prob in finding the size of an pointer?

[AnagJohari]

hello guys
i write an program
void main()
{
char far *s1,*s2;
printf("%d%d",s izeof(s1),sizeo f(s2);
getch();
}

i tell u the ans of this that
42
i m not able to understand why 42 is the ans plz tell the reason of this.....

[newb16]

Because it's the Answer to Life, the Universe, and Everything.
Apparently 'far' modifier is an attribute of '*', so the first pointer is far, and its size is 4, and the second is 'near', with size of 2. Writing the format string as "%d %d" (with space) may help.

[Dheeraj Joshi]

Yep. It must be 4 2 not 42. use newb16's idea of giving the space between the %d's.

Regards
Dheeraj Joshi

[Banfa]

Looks like you are using something like Windows 3.11

That is the only platform I know of that had the concept of far pointers. A near pointer (default for char *) was only capable of addressing memory in the programs data sector which was 64k bytes and only required a 16 bit 2 byte pointer. A far pointer was capable of address any address on the computer and so was 32 bit.

In fact all addresses consisted of a 16 sector and a 16 bit offset in the sector. Sectors could overlap in theory and for the programs own data sector the sector was fixed so it was required to access program memory.