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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; } ?->(Show Answer!)
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1. 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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  • By: guest on 01 Jun 2017 06.01 pm
    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).
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