Thread: INT 15 to return memory size

baseaddr and length are supposed to be 64 bits wide, how do you expect to do that in 16 bit C(I'll give you a clue, those ints are 16 bit.)? In my opinion, this is one of the few screw ups in C, there is no way to define a variable of a specified width.

Originally Posted by User Name:

baseaddr and length are supposed to be 64 bits wide, how do you expect to do that in 16 bit C(I'll give you a clue, those ints are 16 bit.)? In my opinion, this is one of the few screw ups in C, there is no way to define a variable of a specified width.

Um, stdint.h?

Kick ass! I've needed something like that a lot of times, never knew that existed. But he's using an old compiler, stdint.h is C99, will it support that?

Probably not, though just because it is a 16-bit compiler does NOT mean that there aren't 32-bit ints. There where always long int which were supposed to be 32-bit wide, then there was a long long int which was actually ???48 bits wide -- it has been a while and I cannot recall at the moment. I just remember the long long was something screwy.

I got the memory map to print out on the screen finally. I used a structure in my C code and passed it into the assembly like this:

Code:

memmap proc c arg1:word
		
		int 3
		;mov 	di, arg1		;500h	;point this to structure
		
		push	es			;preserve contents of es
		push 	ds			;push current data segment onto stack
		pop 	es			;pop that segment into es for the int 15
		
		mov 	di, arg1	;[bp+6]		;500h	;point this to structure 
                .
                .
                .

now i just need to find the number of unused memory regions in the first megabyte of memory