- y1a = zeros(1,100); offers no advantage. It adds to the execution time.
- "cube operation" is not affected by "variable declaration"
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Brief question: faster to zero before direct computation?
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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
댓글 수: 4
per isakson
2014년 12월 18일
편집: per isakson
2014년 12월 18일
Not just confusing. It was wrong; a "no" was missing. I've edited the comment.
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