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Re: pg_ctl idempotent option

From: Bruce Momjian <bruce(at)momjian(dot)us>
To: Tom Lane <tgl(at)sss(dot)pgh(dot)pa(dot)us>
Cc: Alvaro Herrera <alvherre(at)2ndquadrant(dot)com>,Vik Reykja <vikreykja(at)gmail(dot)com>,Peter Eisentraut <peter_e(at)gmx(dot)net>,PostgreSQL Hackers <pgsql-hackers(at)postgresql(dot)org>
Subject: Re: pg_ctl idempotent option
Date: 2013-01-15 19:28:56
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Lists: pgsql-hackers
On Tue, Jan 15, 2013 at 10:25:23AM -0500, Tom Lane wrote:
> Alvaro Herrera <alvherre(at)2ndquadrant(dot)com> writes:
> > Vik Reykja escribi:
> >> On Mon, Jan 14, 2013 at 4:22 PM, Tom Lane <tgl(at)sss(dot)pgh(dot)pa(dot)us> wrote:
> >>> Idempotent is a ten-dollar word.  Can we find something that average
> >>> people wouldn't need to consult a dictionary to understand?
> >> I disagree that we should dumb things down when the word means exactly what
> >> we want and based on the rest of this thread is the only word or word
> >> cluster that carries the desired meaning.
> I'm not quite convinced that it means *exactly* what we want.  The
> dictionary definition, according to my laptop's dictionary, is "denoting
> an element of a set that is unchanged in value when multiplied or
> otherwise operated on by itself".  I'm well aware that computer people
> often use it to mean "an operation that doesn't change the system state
> if the state is already what's wanted", but I think that's probably an
> abuse of the mathematical usage.  And in any case, I'm not sure that
> non-hackers would immediately recognize the term, nor be enlightened by
> their dictionaries.  But ...

I have heard idempotent used several times by our folks, and I didn't
know what it meant either.  I figured it was a "strong item".  ;-)  I
just looked it up.

  Bruce Momjian  <bruce(at)momjian(dot)us>

  + It's impossible for everything to be true. +

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