Finding odd and even values without functions

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Is it possible to identify if a value is even or odd using for loops instead of using a function (Ex. mod)? If possible, can someone show me?
  2 Commenti
James Tursa
James Tursa il 2 Apr 2020
What have you done so far? What specific problems are you having with your code?
Jose Grimaldo
Jose Grimaldo il 2 Apr 2020
Im trying to separate evenly the array into two variables. The odd index and the even index using for loops. Cant figure it out.
A= % 8x1 array
[r,c]=size(A)
BA=A(1)
CA=A(2)
for i=1:r-1
BA(i+1)=A(i+2)
BC(i+1)=B(i+1)
end

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Risposta accettata

darova
darova il 2 Apr 2020
What about dividing?
while 1
a = a/2;
if abs(a-1) < 0.01 % if very close to '1'
disp('even')
break;
elseif a < 1 % if smaller than '1'
disp('odd')
break;
end
end

Più risposte (2)

per isakson
per isakson il 2 Apr 2020
Try this
>> a=1:12
a =
1 2 3 4 5 6 7 8 9 10 11 12
>> (-1).^[a]==1
ans =
1×12 logical array
0 1 0 1 0 1 0 1 0 1 0 1
>> (-1).^[-a]==1
ans =
1×12 logical array
0 1 0 1 0 1 0 1 0 1 0 1
>> (-1).^[1e9+a]==1
ans =
1×12 logical array
0 1 0 1 0 1 0 1 0 1 0 1
>>
  4 Commenti
darova
darova il 3 Apr 2020
Agree. It was just late. Don't know why i asked it
per isakson
per isakson il 5 Apr 2020
Better safe than sorry; one should be sceptical to the combination of double and ==.

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John D'Errico
John D'Errico il 5 Apr 2020
Modificato: John D'Errico il 5 Apr 2020
A = 1:10;
A == fix(A/2)*2
ans =
1×10 logical array
0 1 0 1 0 1 0 1 0 1
As long as A is composed of integers, this will suffice as a test for even-ness. Yes, I know that is not a word. How about parity instead? ;-) For non-integers of course the concept of even and odd is meaningless.
However, when the target is itself an integer class, then you can be slightly more concise, as the fix is no longer needed. The divide by 2 automatically truncates the fractional part implicitly, casting the division into an integer.
A = uint8(1:10);
A == A/2*2
ans =
1×10 logical array
0 1 0 1 0 1 0 1 0 1

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