# how to find out if a number is even or not

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divya r on 23 Oct 2012
Edited: Matt J on 9 Oct 2020
I know in C language, for any number x using x%2 will calculate the remainder when x is divided by 2, which will help decipher whether its even or not.
How can I do this in matlab?

Jaimin Motavar on 5 Jul 2019
you can do it this in matlab by the very simple way by using implicit function rem(a,b) , where a is devided by b.
for example.
r1=rem(4,2)
r=0;
r2=rem(9,2)
r2=1;
luis fonseca on 9 Oct 2020
This is the easiert way:
N = 1; % number you want to know if even or odd
%% create an expression
s = (-1)^N;
%% if s = -1, the N is odd, else N is even
if s == -1
disp('N is odd')
else
disp('N is even')
end
Steven Lord on 9 Oct 2020
So Inf is even?
>> s = (-1)^Inf
s =
1
>> s = (-1)^NaN
s =
NaN
Does this hold for a complex number as well?
>> N = 3+4i;
>> s = (-1)^N; % Not equal to -1

Walter Roberson on 23 Oct 2012
Edited: MathWorks Support Team on 9 Nov 2018
See mod and rem

#### 1 Comment

Dillen.A on 5 Feb 2020
A quick example:
A = [-2 -1 0 1 2 3 4 5 6]; % A is your value or matrix
IS_EVEN = ~mod(A,2)
Which is the same as
IS_EVEN = ~bitget(abs(A),1)
And the same as
IS_EVEN = ~rem(A,2)
You can use logical() instead of ~ (isnot) for ODD, should you want booleans. Also bitget() does not work for negative integers, hence abs().
A warning though; ONLY bitget() will throw an error if an element in A is not an integer! the others will output 'odd' for fractions.
Unless you will repeat this many many times, the speed is not relevant. Otherwise, you should vectorize.

Jan on 23 Oct 2012
Care for exceptions:
NaN, Inf, 1e54, int8(-128)
There are some FEX submission for this task, e.g. FEX: parity checker.

Matt J on 23 Oct 2012
if bitget(A,1) %odd
else %even
end

Matt J on 23 Oct 2012
Note that solutions based on REM and MOD have certain non-robustness to large numbers, though I never quite understood why:
>> mod(bitmax+[1:8],2) %all are even
ans =
0 0 0 0 0 0 0 0
Josh Meyer on 10 Oct 2018
In more recent versions of MATLAB, bitmax was replaced by flintmax. This is the largest consecutive floating point number. After flintmax, the value of eps is larger than 1 (slowly increasing in powers of 2), so representable numbers larger than flintmax are no longer consecutive.
So, the reason all of those numbers are even is because flintmax is an even number and the spacing between numbers is eps(flintmax) = 2.

Ibn e Adam on 18 Feb 2020
% function to find even/odd
% n is input number for this function
function output=even_or_odd(n)
if rem(n,2)==0
output=even;
else
output=odd;
end
end

Show 1 older comment
Achala Bohra on 24 Feb 2020
why are even and odd (the values of output) not in inverted commas
Walter Roberson on 26 Feb 2020
Ibn e Adam did not define the variables "even" or "odd" so we do not know what datatypes would be returned.
Matt J on 26 Feb 2020
not in inverted commas
I guess inverted commas = single quotes

Anmol singh on 10 Apr 2020
Edited: Anmol singh on 10 Apr 2020
A givennumber is even or odd for this we use & operator.
if any number is odd it must have right most bit 1.
example:
int i=5;
binary form i= 0101
now use & operator
int j=i&1;[0101&1]//
here j have 0001;

#### 1 Comment

Walter Roberson on 10 Apr 2020
This does not work in MATLAB. In MATLAB, the operation
c = A & B
is equivalent to
if A ~= 0
if B ~= 0
c = true;
else
c = false;
end
elseif B ~= 0
c = false;
else
c = false;
end
Yes, this could be made more efficient, but this models the & operator. The more efficient operation is &&
Now notice that this is not a bitwise operation. 5&1 is not binary 0101 & 0001 giving 0001: instead it is (5~=0) and (1 ~= 0)
The MATLAB equivalent to what you are discussing is the bitand() operator
bitand(5,1)
But if you are going to do that, you might as well just ask for the last bit directly:
bitget(5,1) %the 1 is a bit number with LSB being #1

luis fonseca on 9 Oct 2020
This is the easiert way guys, its just math from highschool
N = 1; % number you want to know if even or odd
%% create an expression
s = (-1)^N;
%% if s = -1, the N is odd, else N is even
if s == -1
disp('N is odd')
else
disp('N is even')
end

#### 1 Comment

madhan ravi on 9 Oct 2020
what happens when N is an array?

Matt J on 9 Oct 2020
Edited: Matt J on 9 Oct 2020
One more way to test even-ness in a scalar, s,
isEven=false;
try, validateattributes(s,"numeric","even"); isEven=true; end,