pointer arithmetic

This is a discussion on pointer arithmetic within the C Programming forums, part of the General Programming Boards category; Hi, here is a structure with 2 pointer to integer I want increment the value of x by 1, (tt.c) ...

  1. #1
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    Mar 2004
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    1

    pointer arithmetic

    Hi, here is a structure with 2 pointer to integer

    I want increment the value of x by 1, (tt.c) ++ is not increment to 11 rather it increments to 14. how do i increment the value pointed by tt.c ? any help?
    thx
    Code:
    #include <stdio.h>
    Code:
    main () {typedef struct {
    Code:
    int *c;        int *d; }ABC; ABC  tt;
    Code:
    int x=10, y=11;    tt.c = &x;      tt.d = &y;
    Code:
    (tt.c)++ ;  // lincrementing = 10+1
    Code:
    printf ("value of x %u\n", (tt.c)) ;  }

  2. #2
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    Dec 2001
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    34
    To increment THE VALUE STORED in x: (*tt.c)++

    Also, this
    Code:
    printf("value of x %u/n", (tt.c))
    isn't displaying the value stored in x. It's displaying the address of x. You need to dereference it.

    clu82
    Last edited by clu82; 03-20-2004 at 11:03 PM.

  3. #3
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    Mar 2004
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    Here's a suggestion: when you don't understand program output, sprinkle a few printf()s around.

    consider:
    Code:
    #include <stdio.h>
    
    main () 
    {
      typedef struct {
            int *c;
            int *d; 
      }ABC; 
    
      ABC  tt;
    
      int x=10, y=11;    
      tt.c = &x;      
      tt.d = &y;
    
      printf("tt.c = %p\n", tt.c);
      printf("*tt.c = %d\n", *tt.c);
      printf("tt.d = %p\n", tt.d);
      printf("*tt.d = %p\n", *tt.d);
    
      // the following makes tt.c point to who-knows-where
      (tt.c)++ ;  // lincrementing = 10+1
      printf("After incrementing tt.c = %p\n", tt.c);
      printf("After incrementing tt.c, *tt.c = %d\n", *tt.c);
    
      // tt.c is a pointer; you are printing the value of the pointer,
      // not the value of who-knows-what that tt.c is pointing to
      printf ("value of x %u\n", (tt.c)) ;  
      printf ("value of x %p\n", (tt.c)) ;  
      return 0;
    }
    Dave

  4. #4
    Been here, done that.
    Join Date
    May 2003
    Posts
    1,156

    Re: pointer arithmetic

    The reason (tt.c)++ incremented by 4 is because the value is a pointer and pointers are 4 bytes, to keep them aligned the language will increment by it's size.

    Two other points:

    First, you need code tags around the entire program, not each line

    Second, your formatting is abysmal (based on the removal of code tags). It should be more like:
    Code:
    #include <stdio.h>
    int main () 
    {
        typedef struct 
        {
            int *c;
            int *d; 
        }ABC; 
        ABC  tt;
        int x=10, y=11;
        
        tt.c = &x;
        tt.d = &y; 
        (tt.c)++ ;  // lincrementing = 10+1 
        printf ("value of x %u\n", (tt.c)) ;
        return 0;
    }
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