Function Call

This is a discussion on Function Call within the C Programming forums, part of the General Programming Boards category; function prototype void test (const struct addr*); calling the function test( (struct addr *) &something); The prototype is expecting a ...

  1. #1
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    Function Call

    function prototype

    void test (const struct addr*);


    calling the function
    test( (struct addr *) &something);

    The prototype is expecting a parameter of pointer to the structure addr,
    but the calling function is supplying the address of the pointer of the structure addr.

    is this correct?
    or test((struct addr*) something) is the right way?

  2. #2
    Programming Wraith GReaper's Avatar
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    Is 'something' a pointer to struct or a struct?
    Devoted my life to programming...

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    something maybe is an address of a pointer to the struct.

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    Quote Originally Posted by lilzz View Post
    something maybe is an address of a pointer to the struct.
    So if you don't really know what something is, you can't properly pass it around. If something is an address of a pointer to the struct, here is how I would expect it to be declared/used:
    Code:
    struct addr bar;  // a plain old struct
    struct addr *foo = &bar;  // foo is a pointer to a struct
    struct addr **something = &foo;  // something contains the address of a pointer to a struct
    ...
    test(*something);  // call test with a pointer to a struct (dereferencing the double pointer to a single pointer)
    test(foo);  // no need to do anything, foo is the proper type
    test(&bar);  // bar is a struct, we need the address of (a pointer to) the struct

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    Quote Originally Posted by anduril462 View Post
    Code:
    struct addr bar;  // a plain old struct
    struct addr *foo = &bar;  // foo is a pointer to a struct
    struct addr **something = &foo;  // something contains the address of a pointer to a struct
    ...
    test(*something);  // call test with a pointer to a struct (dereferencing the double pointer to a single pointer)
    test(foo);  // no need to do anything, foo is the proper type
    test(&bar);  // bar is a struct, we need the address of (a pointer to) the struct
    is there a case for
    test(&foo); ?
    because I have seen int * x; some_function(&x) used in a program.

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    Quote Originally Posted by lilzz View Post
    is there a case for
    test(&foo); ?
    because I have seen int * x; some_function(&x) used in a program.
    Yes, if you want to change the address that x points to from within some_function, but have that change be visible outside some_function. E.g.:
    Code:
    void alloc_something(struct something **x)
    {
        *x = malloc(sizeof(**x));  // allocate a struct something, store it's address in *x
    }
    
    struct something *other_alloc_something(void)
    {
        return malloc(sizeof(struct something));  // allocate a struct something, return it's address
    }
    
    int main(void)
    {
        struct something *x, *y;
    
        alloc_something(&x);  // allocates a struct something and puts the address in x
        y = other_alloc_something();  // also allocates a struct something and puts the address in y
    
        free(x);
        free(y);
    }
    EDIT: I prefer the other_alloc_something style, but either way you get the same results. In the first method, you have to check if x is NULL before you can safely do *x.
    Last edited by anduril462; 02-23-2011 at 03:47 PM.

  7. #7
    Registered User TheBigH's Avatar
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    Quote Originally Posted by lilzz View Post
    is there a case for
    test(&foo); ?
    because I have seen int * x; some_function(&x) used in a program.
    Yes, there is. Calling a function with a set of arguments creates COPIES of the arguments that the function then manipulates. Then, when the function ends, the copies are destroyed. So if you wanted to write a program that swaps two numbers and you wrote

    Code:
    void swap( int a, int b ) {
       int temp;
       temp = a;
       a = b;
       b = temp;
       return;
    }
    you would find that nothing happened, because the function swapped COPIES of a and b but left the original a and b untouched. What you would need to do is pass the function the addresses of a and b so that the function can modify the information at those memory locations; those changes will remain in place when you return from the function. Behold:

    Code:
    void swap( int *a, int *b ) {
       int temp;
       temp = *a;
       *a = *b;
       *b = temp;
       return;
    }

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