Question for the Masters......

[Sultan]

Well, I open this thread with the main idea of unload any kind of questions throughout the book reading.

By the way, You know that we don´t have a teacher beside us when we are devoloping the reading and some questions arise always, so that´s why I open this space to ask the Masters about doubts that suddenly appear.

# 1

At page 422 on Chapter two, I compile the prgram StrDemo, and  everything was fine in the assembling process................. But, a doubt sudenlly merge, and is in the moves that  program relates like:

             mov( theString, ebx);              // Get the pointer to the string ( UNDERSTOOD )
   
             mov( [ebx-4], eax );                // Get current length               ( I DON´T KNOW WHY, -4 ???)
             mov( [ebx-8], ecx );                // Get maximum length            ( Same situation here Why -8 ?)

If String length is 19, I don´t understand this movement, may you explain me  more especifically.

Thanks in advance for your kind of attention.

Best regards

[Sevag.K]

Well, I'm not a "Master", but I know my around HLA well enough to answer this.

If you read a little further, you are advised not to use direct offsets as in the example, but to use the str.strvar type instead.
Generally, an HLA string variable is a pointer to character data.  The length and max length are stored at negative offsets to the character data.

Taking a look at the strRec structure:
type
      strRec:   record := -8;

               MaxStrLen:   dword;
               length:      dword;
               strData:   char[12];

            endrecord;

If you take notice, the record offset begins at -8.  A string variable would point to the strData field.  So to access the length field, you would take an offset of -4 from the address of strData.

The correct way to do so:

Code Select Expand


     mov (theString, ebx);                                            // Get the pointer of the string
     mov ( (type str.strRec [ebx]).length, eax);            // Get current length
     mov ( (type str.strRec [ebx]).MaxStrLen, ecx);     // Get maximum length


Using the strRec type, HLA will substitute the correct offsets to get the values you want.

[Sultan]

Understood !!!!!!!!!!!!!!!!!!!

Thanks a lot.

And noooooooooooooooooooo!!!!!!!!!!!!! Don´t tell me that Sevag.K because............................

YOU ARE A MASTER !!!!!!!!!!!!!!!!!!!!!!!!!!!!

HEHEHEHEHEHEHEHEHE !

Thanks again .

Sultan

[Sultan]

Hehehehehehe................. :U

Well I´ve got another one..............Sorry if I bother all of you, but, I need to understand what I´m doing.

Compiling the last programs of Chapter Three arround pages 440 to 460, I found some kind of repetitive syntaxis that make me jump once on awhile because I don´t understand why the writter repeat the procedure with different values of the dwords [], [4], [8] and [12]

This one,  could be a great example of what I´m referring and need to clarify:

You can find it at page 456 of the reading book

begin csIntersection;

    mov( (type dword csetSrc1), eax );              <==
    and( (type dword csetSrc2), eax );
    mov( eax, (type dword csetDest) );

    mov( (type dword csetSrc1 [4]), eax );          <==
    and( (type dword csetSrc2 [4]), eax );
    mov( eax, (type dword csetDest [4]) );

    mov( (type dword csetSrc1 [8]), eax );          <==
    and( (type dword csetSrc2 [8]), eax );
    mov( eax, (type dword csetDest [8]) );

    mov( (type dword csetSrc1 [12]), eax );         <==
    and( (type dword csetSrc2 [12]), eax );
    mov( eax, (type dword csetDest [12]) );


I don´t know why ( for example ) or the reazon why the writter have to repeat as a syntaxis mechanism the same procedure for []  right after for [4], [8] and for last [12].

Randy, Sevag or anybody else would like to clraify my doubt doing a wide explanation of this particular concept please ?

Thanks in advance for your kind of attention

Sultan

[Sevag.K]

A character set is 128 bits

Since we are using 32 bit reigsters, we divide this into 4 dword-sized chunks:

[bytes 0-3]  [bytes 4-7]  [bytes 8-11]  [bytes 12-15]
  dword 1       dword 2       dword 3        dword 4

Notice the offsets of the dwords are at 0, 4, 8 and 12.

The code you posted produces the intersection of 2 character sets, which is basically an AND operation:
csetSource1 AND csetSource2 = csetDest

This is like extended precision arithmatic (you probably didn't get that far yet).
Each of the 4 parts in the source performs a 32-bit chunk of the operation.

Code Select Expand



    mov( (type dword csetSrc1), eax );             // AND the 1st dword of the 2 sources together and copy to dest.
    and( (type dword csetSrc2), eax );
    mov( eax, (type dword csetDest) );

    mov( (type dword csetSrc1 [4]), eax );         // AND the 2nd dword of the 2 sources together and copy to dest.
    and( (type dword csetSrc2 [4]), eax );
    mov( eax, (type dword csetDest [4]) );

    mov( (type dword csetSrc1 [8]), eax );         // AND the 3rd dword of the 2 sources together and copy to dest.
    and( (type dword csetSrc2 [8]), eax );
    mov( eax, (type dword csetDest [8]) );

    mov( (type dword csetSrc1 [12]), eax );        // AND the last dword of the 2 sources together and copy to dest.
    and( (type dword csetSrc2 [12]), eax );
    mov( eax, (type dword csetDest [12]) );


csetSrc1:  [dword 1]   [dword 2]   [dword 3]   [dword 4]
                 AND             AND           AND          AND
csetSrc2:  [dword 1]   [dword 2]   [dword 3]   [dword 4]
                    =                =              =              =
csetDest:  [dword 1]   [dword 2]   [dword 3]   [dword 4]

It's also possible to have a loop repeat 4 times, but that way is more inefficient.

[Sultan]

Ohhhh !!!!!!!!!!! MAN !!!!!!!!!!!!!!!!

What a piece of explanation Sevag !

Thanks a lot.

I wish you the best.

Regards.

Sultan  :cheekygreen: :clap: :U

[Sultan]

Well, I have some more here, between pages 523 and 527, especifically in programs ( SimpleFileOutput, SimpleFileInput, SimppleFileInput2, AppendDemo & EolnDemo ).

For instance, there is writted the following code:

fileio.open( "myfile.txt", fileio.rw ); >>>>>>>>>>> myfile.txt never opened anywhere

Also a banner  with an alert displays: HLA Exception (7)..............File open failure, this happen in all the above mentioned files.

May I know why this kind of behavior is appearing ?

Another doubt merge, when I saw  the AppendDemo & EolnDemo files, particulary in the "While´s" section
the punctuation mark  ( ! ) before fileio.eof( fileHandle ). Can you explain me the reazon of such coding ?

fileio.rewind( fileHandle );
    while( !fileio.eof( fileHandle ) ) do   >>>>>>>>> !

Like always, Thanks in advance for your kind of attention.


Best regards.


Sultan  :green2

[Sevag.K]

fileio.open opens an existing file.  The exception likely indicates that the file was not found.
Use fileio.openNew to create a new file.  See the SimpleFileOutput program pg. 511 to see how this is done.  This program also creates the myfile.txt that the other programs use.

[Sultan]

fileio.open opens an existing file.  The exception likely indicates that the file was not found.
Use fileio.openNew to create a new file.  See the SimpleFileOutput program pg. 511 to see how this is done.  This program also creates the myfile.txt that the other programs use.

Got it and works !

Now I only have the doubt regarding the coding in the while section [ ! ]

Thanks for your quick reply Sevag.K

Sultan

[Sevag.K]

Oh forgot about that part.  The explanation point '!' means NOT.  So in essence the loop is "while not end of file, do..."

[Sultan]

ORALE CAMPEON !!!!!!!!!!!!

Understood !

Thanks Sevag.K for your kind of attention !!!!!!!!

So far I owe you a couple of beers dude, hopefully sometime, we drink some and cheers about HLA hehehehehehehehe.  :dance: :U :cheekygreen:

Regards.


Sultan  :green2

[Sultan]

Page 565

program QSDemo;

Line 36

shr( l, eax );

Encountered an error and compiler tells me:

C:\ HLA\ Projects\ QSDemo>hla QSDemo.hla
Error in file "QSDemo.hla" at line 36 [errid:41506/hlaparse.bsn]:
Undefined symbol.
Near: << l >>

So, What do I have to do ?

Regards.


Sultan  :red

[Sevag.K]

Page 565

program QSDemo;

Line 36

shr( l, eax );

Encountered an error and compiler tells me:

C:\ HLA\ Projects\ QSDemo>hla QSDemo.hla
Error in file "QSDemo.hla" at line 36 [errid:41506/hlaparse.bsn]:
Undefined symbol.
Near: << l >>

So, What do I have to do ?

Regards.


Sultan  :red

change the capital i to the number 1

:)

[Sultan]

Got it !!!!!!!!!!!!!!

Thanks again.

Sultan  :U :clap: :dance: