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Home/ Questions/Q 6709563
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Editorial Team
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Editorial Team
Asked: May 26, 20262026-05-26T07:53:49+00:00 2026-05-26T07:53:49+00:00

I’ve been trying to implement a bst, in C. I think I’m almost there,

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I’ve been trying to implement a bst, in C. I think I’m almost there, but in my add node function, I create a temporary node called current to store the current node which is visited in the tree. Then when I modify the current node, my orignal pointer is not modified after the function finishes.

I’ve read up about this, and I think I may need a pointer of a pointer, but I stil don’t quite know how to updated original struct.

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  1. Editorial Team
    Editorial Team
    2026-05-26T07:53:49+00:00Added an answer on May 26, 2026 at 7:53 am

    You are right that the problem has to do with the pointer to a pointer in bstlist_add. Here’s an example that should help you figure out what you need to change in your code.

    int a=10;
    int b=20;
    
    void noChange(int * pSomeInt);
    void change(int ** ppSomeInt);
    
    int main(int argc,char * argv[])
    {
        int * pMainInt=&a;
    
        noChange(pMainInt);
        //pMainInt will still point to a
    
        //since the parameter to change is int **, we have to use & here
        change(&pMainInt);
        //pMainInt now points to b
    
        return 0;
    }
    
    void noChange(int * pSomeInt)
    {
        //while pSomeInt is a pointer, it is a copy of pMainInt, not a pointer to it 
        //so this creates a pointer to the parameter, pSomeInt, itself
        int ** ppSomeInt=&pSomeInt;
    
        //so this changes the parameter, pSomeInt
        *ppSomeInt=&b;
    }
    
    void change(int ** ppSomeInt)
    {
        //ppSomeInt is a pointer to pMainInt, which is itself an int *
        //so *ppSomeInt is pMainInt and not a copy of it
        *ppSomeInt=&b;
    }
    
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