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Bug Reports

Author Thread: Scan operator
Alain.AMIOUNI
Scan operator
Posted: Friday, June 17, 2005 11:04 AM (EST)

 This is a glitch likely to cause real problems

 in a program expecting a scan derived function

 to return its argument if empty.

 Which should be the case.

 

 Here everything is OK.

 

       A„2 0 0 2½0


      A­(match)B„×™A ª A(match)­×\[2]A ª A­(match)×\[3]A ª A(match)­×\A


 1
 1
 1
 1
      0(match)­†B
 1
     

  Here it is not anymore.

      

        A„2 0 0 2½›¼3


      A­(match)B„×™A ª A­(match)×\[2]A ª A­(match)×\[3]A ª A(match)­×\A


 0
 1
 1
 0
      ' '­(match)†B
 1


Comments:

Author Thread:
Support
Scan operator
Posted: Friday, June 17, 2005 12:45 PM (EST)

Alain,

The APL symbols don't show up in your message making it hard to read.  Please repost your message after having used the Preview option to verify the APL symbols appear correctly in your message.

APL2000 Support

     

brent.hildebrand
Scan operator
Posted: Friday, June 17, 2005 12:53 PM (EST)
If you copy and paste the text into an APL session, and remove the "(match)" text, you can see the code. The problem with APLDN is that the Match character does not appear in the Browser. But if you copy and paste the text into APL, the match character magically reappears. I believe the original poster put the text "(match)" to show where that character was located.

     

Alain.AMIOUNI
Scan operator
Posted: Friday, June 17, 2005 1:30 PM (EST)

    
       © There is a glitch likely to    cause real problems in a program
       © expecting a Scan derived function to return its argument if empty.
       © Which should be the case.


       © Here everything is OK.

 

         A„2 0 0 2½0


      A(match)­B„×™A ª A­(match)×\[2]A 

      A­(match)×\[3]A ª A­(match)×\A


       1
       1
       1
       1


 0(match)­†B


 1

 

       © In what follows it is not anymore.
      
         A„2 0 0 2½›¼3

 
       A­(match)B„×™A ª A(match)­×\[2]A 

       A(match)­×\[3]A ª A(match)­×\A


       0
       1
       1
       0


 ' '(match)­†B       © Instead of 0 0 0


 1    
    

     

Alain.AMIOUNI
Scan operator
Posted: Friday, June 17, 2005 1:33 PM (EST)

    
       © There is a glitch likely to    cause real problems in a program
       © expecting a Scan derived function to return its argument if empty.
       © Which should be the case.


       © Here everything is OK.

 

         A„2 0 0 2½0


      A(match)­B„×™A ª A­(match)×\[2]A 

      A­(match)×\[3]A ª A­(match)×\A


       1
       1
       1
       1


 0(match)­†B


 1

 

       © In what follows it is not anymore.
      
         A„2 0 0 2½›¼3

 
       A­(match)B„×™A ª A(match)­×\[2]A 

       A(match)­×\[3]A ª A(match)­×\A


       0
       1
       1
       0


 ' '(match)­†B     © Instead of 0 0 0


 1    
    

     

Alain.AMIOUNI
Scan operator
Posted: Friday, June 17, 2005 2:30 PM (EST)

I must have clicked twice on the "Submit" button.

Font I used is a bit large. Sorry.

     



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