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Re: how does it evaluate: y * w # A where: [message #36321]

Wed, 10 September 2003 02:12

iqbal_hassan
Messages: 6
Registered: August 2003

Junior Member

David Fanning <david@dfanning.com> wrote in message news:<MPG.19c6d067e7fa78f59896e1@news.frii.com>...
> Craig Markwardt writes:
>
>> Hmm, David, in the manuals I've looked at, the # (matrix multiply) and
>> * (multiply) operators have the *same* precedence.
>
> Oh, whoops! I must have been looking at pointer de-referencing.
> The font is a bit tiny in my on-line help, and I had misplaced
> my good glasses this morning (again!). :-)
>
> Cheers,
>
> David

Thanks Craig and David. If * and matrix # operators have the same
precedence and if operators with same precedence are evaluated left to
right then I am able to understand the things with the help from you
guys.
Actually my online help on operators also shows * as having third
highest precedence and # as having fourth. That is why I was having
this whole lot of confusion.
Thanks once again.
Hassan

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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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