Brief question: faster to zero before direct computation?

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 comentarios

per isakson
per isakson el 18 de Dic. de 2014
Editada: per isakson el 18 de Dic. de 2014
  • y1a = zeros(1,100); offers no advantage. It adds to the execution time.
  • "cube operation" is not affected by "variable declaration"
Miguel
Miguel el 18 de Dic. de 2014
this reply is a bit confusing, you are saying it offers no advantage, correct?
per isakson
per isakson el 18 de Dic. de 2014
Editada: per isakson el 18 de Dic. de 2014
Not just confusing. It was wrong; a "no" was missing. I've edited the comment.
Miguel
Miguel el 18 de Dic. de 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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