How do I modify a pointer that has been passed into a function in C?

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予麋鹿
予麋鹿 2020-11-21 05:25

So, I have some code, kind of like the following, to add a struct to a list of structs:

void barPush(BarList * list,Bar * bar)
{
    // if there is no move t         


        
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  • 2020-11-21 05:58

    Remember, in C, EVERYTHING is passed by value.

    You pass in a pointer to a pointer, like this

    int myFunction(int** param1, int** param2) {
    
    // now I can change the ACTUAL pointer - kind of like passing a pointer by reference 
    
    }
    
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  • 2020-11-21 05:59

    Yes, you have to pass in a pointer to the pointer. C passes arguments by value, not by reference.

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  • 2020-11-21 06:06

    Generic answer: Pass a pointer to the thing you want to change.

    In this case, it would be a pointer to the pointer you want to change.

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  • 2020-11-21 06:06

    This is a classic problem. Either return the allocated node or use a pointer of pointer. In C, you should pass a pointer to a X to a function where you want your X to be modified. In this case, since you want a pointer to be modified, you ought to pass a pointer to a pointer.

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  • 2020-11-21 06:13

    You need to pass in a pointer to a pointer if you want to do this.

    void barPush(BarList ** list,Bar * bar)
    {
        if (list == NULL) return; // need to pass in the pointer to your pointer to your list.
    
        // if there is no move to add, then we are done
        if (bar == NULL) return;
    
        // allocate space for the new node
        BarList * newNode = malloc(sizeof(BarList));
    
        // assign the right values
        newNode->val = bar;
        newNode->nextBar = *list;
    
        // and set the contents of the pointer to the pointer to the head of the list 
        // (ie: the pointer the the head of the list) to the new node.
        *list = newNode; 
    }
    

    Then use it like this:

    BarList * l;
    
    l = EMPTY_LIST;
    barPush(&l,&b1); // b1 and b2 are just Bar's
    barPush(&l,&b2);
    

    Jonathan Leffler suggested returning the new head of the list in the comments:

    BarList *barPush(BarList *list,Bar *bar)
    {
        // if there is no move to add, then we are done - return unmodified list.
        if (bar == NULL) return list;  
    
        // allocate space for the new node
        BarList * newNode = malloc(sizeof(BarList));
    
        // assign the right values
        newNode->val = bar;
        newNode->nextBar = list;
    
        // return the new head of the list.
        return newNode; 
    }
    

    Usage becomes:

    BarList * l;
    
    l = EMPTY_LIST;
    l = barPush(l,&b1); // b1 and b2 are just Bar's
    l = barPush(l,&b2);
    
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