What is the output of the program in Turbo C (in DOS 16-bit OS)? #include int main() { char *s1; char far *s2; char huge *s3; printf("%d, %d, %d ", sizeof(s1), sizeof(s2), sizeof(s3)); return 0; }

Question

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
", sizeof(s1), sizeof(s2), sizeof(s3));
    return 0;
}

Accepted Answer

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 valu…

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; }

Discussion & Comments

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; }

Discussion & Comments

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; }