[YLUG] Help with c++ pointers - Please

[Dominic Hibbs]

The trouble with assembly language is its built in obsolescence.

On 13/01/2008, Nicholas Thomas <nick at lupine.me.uk> wrote:
>
> Dominic Hibbs wrote:
> > POINTER HELP REQUESTED
> >
> > I am an assembly language programmer - I also program in many other
> > languages BUT, as many of you youngsters may not yet have learned,
> > Knowledge has a short shelf life and my C/C++ has been on the shelf
> > too long.
> > In this - "long int" means 64 bit.
> >
> > In assembly language there is no typing at all.  A 32 bit quantity may
> be
> > 1. an integer,
> > 2. a sixteen bit number followed by a byte with no defined value and a
> >     character,
> > 3. half of a double,
> > 4. four bytes of program.
> > 5. a pointer to something,
> > 6. or a pointer to a pointer
> > and it is up to the programmer how something is interpreted and what
> > alignment is used.
> >
> > I am converting a program, written in assembler to C/C++ and I am
> > stuck over pointers. This data structure is a linked list of variables.
> >
> > A block of memory consists of records made up of
> >
> > Byte No. Content
> > (0..7)     a 32 bit pointer to
> >                 a long int (8 Byte)
> >                 or a double
> >         or (32 bit pointer and two 16 bit integers)
> > (8..15)     followed by one byte made up of
> >         2 bits of flags, 4 bits of a number and two more bit flag
> >         followed by an ASCIZ string of 4 n + 3 bytes long (n = 0..30)
> > (16..23) followed by a long int or Double
> >         or (32 bit pointer and two 16 bit integers)
> >
> > Bits 8..15 are anded with  & 0x00 00 00 3c and then added to the
> > address of this record to get the next record.
> >
> > First variable is a long int called "ten" and has the value 10 so the
> > first two bits of byte 8 hav the value 0b01 and the record length is 20
> > bytes the other two bit flags have the value 0b01 (0 means Not an array
> > and the 1 bit says this variable has been initialised).
> >
> > Abit of (obviously incorrect) code follows:
> >
> >     char * s = "ten";
> >     void * blockaddr = malloc(varspace);  // Check return value for NULL
> >     recptr = blockaddr;       // get address of first record
> >     * recptr = * recptr + 16; // insert into memory at blockaddr
> >                       // blockaddress + 16
> >     recptr += 8;          // pointer arithmetic point to byte 8
> >
> >     * recptr = 1 | 20 | 0x80  // two bit flags | record length |
> >                                   // initialised as a byte value
> >     recptr++;
> >     *recptr++ = *s;      // recptr is a char *
> >     char ch;
> >     do {
> >                *recptr++ = ch = * (++ s);
> >     } while( ch);
> >     recptr +=3;               // align on four byte -
> >                                   // we are at the start of the next
> > record.
> >
> >
> Urgh. Ouch. Eeep. Tell me, is there any particular reason you need to
> emulate the program's memory allocation behaviour so precisely? If this
> were me, I'd be looking at using (probably) std::list, containing either
> a (fairly) simple class or a struct/union thingy (which I haven't used
> for ages). Trying to literally convert an ASM program to C/C++ sounds
> like a lesson in pain (why not just use asm { ... } in that case? ;) ).
>
> Also, you can't guarantee that long int is 64 bits. As an example:
> <lupine_85> geordi: {long int i; cout << sizeof(i); }
> <geordi> 4
> <lupine_85> geordi: {long long i; cout << sizeof(i); }
> <geordi> ISO C++ does not support 'long long'
> <lupine_85> geordi: {long long int i; cout << sizeof(i); }
> <geordi> ISO C++ does not support 'long long'
>
> If you have <stdint.h>, you can use uint64_t (or int64_t) instead.
>
> /Nick
>
>
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