C query

Replies

• 

  ameyaamu

  ce_neha

  Hi ,
  
  Please make me understand the output of the following code:-
  
  #include
  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
• 

  M.Pardhu

  THE OUTPUT OF PROGRAM IS
  
  4 4 4
  
  
  
  😎PARDHU😎
• 

  gaurav.bhorkar

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

  devil_rads

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

  gaurav.bhorkar

  devil_rads

  I 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
  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)
• 

  brahmaasthra

  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.
• 

  gaurav.bhorkar

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

  What is far and huge ?
• 

  Sahithi Pallavi

  ce_neha

  Hi ,
  
  Please make me understand the output of the following code:-
  
  #include
  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....
• 

  pdpatel

  ce_neha

  Hi ,
  
  Please make me understand the output of the following code:-
  
  #include
  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.
• 

  Saandeep Sreerambatla

  brahmaasthra

  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.

  
  I guess this is the correct answer!!!