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Brief question: faster to zero before direct computation?

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Miguel
Miguel on 18 Dec 2014
Closed: Miguel on 18 Dec 2014
Hello all, quick question,
in the simplest of examples,
x1a = linspace(-1,1,100);
y1a = zeros(1,100);
y1a = x1a.^3;
Is it a clear computational speed and economy advantage to declare y1a first with zeros, or not, in this simple case?
Is the cube operation one that does not require nor benefit from variable declaration?
Cheers
  4 Comments
per isakson
per isakson on 18 Dec 2014
Edited: per isakson on 18 Dec 2014
Not just confusing. It was wrong; a "no" was missing. I've edited the comment.
Miguel
Miguel on 18 Dec 2014
Ok then, I reckon this answers it, I would like to know, however, if the statement y1a = x1a.^3 is something that builds the y1a variable one by one increasing its size after each value is computed, or, what I am suspecting is the case, applies the operation first to the x1a matrix, and then assigns this to variable y1a?

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