Question
#include<stdio.h>
int main()
{
char *s1;
char far *s2;
char huge *s3;
printf("%d, %d, %d\n", sizeof(s1), sizeof(s2), sizeof(s3));
return 0;
}
What is the output of the program in Turbo C (in DOS 16-bit OS)?
#include<stdio.h>
int main()
{
char *s1;
char far *s2;
char huge *s3;
printf("%d, %d, %d\n", sizeof(s1), sizeof(s2), sizeof(s3));
return 0;
}
Answer: Option C
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Any pointer size is 2 bytes. (only 16-bit offset)
So, char *s1 = 2 bytes.
So, char far *s2; = 4 bytes.
So, char huge *s3; = 4 bytes.
A far, huge pointer has two parts: a 16-bit segment value and a 16-bit offset value.
Since C is a compiler dependent language, it may give different output in other platforms.
The above program works fine in Window (TurboC), but error in Linux (GCC Compiler).
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