Question

//The output for this code is observed to be 4 4 2 ? I dont get why is it 2 for ptr3?please help

#include<stdio.h>

int main()
{
char huge *near *far *ptr1;
char near *far *huge *ptr2;
char far *huge *near *ptr3;
printf("%d, %d, %d\n", sizeof(ptr1), sizeof(ptr2), sizeof(ptr3)); 
return 0;
}

Answer 1

near, far, and huge pointers are non-standard.

In implementations that support them (usually for very old x86 systems), a specific location in memory can be specified by a segment and an offset. A near pointer contains only an offset, leaving the segment implicit. Thus it requires less space than a far pointer.

Modern compilers do not use any of these keywords. I suggest you find yourself a newer compiler; there are plenty of free ones available.

See this question for more information.

Incidentally, printf’s "%d" format requires an argument of type int; sizeof yields a result of type size_t, which is a different type. You can change your printf call to:

printf("%d, %d, %d\n", (int)sizeof ptr1, (int)sizeof ptr2, (int)sizeof ptr3);

(The modern way to do that is:

printf("%zu, %zu, %zu\n", sizeof ptr1, sizeof ptr2, sizeof ptr3);

but an implementation that supports near and far pointers isn’t likely to be modern enough to support "%zu", which was introduced in C99.)