Re: [PD] Expr does not play well with ints?

2007-07-25 Thread Frank Barknecht
Hallo,
Mathieu Bouchard hat gesagt: // Mathieu Bouchard wrote:

 On Wed, 25 Jul 2007, Frank Barknecht wrote:
 Frank Barknecht hat gesagt: // Frank Barknecht wrote:
 B. Bogart hat gesagt: // B. Bogart wrote:
 It does work, unless you're using an old, pre-0.40 version of Pd.
 Ah, sorry, I should have tested it: [expr $1/$2] doesn't work. :(
 
 Huh? [expr $1/$2] works here on Miller's 0.40.

I mean, it doesn't create all the time, see attached.

Ciao
-- 
 Frank Barknecht _ __footils.org_ __goto10.org__


expr-div-help.pd
Description: application/puredata


expr-div.pd
Description: application/puredata
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Re: [PD] Expr does not play well with ints?

2007-07-25 Thread Mathieu Bouchard

On Wed, 25 Jul 2007, Frank Barknecht wrote:

Mathieu Bouchard hat gesagt: // Mathieu Bouchard wrote:

Huh? [expr $1/$2] works here on Miller's 0.40.

I mean, it doesn't create all the time, see attached.


It won't expand $1/$2 to 0/0 or anything else that [expr] could make sense 
of, because it's a DOLLSYM. If you want 0.0/0.0 you need to use spaces 
because then each becomes a DOLLAR. That is something that I explain in a 
previous mail of this thread.


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Re: [PD] Expr does not play well with ints?

2007-07-24 Thread Frank Barknecht
Hallo,
B. Bogart hat gesagt: // B. Bogart wrote:

 Ah, I see now.
 
 I could not use that method, since 8/6 was an example, and I wanted to
 do $8 / $6, so treating $8 as a symbol would not have helped.

It does work, unless you're using an old, pre-0.40 version of Pd.

Ciao
-- 
 Frank Barknecht _ __footils.org_ __goto10.org__

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Re: [PD] Expr does not play well with ints?

2007-07-24 Thread Mathieu Bouchard

On Tue, 24 Jul 2007, B. Bogart wrote:


Ah, I see now.
I could not use that method, since 8/6 was an example, and I wanted to
do $8 / $6, so treating $8 as a symbol would not have helped.


Ah, caution with that: literals like 8 6 8.0 6.0 and pd's substitutors ($1 
$2 $3) work in the same way in this case, but [expr]'s own substitutors 
($f1 $f2 $f3) don't have that int-vs-float confusion problem because then 
[expr] does not attempt to read those floats as being text.


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Re: [PD] Expr does not play well with ints?

2007-07-24 Thread Frank Barknecht
Hallo,
Frank Barknecht hat gesagt: // Frank Barknecht wrote:

 Hallo,
 B. Bogart hat gesagt: // B. Bogart wrote:
 
  Ah, I see now.
  
  I could not use that method, since 8/6 was an example, and I wanted to
  do $8 / $6, so treating $8 as a symbol would not have helped.
 
 It does work, unless you're using an old, pre-0.40 version of Pd.

Ah, sorry, I should have tested it: [expr $1/$2] doesn't work. :(

Ciao
-- 
 Frank Barknecht _ __footils.org_ __goto10.org__

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Re: [PD] Expr does not play well with ints?

2007-07-24 Thread Mathieu Bouchard

On Wed, 25 Jul 2007, Frank Barknecht wrote:

Frank Barknecht hat gesagt: // Frank Barknecht wrote:

B. Bogart hat gesagt: // B. Bogart wrote:
It does work, unless you're using an old, pre-0.40 version of Pd.

Ah, sorry, I should have tested it: [expr $1/$2] doesn't work. :(


Huh? [expr $1/$2] works here on Miller's 0.40.

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[PD] Expr does not play well with ints?

2007-07-23 Thread B. Bogart
Hey all,

So I've been doing lots of calculations with expr and creation
arguments, and it seems to send an int when it should send a float way
too often:

[expr 8 / 20]

returns 0!!!

See attached patch.

Have I lost my mind? Is my machine going crazy? or is expr actually
doing what I see here?

I cringe at the idea of doing these calculations with separate [*] [/]
[+] [-] objects...

Thanks,
B. Bogart

#N canvas 0 0 450 300 10;
#X obj 74 90 expr 8 / 20;
#X obj 142 24 bng 15 250 50 0 empty empty empty 0 -6 0 8 -262144 -1
-1;
#X floatatom 93 128 5 0 0 0 - - -;
#X floatatom 213 133 5 0 0 0 - - -;
#X msg 191 47 1;
#X obj 194 95 expr $f1 * 8 / 20;
#X connect 0 0 2 0;
#X connect 1 0 0 0;
#X connect 1 0 4 0;
#X connect 4 0 5 0;
#X connect 5 0 3 0;

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Re: [PD] Expr does not play well with ints?

2007-07-23 Thread Mathieu Bouchard

On Mon, 23 Jul 2007, B. Bogart wrote:


So I've been doing lots of calculations with expr and creation
arguments, and it seems to send an int when it should send a float way
too often:
[expr 8 / 20]
returns 0!!!
Have I lost my mind? Is my machine going crazy? or is expr actually
doing what I see here?


expr supports the int type because [expr] comes from jMax/FTS/Max/etc. So 
if you put a number without a dot it's parsed as an int. Then the main 
difference between ints and floats is that an int divided by an int is 
rounded towards zero (many languages do it like that but some others round 
it always downwards instead; some others instead force the result to be a 
float or a rational)


What you can do is either [expr 8 / 20.0] or [expr 8.0 / 20] or obviously 
[expr 8.0 / 20.0]


You only need to make one of the two as float because float has priority 
on int, so when a float comes into contact with an int, the int is 
converted, even though float can't represent everything that int can, e.g. 
compare


  [expr 123456789%10] finds last digit of 123456789
  [expr int(123456789.0)%10] fails because float isn't that precise

do not put a space before % in that first example! if you do, pd will 
parse it as a number, turn it to float, then give it to [expr], which 
determines that float % int = syntax error.


cool, eh?

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Re: [PD] Expr does not play well with ints?

2007-07-23 Thread B. Bogart
Mathieu, are you really able to put 8.0 in an expr argument? In my PD
(.39 ubuntu package) the 8.0 gets turned into 8 and remains an int.

Hmm?

B. Bogart

Mathieu Bouchard wrote:
 On Mon, 23 Jul 2007, B. Bogart wrote:
 
 So I've been doing lots of calculations with expr and creation
 arguments, and it seems to send an int when it should send a float way
 too often:
 [expr 8 / 20]
 returns 0!!!
 Have I lost my mind? Is my machine going crazy? or is expr actually
 doing what I see here?
 
 expr supports the int type because [expr] comes from jMax/FTS/Max/etc.
 So if you put a number without a dot it's parsed as an int. Then the
 main difference between ints and floats is that an int divided by an int
 is rounded towards zero (many languages do it like that but some others
 round it always downwards instead; some others instead force the result
 to be a float or a rational)
 
 What you can do is either [expr 8 / 20.0] or [expr 8.0 / 20] or
 obviously [expr 8.0 / 20.0]
 
 You only need to make one of the two as float because float has priority
 on int, so when a float comes into contact with an int, the int is
 converted, even though float can't represent everything that int can,
 e.g. compare
 
   [expr 123456789%10] finds last digit of 123456789
   [expr int(123456789.0)%10] fails because float isn't that precise
 
 do not put a space before % in that first example! if you do, pd will
 parse it as a number, turn it to float, then give it to [expr], which
 determines that float % int = syntax error.
 
 cool, eh?
 
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