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 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 Windows (TurboC), but error in Linux (GCC Compiler).

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; }

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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; }

More Questions from Declarations and Initializations

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; }