C query

ce_nehaHi ,

Please make me understand the output of the following code:-

#include<stdio.h>
void main()
{
char *s1;
char far *s2;
char huge *s3;
printf("%d %d %d ",sizeof(s1),sizeof(s2),sizeof(s3));
}

i know answer of 1st 1 byte,4 byte,4byte

THE OUTPUT OF PROGRAM IS

4 4 4



😎PARDHU😎

The output is 4 4 4 because every pointer is 4 bytes long irrespective of what type it is pointing.

I just ran the program and it is showing the output as 2 4 4....
Can u please elaborate your ans...

devil_radsI just ran the program and it is showing the output as 2 4 4....
Can u please elaborate your ans...

The program gives errors on compiling.

It runs only after modifying in this way

#include<stdio.h>
void main()
{
     char *s1;
     char *s2;
     char *s3;
     printf("%d %d %d ",sizeof(s1),sizeof(s2),sizeof(s3));
     getch();
}

Output: 4 4 4
Using dev-c++ compiler. (or gcc)

First one char *str : str is a pointer variable which stores an address of a char datatype. So the size is 2 bytes (default). In which you can address from (0000)hexa to (FFFF)hexa. you cant go beyond that.

Second one is char far *str: far pointer increases the addressable locations. where you can give address ranges from (00000000)hexa to (FFFFFFFF)hexa. Ex:- for the monitor output the address is 0800 0000 hexa.

Third one is char huge *str: when far pointer overflows ie gets incremented from (FFFFFFFF) + 1 it again goes to (00000000) hexa. The address is inside the same process.
But in the case of huge if it gets overflow (FFFFFFFF)+1 it goes to the next process 1 0000 0000 hexa.

char *s1;
char far *s2;
char huge *s3;

What is far and huge ?

ce_nehaHi ,

Please make me understand the output of the following code:-

#include<stdio.h>
void main()
{
char *s1;
char far *s2;
char huge *s3;
printf("%d %d %d ",sizeof(s1),sizeof(s2),sizeof(s3));
}

hey first can you tell me what is far and huge in your program.?





WINNERS DONT DO DIFFERENT THINGS....THEY DO THINGS DIFFERENTLY....

ce_nehaHi ,

Please make me understand the output of the following code:-

#include<stdio.h>
void main()
{
char *s1;
char far *s2;
char huge *s3;
printf("%d %d %d ",sizeof(s1),sizeof(s2),sizeof(s3));
}

As per my knowledge:

This is happen because simple pointer store in ram and far pointer use video memory.

brahmaasthraFirst one char *str : str is a pointer variable which stores an address of a char datatype. So the size is 2 bytes (default). In which you can address from (0000)hexa to (FFFF)hexa. you cant go beyond that.

Second one is char far *str: far pointer increases the addressable locations. where you can give address ranges from (00000000)hexa to (FFFFFFFF)hexa. Ex:- for the monitor output the address is 0800 0000 hexa.

Third one is char huge *str: when far pointer overflows ie gets incremented from (FFFFFFFF) + 1 it again goes to (00000000) hexa. The address is inside the same process.
But in the case of huge if it gets overflow (FFFFFFFF)+1 it goes to the next process 1 0000 0000 hexa.

I guess this is the correct answer!!!