Ah, now explained like that with all the maths makes it much clearer
(yes, really it does :) so when we want to find out the exponent of the
number whose common logarithm is the argument to INT() then we have a
useful function that does just that.
> This is one case where the manual gets it wrong; it states: "INT will
return the
> integer part of the specified floating point expression."
I still think an INT function that does just that, i.e. round towards
zero, would be more useful more often. However, once you know that the
behaviour has a logical explanation and is predictable, it is easier to
accept the limitations and work around them.
Ian.
> -----Original Message-----
> From: cigmorfil
> Sent: 19 June 2001 21:20
> To: ql-users
> Cc: cigmorfil
> Subject: Re: [ql-users] NEXT in FOR-loop
>
>
> > Hmmm - the odd behaviour with negative numbers is not
> unique to SBASIC.
> > I've just tried it with QBASIC in NT and it does exactly the same.
> > I'd be interested to read the ISO or ANSI standard on this for the
> > justification. Curiously, the 68040 provides two instructions for
> > getting an integer from a float: FINT allows user
> specified rounding,
> > e.g. 'round-to-nearest', while FINTRZ always does
> 'round-to-zero'. I
> > still can't see the usefulness in having INT(-3.01)=-4 though. Any
> > mathematicians care to comment?
>
> When I first learnt BASIC (all those years ago, when you
> could buy x, y
> and z, the bus fare home and still have change from a �50 pound note
> <grin>) INT was described as returning the largest integer not greater
> than its argument. [GeeWizz-BASIC that came with PC's ('scuse the
> language) defined INT as "Returns the largest integer less
> than or equal
> to x. Examples INT(99.89) -> 99, INT(-12.11) -> -13.]
>
> The only [vague] usefulness of this that I can think of (off hand) is
> when I learnt to use (base 10) log tables (23+ years ago, BC - Before
> Computers [well, ok, before micros were available] - when a pocket
> calculator, needing B6 size pockets, only did basic maths):
>
> 1: convert the number to standard (scientific) format (y.yy x 10^n)
> 2: look up the log of y.yy (0.lll)
> 3: add the exponent n to give n.lll
>
> eg: lg(202) = lg(2.02 x 10^2) = 0.305 + 2 = 2.305
>
> INT(2.305) => 2 implies the exponent(/size?) of the original number is
> 2.
>
> However, if n is negative, the actual number will be -m.kkk.
>
> eg: lg(0.00202) = lg(2.02 x 10^-3) = 0.305 + -3 = -2.695.
>
> So, using INT(-2.695) = -3 implies that the exponent(/size?) of the
> original number is -3.
>
> BTW, my (Classic) JM QL gives INT(-3.1) => -4. This is one case where
> the manual gets it wrong; it states: "INT will return the integer part
> of the specified floating point expression."
>
> Robert
>
>
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