Call by reference question

This is a discussion on Call by reference question within the C Programming forums, part of the General Programming Boards category; I have a question regarding call by reference - essentially using a function i made and calling it in the ...

  1. #1
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    Call by reference question

    I have a question regarding call by reference - essentially using a function i made and calling it in the main function. I'm passing 2 variables (pointers) into it and the value is copied in the corresponding function but the variable itself is not changed in the calling environment (main).

    ---

    Code:
    void somefunction(arg1, arg2, arg3, arg4, *ans1, *ans2)
    // function does some arithmetic and spits out 2 output values/variables *ans1 and *ans2
    
    ---
    
    NOT SHOWED is the header file that has the function prototype etc
    
    int main(void)
    {
    int = arg1, arg2, arg3, *ans1, **ans11;
    double = arg4, *ans2; **ans22
    
    ans11 = &ans1
    ans22 = &ans2
    
    timediff(arg1, arg2, arg3, arg4, *ans1, *ans2);
    
    printf("answers are %d and %d", ans11, ans22);
    
    return 0;
    }
    ---

    I didn't copy my program in but made a close generic approximation version of it... hopefully that can help a little...

    Thought the best way to do it is try and use a pointer to a pointer...

    Any thoughts?

  2. #2
    C++ Witch laserlight's Avatar
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    What is your question? You might want to show us the smallest and simplest (compilable) program that demonstrates the problem instead of showing remarkably C-like pseudocode.
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  3. #3
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    You can NEVER change anything inside a function that you haven't got the address of. In other words, if what you pass to a function is not a pointer to what you are trying to modify, you will only modify the variable that is local to the function.

    For example:
    pass a variable to modify an integer:
    Code:
    int func(int *p)
    {
       *p = 7;
    }
    
    int main()
    {
    ...
       int x = 3;
       func(&x);
    ...
    }
    Note that the same applies to pointers:
    Code:
    void func2(const char **ptr)
    {
       if (x % 2)   // odd
       {
           *ptr = "Odd";
       }
       else
       {
           *ptr = "Even";
       }
    }
    --
    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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    Quote Originally Posted by matsp View Post
    You can NEVER change anything inside a function that you haven't got the address of. In other words, if what you pass to a function is not a pointer to what you are trying to modify, you will only modify the variable that is local to the function.

    For example:
    pass a variable to modify an integer:
    Code:
    int func(int *p)
    {
       *p = 7;
    }
    
    int main()
    {
    ...
       int x = 3;
       func(&x);
    ...
    }
    Note that the same applies to pointers:
    Code:
    void func2(const char **ptr)
    {
       if (x % 2)   // odd
       {
           *ptr = "Odd";
       }
       else
       {
           *ptr = "Even";
       }
    }
    --
    Mats
    Thank you.

    Are you saying that I am on the right track with a pointer to a pointer?

  5. #5
    C++ Witch laserlight's Avatar
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    Quote Originally Posted by badtwistofate
    Are you saying that I am on the right track with a pointer to a pointer?
    Yes, if your intention is really to change the pointer, not just what it points to.
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