C Programming - Declarations and Initializations - Discussion
Discussion Forum : Declarations and Initializations - Find Output of Program (Q.No. 4)
4.
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
Explanation:
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).
Discussion:
55 comments Page 4 of 6.
Kim Joe said:
1 decade ago
These concepts come into picture when working with DOS where memory is limited to 1 MB and CPU registers are 16bit wide,so to access more than 16bit data the memory was accessed with offset:segment combination using far and huge,but in 32 bit compilers these keywords are obsolete .
Gangadhararao said:
1 decade ago
#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;
}
How is the answer is 2,4,4 explain in depth?
int main()
{
char *s1;
char far *s2;
char huge *s3;
printf("%d, %d, %d\n", sizeof(s1), sizeof(s2), sizeof(s3));
return 0;
}
How is the answer is 2,4,4 explain in depth?
Bharadwaj said:
1 decade ago
#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;
}
How the answer is 2,4,4?
Actually I compiled and I got the output as 4,4,4
So clarify this once
int main()
{
char *s1;
char far *s2;
char huge *s3;
printf("%d, %d, %d\n", sizeof(s1), sizeof(s2), sizeof(s3));
return 0;
}
How the answer is 2,4,4?
Actually I compiled and I got the output as 4,4,4
So clarify this once
Sundar said:
1 decade ago
@All
sizeof(int) = 2 bytes in 16 bit platform. (Turboc Under DOS)
sizeof(int) = 4 bytes in 32 bit platform. (GCC under Linux, C++ under Windows)
Like the same lot of things to be considered depending upon the platform.
All the answers given on this website are based on 16-bit platform (Turbo C under DOS).
sizeof(int) = 2 bytes in 16 bit platform. (Turboc Under DOS)
sizeof(int) = 4 bytes in 32 bit platform. (GCC under Linux, C++ under Windows)
Like the same lot of things to be considered depending upon the platform.
All the answers given on this website are based on 16-bit platform (Turbo C under DOS).
Kavita.C.Karjagar said:
1 decade ago
1.Explain what is far and huge?
2.were it is used?
2.were it is used?
Rovin varshney said:
1 decade ago
char *s1;
char far *s2;
char huge *s3;
Can anyone explain these three line, whether it is pointer or simple variable declaration. Its so much confusing.
char far *s2;
char huge *s3;
Can anyone explain these three line, whether it is pointer or simple variable declaration. Its so much confusing.
Vinu said:
1 decade ago
What is far and huge?
Reddy said:
1 decade ago
How they are using "*" symbol for variable declaration? i.e. *s1, *s2 n *s3
Any one answer me?
Any one answer me?
Neethu said:
1 decade ago
Explain, in which cases we were used these far and huge ?
Dhiraj said:
1 decade ago
@Reddy
Here *s1 means its a pointer variable which is of character type, means the pointer s1 points to some other variable which is of character type and can access that character variable.
Here *s1 means its a pointer variable which is of character type, means the pointer s1 points to some other variable which is of character type and can access that character variable.
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