italian for even}

Is this for C or C++? In C++, that code is invalid. I believe in C as well.\ — Luchian Grigore
– Luchian Grigore, Commented Oct 9, 2012 at 19:56
@LuchianGrigore, you can use both C and C++. The choice depends on the language that lets you write the shortest code. The code above compiles correctly with gcc, maybe I should add that as a requirement. — Vincenzo Pii
– Vincenzo Pii, Commented Oct 9, 2012 at 20:01
Sorry, but your best example doesn't compile with gcc 4.6.3 (Edit: it does) — Pieter Bos
– Pieter Bos, Commented Oct 9, 2012 at 20:06
Slight variation (untested): use return value of scanf to shorten into main(i){puts("dispari"+3*(scanf("%d",&i)&~i));}. — Howard
– Howard, Commented Oct 9, 2012 at 20:12
You can't legally and portably perform output without either #include <stdio.h> (which is a preprocessor directive) or your own declaration of the routines you're using. As of C90, calling an undeclared function has defined behavior only in narrow circumstances; as of C99, it's a constraint violation (C's version of "illegal"). — Keith Thompson
– Keith Thompson, Commented Oct 10, 2012 at 0:40

ugoren · Accepted Answer · 2012-10-10 08:38:37Z

8

47 bytes (2 less)

main(i){scanf("%d",&i);puts("dispari"-~i%2*3);}

For a positive i, ~i is negative, so ~i%2 is 0 or -1.

answered Oct 9, 2012 at 21:49

17.6k5 gold badges54 silver badges75 bronze badges

baby-rabbit · Accepted Answer · 2012-10-10 01:42:45Z

Given you didn't specified the range of the number input, then here's a solution that handles 0..9 only:

main(){puts("dispari"-~getchar()%2*3);}

Community · Accepted Answer · 2020-06-17 09:04:33Z

3

main(i){scanf("%d",&i);puts("dispari"+3-i%2*3);}

1

answered Oct 9, 2012 at 20:12

1314 bronze badges

peter ferrie · Accepted Answer · 2017-11-26 21:02:16Z

Build with

gcc -nostartfiles -Wl,-ea odd.c

Works in GCC 6.3.0.

a(){puts("dispari"-~getch()%2*3);}