🚀 go-pugleaf

davis@space.mit.edu (John E. Davis) writes:

>So, does this fail:

>   double *d;
>   char *p;
>   unsigned int size;
>   
>   size = sizeof (double);
>   d = malloc (size);
>   p = malloc (size);

No.

>What is the difference between d and p?

Well, to start with, they must point to different objects.  (:-)

But, to take a real example, 
	d might be represented as
		indirect flag in sign bit
		31 bit "word" address, where a "word" is 16 bits
	p might be represented as
		31 bit "word" address (starting in sign bit)
		1 bit byte-within-word offset

>Is malloc smart enough to know which representation to return?

malloc() _always_ returns a void*.  It doesn't know, can't tell, and
doesn't care, what you assign that to.  It is the compiler-generated
code "convert-void*-to-double*" and "convert-void*-to-char*" that is
responsible for changing the representations.

For example, you might get
	LDW	%A,=8
	STW	%A,size
	PUSHW	%A
	CALL	malloc,1
	ROTR	%A		; right rotate		; cast void*
	TRAPN			; trap if sign bit set	; to double*
	STW	%A,d

	LDW	%A,size
	PUSHW	%A
	CALL	malloc,1
	STW	%A,p		; void* and char* have same representation

(The assembler here is imaginary, the encoding most certainly is not.)

Then I could tell you about the machine I used where double* pointers were
32 bits and char* pointers were 48 (and had to be loaded into one of two
very special registers).

-- 
Result of state election = 50% Victorians didn't get MP of their choice.
Richard A. O'Keefe; http://www.cs.rmit.edu.au/~ok; RMIT Comp.Sci.