pointers to arrays

[Robert Forsman]

in article 13339 guy at auspex.UUCP writes
>
>  Pointers are usually used to modify things, and you can definitely
>  modify an array by assigning to elements of that array:
>
>	int (*foo)[13];
>	int bar[13];
>
>	foo = &bar;
>	(*foo)[7] = 666;
>
>  should work in an ANSI C program once ANSI C exists and programs can be
>  written in it and compiled by ANSI C-conformant compilers.
>

AUUUGH!  what happens if you say (*foo) = muck (where muck is int muck[13])
Help! somebody's screwing with my stack.  OK, lets assume this is a C 
compiler of the usual flavor (meaning one I can understand) and that foo
and bar are local variables.  For illustration purposes let's declare another
pointer to int (int *intptr;)  Now the stack might look like this.

+-----+-----+-----+----------------+-----+-----+-----+
| NULL|  0  |  0  |  (9 more ints) |  0  |  0  | NULL|
+-----+-----+-----+----------------+-----+-----+-----+
 foo    ^--bar-------------------------------^   intptr

When we modify foo the compiler generates code like

MOVE_INTO_MEMORY_ADDR     BasePointer+0,   424242

When we modify an element of bar it generates

MOVE_INTO_MEMORY_ADDR     BasePointer+8,    666		/* bar[7]=666; */

When we modify some element that intptr (it could point to an array) points to

MOVE_MEMORY_INTO_REG	  REG,     BasePointer+14	/* grab intptr    */
MOVE_INTO_MEMORY	  REG+7,  666			/* intptr[7]=666; */

Note that the compiler knows where the array bar is -vs- it has to figure out
where intptr is.  You can say bar[7] and intptr[7] but the code is slightly
different.  bar is a CONSTANT (BasePointer+1), intptr is a VARIABLE! (data at
memory location BasePointer+14).  Using *(&intptr) you can modify the memory
at Basepointer+14 but trying *(&bar) would be trying to modify the constant
BasePointer+1.

  When you say bar the comiler knows EXACTLY where the elements are.
When you use intptr the code has to go into memory to get the address
of the elements.  Since this address is stored in memory you can take
its address and modify it.  You take &bar 'cause you CANNOT modify bar
AT ALL.. You can say bar[7]=4 all you want but if you say bar=intptr
I'M GOING TO SHOOT YOU! :)-
	 this is a tongue ^

  Now messing with pointers can screw with your mind (I claim immunity)
but the point (and I know I'm going to get flamed on this) is that bar is
a constant and intptr is a variable.  I KNOW bar[7] IS A VARIABLE, BUT
bar IS NOT.  Try changing it sometime.
  Any of you wizards out there who have managed to make it this far PLEASE
support me and help stomp out the pointer/array debate, otherwise I'll
gladly let you guys rot in hell and occasionally send you a letter with
SEGMENTATION VIOLATION written on it.

  One of these days I'm going to try some of this stuff and really look at the
assembler it generates to see how correct I am ( 0, 1, .643 who knows?).

/*  Sorry if there were any typos 'cause I don't know how to use emacs-yank-
news-posting or whatever and had to copy it by hand.  Frankly I think they
should invent a version of C with training wheels, no pointers, no arrays
just ints, chars and reals and say UNSUPPORTED on the other version. */

/*  Of course my opinions are my own, I haven't recieved a word of support
	yet on this issue */   Just say maybe to .signature