Finding odd and even values without functions
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Jose Grimaldo
am 1 Apr. 2020
Bearbeitet: John D'Errico
am 5 Apr. 2020
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 Kommentare
James Tursa
am 2 Apr. 2020
What have you done so far? What specific problems are you having with your code?
Akzeptierte Antwort
darova
am 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
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Weitere Antworten (2)
per isakson
am 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 Kommentare
per isakson
am 5 Apr. 2020
Better safe than sorry; one should be sceptical to the combination of double and ==.
John D'Errico
am 5 Apr. 2020
Bearbeitet: John D'Errico
am 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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