Finding odd and even values without functions

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Jose Grimaldo
Jose Grimaldo 2020-4-1
编辑： John D'Errico 2020-4-5
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 个评论
James Tursa
James Tursa 2020-4-2
What have you done so far? What specific problems are you having with your code?
Jose Grimaldo
Jose Grimaldo 2020-4-2
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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采纳的回答

darova
darova 2020-4-2
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

更多回答（2 个）

per isakson
per isakson 2020-4-2
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 个评论
显示 2更早的评论
darova
darova 2020-4-3
Agree. It was just late. Don't know why i asked it
per isakson
per isakson 2020-4-5
Better safe than sorry; one should be sceptical to the combination of double and ==.

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John D'Errico
John D'Errico 2020-4-5
编辑：John D'Errico 2020-4-5
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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