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Re: how does it evaluate: y * w # A where: [message #36341 is a reply to message #36340]

Mon, 08 September 2003 16:28

Craig Markwardt
Messages: 1869
Registered: November 1996

Senior Member

David Fanning <david@dfanning.com> writes:
> You find that the * operator has the third highest
> precedence, whereas the # operator has the fourth
> highest precedence. Thus, multiplication is done
> before matrix operations. If operators have the
> same order of precedence, then they are performed
> left to right across the expression.

Hmm, David, in the manuals I've looked at, the # (matrix multiply) and
* (multiply) operators have the *same* precedence.

Craig

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Craig B. Markwardt, Ph.D. EMAIL: craigmnet@cow.physics.wisc.edu
Astrophysics, IDL, Finance, Derivatives | Remove "net" for better response
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[Message index]

		Re: how does it evaluate: y * w # A where: By: iqbal_hassan on Wed, 10 September 2003 02:12
		Re: how does it evaluate: y * w # A where: By: David Fanning on Mon, 08 September 2003 17:44
		Re: how does it evaluate: y * w # A where: By: Craig Markwardt on Mon, 08 September 2003 16:28
		Re: how does it evaluate: y * w # A where: By: David Fanning on Mon, 08 September 2003 07:43
		Re: how does it evaluate: y * w # A where: By: David Fanning on Mon, 08 September 2003 07:36
		Re: how does it evaluate: y * w # A where: By: Craig Markwardt on Mon, 08 September 2003 07:26

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