Even or odd
Test whether an integer is even or odd.
There is more than one way to solve this task:
- Use the even and odd predicates, if the language provides them.
- Check the least significant digit. With binary integers, i bitwise-and 1 equals 0 iff i is even, or equals 1 iff i is odd.
- Divide i by 2. The remainder equals 0 iff i is even. The remainder equals +1 or -1 iff i is odd.
- Use modular congruences:
- i ≡ 0 (mod 2) iff i is even.
- i ≡ 1 (mod 2) iff i is odd.
bc
There are no bitwise operations, so this solution compares a remainder with zero. Calculation of i % 2 only works when scale = 0.
<lang bc>i = -3
/* Assumes that i is an integer. */ scale = 0 if (i % 2 == 0) "i is even " if (i % 2) "i is odd "</lang>
C
Test by bitwise and'ing 1, works for any builtin integer type as long as it's 2's compliment (it's always so nowadays): <lang c>if (x & 1) {
/* x is odd */
} else {
/* or not */
}</lang>
If using long integer type from GMP (mpz_t), there are provided macros:
<lang c>mpz_t x;
...
if (mpz_even_p(x)) { /* x is even */ }
if (mpz_odd_p(x)) { /* x is odd */ }</lang>
The macros evaluate x more than once, so it should not be something with side effects.
Common Lisp
Standard predicates: <lang lisp>(if (evenp some-var) (do-even-stuff)) (if (oddp some-other-var) (do-odd-stuff))</lang>
Factor
The math vocabulary provides even? and odd? predicates. This example runs at the listener, which already uses the math vocabulary.
( scratchpad ) 20 even? . t ( scratchpad ) 35 even? . f ( scratchpad ) 20 odd? . f ( scratchpad ) 35 odd? . t
Haskell
even and odd functions are already included in the standard Prelude.
<lang haskell>Prelude> even 5 False Prelude> even 42 True Prelude> odd 5 True Prelude> odd 42 False</lang>
J
Modulo:
<lang j> 2 | 2 3 5 7 0 1 1 1
2|2 3 5 7 + (2^89x)-1
1 0 0 0</lang>
Remainder:
<lang j> (= <.&.-:) 2 3 5 7 1 0 0 0
(= <.&.-:) 2 3 5 7+(2^89x)-1
0 1 1 1</lang>
Last bit in bit representation: <lang j> {:"1@#: 2 3 5 7 0 1 1 1
{:"1@#: 2 3 5 7+(2^89x)-1
1 0 0 0</lang>
Bitwise and:
<lang j> 1 (17 b.) 2 3 5 7 0 1 1 1</lang>
Note: as a general rule, the simplest expressions in J should be preferred over more complex approaches.
Java
Bitwise and: <lang java>public static boolean isEven(int i){
return (i & 1) == 0;
}</lang> Modulo: <lang java>public static boolean isEven(int i){
return (i % 2) == 0;
}</lang> Arbitrary precision bitwise: <lang java>public static boolean isEven(BigInteger i){
return i.and(BigInteger.ONE).equals(BigInteger.ZERO);
}</lang>
Arbitrary precision bit test (even works for negative numbers because of the way BigInteger represents the bits of numbers):
<lang java>public static boolean isEven(BigInteger i){
return !i.testBit(0);
}</lang> Arbitrary precision modulo: <lang java>public static boolean isEven(BigInteger i){
return i.mod(BigInteger.valueOf(2)).equals(BigInteger.ZERO);
}</lang>
OCaml
Modulo:
<lang ocaml>let is_even d =
(d mod 2) = 0
let is_odd d =
(d mod 2) <> 0</lang>
Bitwise and:
<lang ocaml>let is_even d =
(d land 1) = 0
let is_odd d =
(d land 1) <> 0</lang>
Pike
<lang Pike>> int i = 73; > (i&1); Result: 1 > i%2; Result: 1</lang>
Python
<lang python>>>> def is_odd(i): return bool(i & 1)
>>> def is_even(i): return not is_odd(i)
>>> [(j, is_odd(j)) for j in range(10)] [(0, False), (1, True), (2, False), (3, True), (4, False), (5, True), (6, False), (7, True), (8, False), (9, True)] >>> [(j, is_even(j)) for j in range(10)] [(0, True), (1, False), (2, True), (3, False), (4, True), (5, False), (6, True), (7, False), (8, True), (9, False)] >>> </lang>
Ruby
<lang ruby>print "evens: " p -5.upto(5).select {|n| n.even?} print "odds: " p -5.upto(5).select {|n| n.odd?}</lang>
outputs
evens: [-4, -2, 0, 2, 4] odds: [-5, -3, -1, 1, 3, 5]
Tcl
<lang tcl>package require Tcl 8.5
- Bitwise test is the most efficient
proc tcl::mathfunc::isOdd x { expr {$x & 1} } proc tcl::mathfunc::isEven x { expr {!($x & 1)} }
puts " # O E" puts 24:[expr isOdd(24)],[expr isEven(24)] puts 49:[expr isOdd(49)],[expr isEven(49)]</lang> Output:
# O E 24:0,1 49:1,0