void main() { char far *farther,*farthest; printf("%d..%d",si

void main()
{
char far *farther,*farthest;

printf("%d..%d",sizeof(farther),sizeof(farthest));

}

Question Posted / surenda pal singh chouhan

3 Answers
30878 Views
ME, pspl, I also Faced
E-Mail Answers

Answers were Sorted based on User's Feedback

void main() { char far *farther,*farthest; printf("%d..%d",sizeof(farther),sizeof..

Answer / surenda pal singh chouhan

4..2

Explanation:
the second pointer is of char type and not a
far pointer

Is This Answer Correct ?

94 Yes

11 No

void main() { char far *farther,*farthest; printf("%d..%d",sizeof(farther),sizeof..

Answer / jaroosh

Some explanation first:
far pointers are outdated concepts from C, and are actually
compiler extentions, so you might get compiler errors trying
to use far pointers on some compilers.
In the program above,
farther's type - FAR pointer to char
farthest's type - near pointer to char
now, the difference in size of those stems from the fact
that far pointers consist of the segment and offset
together, while near pointers just have the offset.
Near pointers thus have size of 2 (just the offset), while
far pointers - size of 4 bytes.

Is This Answer Correct ?

64 Yes

4 No

void main() { char far *farther,*farthest; printf("%d..%d",sizeof(farther),sizeof..

Answer / basha

syntax error before '*' token
error: `farther' undeclared (first use in this function)
error: `farthest' undeclared (first use in this function)

I am getting these types of errors how u get 4 and 2 i
didn't understood

Is This Answer Correct ?

15 Yes

30 No

Post New Answer