why same address? Need help...

This is a discussion on why same address? Need help... within the C Programming forums, part of the General Programming Boards category; Code: #include <stdio.h> int main() { int i; char a[]="abc"; printf("%p %p\n",a[1],i); //different address i=a[1]; printf("%c\n",a[1]); printf("%c\n",i); printf("%d\n",i); printf("%p %p\n",a[1],i); ...

  1. #1
    Registered User
    Join Date
    Jun 2010
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    3

    why same address? Need help...

    Code:
    #include <stdio.h>
    int main()
    {
      int i;
      char a[]="abc";
      printf("%p %p\n",a[1],i); //different address
      i=a[1];
    
      printf("%c\n",a[1]);
      printf("%c\n",i);
      printf("%d\n",i);
      printf("%p %p\n",a[1],i); //same address
    }
    does anybody can explain this? Thx...

  2. #2
    Registered User
    Join Date
    Nov 2008
    Location
    INDIA
    Posts
    64
    Actually you have to print the address , is it .If you want the print the address of the variables you have to use '&' since you are trying to get the variables address.If you give the pointer there there there is no need for '&'.

    Code:
    #include <stdio.h>
    main()
    {
      int i;
      char a[]="abc";
      printf("%p %p\n",&a[1],&i);  // use &
      i=a[1];
    
      printf("%c\n",a[1]);
      printf("%c\n",i);
      printf("%d\n",i);
      printf("%p %p\n",&a[1],&i);  //use &
    }
    Result is,

    Code:
    0xbfd70d9d 0xbfd70da0
    b
    b
    98
    0xbfd70d9d 0xbfd70da0

    The following is the correct way with out '&'.
    Code:
    int *p;
    printf("%p",p); // No need for & since it is a pointer
    Last edited by karthigayan; 06-03-2010 at 07:47 AM.

  3. #3
    Registered User
    Join Date
    Jun 2010
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    3
    Quote Originally Posted by karthigayan View Post
    Actually you have to print the address , is it .If you want the print the address of the variables you have to use '&' since you are trying to get the variables address.If you give the pointer there there there is no need for '&'.

    Code:
    #include <stdio.h>
    main()
    {
      int i;
      char a[]="abc";
      printf("%p %p\n",&a[1],&i);  // use &
      i=a[1];
    
      printf("%c\n",a[1]);
      printf("%c\n",i);
      printf("%d\n",i);
      printf("%p %p\n",&a[1],&i);  //use &
    }
    Result is,

    Code:
    0xbfd70d9d 0xbfd70da0
    b
    b
    98
    0xbfd70d9d 0xbfd70da0

    The following is the correct way with out '&'.
    Code:
    int *p;
    printf("%p",p); // No need for & since it is a pointer
    I think I know what you mean.
    But at first I did't add & in front of the variable, it also worked. what's that address?

  4. #4
    Registered User
    Join Date
    Jun 2010
    Posts
    3
    I totally understand.
    if i don't add & in front of the variable, it prints the value of this variable.
    and 62 is binary, change it into decimal. it's 98, it's 'b'.
    thank you so much.

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