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Pointers

Discussion in 'C' started by swapnaoe, Mar 12, 2007.

  1. swapnaoe

    swapnaoe New Member

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    Hi ALL,
    When I run the follwoing code, I get the follwoing
    1245048
    1245044
    1245052
    6
    I tried changing the value of 'a' and the last line of the code, the printf prints whatever is assigned to 'a'. Please explain me this behaviour.
    Code:
    void main()
    {
    	int a=6;
    	int b=1024;
    	int *swap=(&b-1);
    	printf("%u\n",&b);
    	printf("%u\n",swap);
    	swap+=2;
    	printf("%u\n",swap);
    	printf("%d\n",*swap);
    }
    Regards,
    Swapna
     
    Last edited by a moderator: Mar 12, 2007
  2. shabbir

    shabbir Administrator Staff Member

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    You are printing the address of b which is in your system 1245048 and swap which is 4 bytes less which should be 1245044 and swap after adding 2 should be 1245044 + 2 = 1245052 which is what is expected.
     
  3. swapnaoe

    swapnaoe New Member

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    Yes u r right.. then
    what is *swap in the last line???
    The value present in 1245052 at that point of time.. right???
    If u can run the program in ur system.. u can find that, the value is nothing but 'a', in this case 6.
    If u change the value of 'a' then the output also changes..
    Any idea fro this type of behaviour??
     
  4. DaWei

    DaWei New Member

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    swap is set to the address of the integer before b (&b minus 1), in this case, a. Therefore, when you print what swap points to (a), you get the contents (a). This only works because your machine happens to have a common implementation of the stack. There is no requirement of the language that requires it to be so.
     
  5. swapnaoe

    swapnaoe New Member

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    Mr DaWei
    Let us consider the follwoing example:
    &a=1245044
    &b=1245048
    Now,
    swap=&b-1, swap=1245044(According to ur explanation)
    swap+=2, swap=swap+2, swap=1245052(if int occupies 4 bytes)
    *swap= contents(1245052), swap= unknown value not 'a'.
    Please correct me if i am wrong

    Regards,
    Swapna
     
  6. wrecker

    wrecker New Member

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    Quite simple, should be no confusion. Just pointer manipulation. Understand whats actually procedure of pointer storage and you will make it. Or take a K&R and make your task simple.....
     
  7. swapnaoe

    swapnaoe New Member

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    Wrecker,
    Could you please go through the discussion i had with Dawei!!!
    Swapna
     
  8. DaWei

    DaWei New Member

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    It's almost certainly the case that your system is implemented in such a way that locals and automatic variables are allocated on the stack, and this in a system, such as the x86, where the stack grows downward. Presuming that 'a' is located at 1245412, in your case, then the variables are at the following addresses:
    Code:
    1245412 a	6
    1245408 b	1024
    1245404 swap	1245404 (&b - 1*sizeof int)
    
    swap then becomes, after += 2*sizeof int, 1245404 + 8, 1245412, which is the address of a
    when dereferenced, then, the value of a (6) is printed. Your number, 1245052, seems to be a typo.

    Again, this is not only implementation dependent, hardware-wise, but also, software-wise. Exercises like this are only useful when produced on the same system with the same OS and the same compiler. Just a caveat.
     
  9. swapnaoe

    swapnaoe New Member

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    Thank you guys.. its clear now.. :)
    Swapna
     

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