Question about pointers

This is a discussion on Question about pointers within the C Programming forums, part of the General Programming Boards category; today i was trying some code in pointers whish really amaze me for incrementing but some variable i get weird ...

  1. #1
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    Question about pointers

    today i was trying some code in pointers whish really amaze me for incrementing but some variable i get weird results
    Code:
    #include <stdio.h>
    int increment_it(int *num)
    {
        int result;
        (*num)++;//this works
        *num=*num+1;//works
    #ifdef USE_NON_WORKING
        *num++;//this doesnt
    #endif
        return result;
    }
    int main(void)
    {
        int x=0;
        increment_it(&x);
        printf("%d",x);
        getchar();
        return 0;
    }
    first i thought *num++; would just increment the address itself but it doesnt even increment anything so any idea what it doess coz this is really weird.

  2. #2
    ATH0 quzah's Avatar
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    It dereferences the pointer. Then it increments the pointer.


    Quzah.
    Hope is the first step on the road to disappointment.

  3. #3
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    Quote Originally Posted by quzah View Post
    It dereferences the pointer. Then it increments the pointer.
    ...and num is a local variable, so you won't notice any change in main().

    *num++ increments num, not *num. Have a look at the operator precedence rules.

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    Philip
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  4. #4
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    yah i know i made before that return result=*num; i just forgot to put it. but it
    Code:
    #include <stdio.h>
    int increment_it(int *num)
    {
        int result;
        (*num)++;//this works
        *num=*num+1;//works
    #ifdef USE_NON_WORKING
        *num++;//this doesnt
    #endif
        return result=*num;
    }
    int main(void)
    {
        int x=0;
        increment_it(&x);
        printf("%d",x);
        getchar();
        return 0;
    }
    but isnt *num++ same as *num=*num+1; ?

  5. #5
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    Quote Originally Posted by Snafuist View Post
    ...and num is a local variable, so you won't notice any change in main().

    *num++ increments num, not *num. Have a look at the operator precedence rules.

    Greets,
    Philip
    num is a prameter in the function its not a local variable

  6. #6
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    Quote Originally Posted by elwad View Post
    num is a prameter in the function its not a local variable
    Yes, but it has the same effect - it is a local copy of the variable used in the caller.

    --
    Mats
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  7. #7
    C++ Witch laserlight's Avatar
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    Quote Originally Posted by elwad
    but isnt *num++ same as *num=*num+1; ?
    No, as has been pointed out in this thread, it has net effect equivalent to:
    Code:
    *num;
    num++;
    Quote Originally Posted by elwad
    num is a prameter in the function its not a local variable
    A parameter is a local variable.
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  8. #8
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    yah i forgot x=increment_it(&x);
    Code:
    #include <stdio.h>
    int increment_it(int *num)
    {
        int result;
        (*num)++;//this will increment it to 1
        *num=*num+1;//this will increment it to 1
    #ifdef USE_NON_WORKING
        *num++;//this wont increment it to 1
    #endif
        return result=*num;
    }
    int main(void)
    {
        int x=0;
        x=increment_it(&x);
        printf("%d",x);//result 2 coz i increment it twice
        getchar();
        return 0;
    }
    but why does *num++; deference is since i alrdy did in my function x=increment_it(&x); ?

  9. #9
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    1. You do not need the variable result - all you do with it is assign it a value in the same statement as the return statement.

    2. *num++ does a dereference because you have * in the statement. It does this first, then increments num to point to the next integer element after what it originally pointed to. Since you are not actually using the dereferenced value, the compiler just throws it away again (it may not even dereference it at all if you enable enough optimization in the compiler).

    3. The purpose of calling increment_it() with a pointer value seems lost if you also return the new value and assign it to the original value passed in. After all, the function itself has already modified the value pointed to by the pointer.

    --
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  10. #10
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    oh i get it now also yah your right i didnt even need to return anything since its alrdy incremented it

  11. #11
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    I am finding little hard to comprehend things

    Hope i am not hijacking the thread.

    But in the following case:

    Code:
    int main()
    {
       int i[] = {3, 5}; 
       int *p = i; 
       int j = --*p++; 
       
       printf("j = %d, p= %d\n\n", j,*p);
       //system("pause");
       return 0;
    }
    Things look little different

    j = 2 and *p = 5

    They say that Postfix (++ and --)has an associativity of left to right ,
    but at the same time we can see a Prefix associated above, having an associativity of right to left.

    Can explain how to comprehend the above line of code. how does the compiler preceive the above statement.

    Thanks to all in advance

  12. #12
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    can someone explain

  13. #13
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    If we split it and add parenthesis to show what is going on:

    Code:
       int *p = i; 
       int j = --(*p);
       p++;
    So, j is the *p, which is the first element of array i, after it has had 1 subtracted from it (the -- part does that).
    Then p is incremented.

    Writing code like this is VERY bad, and it's unlikely to give any better code generated by the compiler than splitting it to several steps [which then becomes much easier to understand]. Using parenthesis to explain what you are doing, etc, will also help.

    --
    Mats
    Compilers can produce warnings - make the compiler programmers happy: Use them!
    Please don't PM me for help - and no, I don't do help over instant messengers.

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