Announcement

**Eddy Van Esch** · 2 May 2007, 08:11 AM

Heinz,

Have you tried to specify the number of digits, like this?
PRINT BIN$(-1&, 32)

Kind regards

------------------
Eddy
www.devotechs.com -- HIME Huge Integer Math and Encryption library--

**Heinz Salomon** · 2 May 2007, 12:50 PM

Eddy,

Alas, PB/DOS's BIN$() syntax doesn't allow you to specify the number of digits of the result string:

BIN$ function
─────────────
[...]
Syntax: s$ = BIN$(numeric expression)

Regards,
Heinz Salomon

------------------

**Michael Mattias** · 2 May 2007, 05:21 PM

>Alas, PB/DOS's BIN$() syntax doesn't allow you to
>specify the number of digits of the result string

Well, damn.

Too bad there is no way to add enough zeroes to the start of whatever BIN$ returns to make it 32 characters long.

Damn.

**Eric Pearson** · 2 May 2007, 05:31 PM

> Too bad there is no way to add enough zeroes

Unless I read the first post incorrectly, that's not a solution.

-- Eric

**Mel Bishop** · 2 May 2007, 05:51 PM

Code:

te$ = bin$(x)
do until len(te$) = 32
te$ = "0" + te$
loop

check my code at: http://www.powerbasic.com/support/pb...ad.php?t=24418

binari$ will return an unsigned binary string up to the capactiy
of your computer.
------------------

[this message has been edited by mel bishop (edited may 02, 2007).]

**Eddy Van Esch** · 2 May 2007, 05:56 PM

Indeed. Adding zeroes does not help .. Read first post ...

Kind regards

------------------
Eddy
www.devotechs.com -- HIME Huge Integer Math and Encryption library--

**Dave Stanton** · 2 May 2007, 06:01 PM

If I remember correctly from my PBDOS days.

Deleted

Edit:- I did NOT remember correctly

For non-negative integers

FUNCTION BIN16$(Q??)
FUNCTION = RIGHT$("0000000000000000"+BIN$(Q??),16)
END FUNCTION

FUNCTION Bin32$(Q???)
FUNCTION = BIN16$(Q??? DIV 65536)+ BIN16$(Q??? MOD 65536)
END FUNCTION

------------------
Dave

For a real Nightmare

[This message has been edited by Dave Stanton (edited May 02, 2007).]

**Robert DeBolt** · 2 May 2007, 06:51 PM

Thanks, Mike and Dave!

Code:

printmystuff(0)
printmystuff(1)
printmystuff(-1)
printmystuff(65535)
printmystuff(65536)
printmystuff(-65535)
printmystuff(-65536)
printmystuff(2147483647)
printmystuff(-2147483648)
 
  FUNCTION printmystuff(a&)
    print "Number =" + str$(a&)
    print bin$(a&)
    print binfmt$(a&,32)  ' From help screen.
    print bin32$(a&)      ' From Dave Stanton
  END FUNCTION
 
  FUNCTION BinFmt$(value&,size%) ' returns a binary string of size% digits
    if value& < 0 then
      BinFmt$ = RIGHT$("111111111111111"+BIN$(value&),size%)
    else
      BinFmt$ = RIGHT$("000000000000000"+BIN$(value&),size%)
    end if
  END FUNCTION
 
 
  FUNCTION Bin32$(N&)
  FUNCTION = BIN$(N& \ 65536)+ BIN$(n& MOD 65536)
  END FUNCTION
 
 
Number = 0
0
0000000000000000
00
Number = 1
1
0000000000000001
01
Number =-1
1111111111111111
1111111111111111111111111111111
01111111111111111
Number = 65535
1111111111111111
0000000000000001111111111111111
01111111111111111
Number = 65536
10000000000000000
00000000000000010000000000000000
10
Number =-65535
11111111111111110000000000000001
11111111111111110000000000000001
011111111111111110000000000000001
Number =-65536
11111111111111110000000000000000
11111111111111110000000000000000
11111111111111110
Number = 2147483647
1111111111111111111111111111111
01111111111111111111111111111111
1111111111111111111111111111111
Number =-2147483648
10000000000000000000000000000000
10000000000000000000000000000000
10000000000000000

------------------
Regards,
Bob
--------
Composed with EZ-Post!

**Heinz Salomon** · 3 May 2007, 03:53 AM

Thanks to all for your replies!

All I wanted to know was if BIN$() accepts numbers which fall within the long integer range - as described in the online help. Obviously, you only get correct results as long as you pass (short) integers to this function. So if I want to get the binary representation of a number which falls within the long integer range, I have to get by with some workaround (at the expense of "programming comfort" and decreased execution speed).

Summary:
Since the BIN$() function doesn't work as described in the documentation, this behaviour is either a bug or the documentation is in error.

Regards
Heinz Salomon

------------------

**Bob Zale** · 3 May 2007, 06:40 PM

Unfortunately, both of your inferences are remarkably false. If "All you wanted to know was if BIN$() accepts numbers which fall within the long integer range...", it might have been best to actually try it and then just look at the result?

Perhaps something like...

PRINT BIN$(65536)

Best regards,

Bob Zale
PowerBASIC Inc.

------------------

**Heinz Salomon** · 4 May 2007, 01:26 AM

For me, the problem lies in the inconsistency regarding the results of the BIN$() function. If I pass a long integer value to the BIN$() function, I'd expect the result to display the correct 32-bit binary representation of this number, even if the passed value actually falls within the range of a (short) integer. If - in some cases - I have to expand the results "manually" to match the number of bits of the passed argument, I could as well program my own BIN$() function.

Regards
Heinz Salomon

(/edit typo)

[This message has been edited by Heinz Salomon (edited May 04, 2007).]

**Bob Zale** · 4 May 2007, 01:47 AM

There's nothing at all inconsistent about it. The result is not returned in a fixed length field, but rather the size "floats" based upon the value. This is consistent with the RADIX functions (such as HEX$) found in both PowerBASIC and Microsoft compilers for the past twenty (20) years.

"If I pass a long integer value to the BIN$() function, I'd expect the result to display the correct 32-bit binary representation of this number, even if the passed value actually falls within the range of a (short) integer."

The PowerBASIC BIN$() function does precisely that.

"If - in some cases - I have to expand the results "manually" to match the number of bits of the passed argument, I could as well program my own BIN$() function."

That's the beauty of a programming language product like PowerBASIC! It presumes that you are a programmer! If you'd like to receive results in a format other than ours, you only need to write your own simple function to do just that! Wouldn't it be more productive to just try that approach?

Best regards,

Bob Zale
PowerBASIC Inc.

------------------

**Heinz Salomon** · 4 May 2007, 04:06 AM

I understand the point *smile*.

Code:

FUNCTION Long2Bin(BYVAL rlngNumber AS LONG) AS STRING
  DIM i AS LOCAL INTEGER
  DIM ptrDigit AS LOCAL BYTE PTR     
  DIM strResult AS STATIC STRING * 32  
  LSET strResult = "00000000000000000000000000000000"
  ptrDigit = VARPTR32(strResult) + 31
  FOR i = 1 TO 32
    IF rlngNumber& MOD 2 THEN
      @ptrDigit = &H31
    END IF
    SHIFT RIGHT rlngNumber&, 1
    DECR ptrDigit
  NEXT
  FUNCTION = strResult
END FUNCTION

/EDIT
still better:

Code:

FUNCTION Long2Bin(BYVAL rlngNumber AS LONG) AS STRING
  DIM ptrDigit AS LOCAL BYTE PTR     
  DIM strResult AS STATIC STRING * 32  
  LSET strResult = "00000000000000000000000000000000"
  ptrDigit = VARPTR32(strResult) + 32
  WHILE rlngNumber&
    DECR ptrDigit
    IF rlngNumber& MOD 2 THEN
      @ptrDigit = &H31
    END IF
    SHIFT RIGHT rlngNumber&, 1
  WEND
  FUNCTION = strResult
END FUNCTION

Regards
Heinz Salomon

[This message has been edited by Heinz Salomon (edited May 04, 2007).]

**Dave Stanton** · 4 May 2007, 04:21 AM

What's wrong with:-

FUNCTION Bin32$(Q???)
FUNCTION=RIGHT$("00000000000000000000000000000000"+BIN$(Q???),32)
END FUNCTION

------------------
Dave

For a real Nightmare

**Heinz Salomon** · 4 May 2007, 05:06 AM

Dave,

your function wouldn't work for negative arguments between -1 and -32,768:

Code:

Bin32$(-1???) = 00000000000000001111111111111111 = 65,535 instead of
                11111111111111111111111111111111 = -1

Regards
Heinz Salomon

------------------

**Steve Miltiadous** · 4 May 2007, 05:43 AM

How about

macro Bin32(Q ) = RIGHT$(iif$(Q>0,"00000000000000000000000000000000","11111111111111111111111111111111")+BIN$(Q),32)

------------------

**Bob Zale** · 4 May 2007, 07:58 AM

Precisely the right idea Steve. Very good coding... likely the most efficient possible. But you probably want to make a very small change to:

macro Bin32(Q ) = RIGHT$(iif$(Q>=0,"00000000000000000000000000000000","11111111111111111111111111111111")+BIN$(Q),32)

Best regards,

Bob Zale
PowerBASIC Inc.

------------------

**Arthur Gomide** · 4 May 2007, 09:52 AM

Macro? IIF$? PB35?

------------------
Arthur Gomide
"If you think good architecture is expensive, try bad architecture." -- Brian Foote and Joseph Yoder

**Bob Zale** · 4 May 2007, 09:59 AM

Function Bin32(q&)AS STRING
IF q& >= 0 THEN
FUNCTION = RIGHT$("00000000000000000000000000000000" + BIN$(q&),32)
ELSE
FUNCTION = RIGHT$("11111111111111111111111111111111" + BIN$(q&),32)
END IF
END FUNCTION

------------------

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BIN$() function

BIN$() function

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