"duplicating" an array of pointers

This is a discussion on "duplicating" an array of pointers within the C Programming forums, part of the General Programming Boards category; I have an array of pointers to a struct called p. Code: typedef struct node { int exp; float koef; ...

  1. #1
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    "duplicating" an array of pointers

    I have an array of pointers to a struct called p.

    Code:
    typedef struct node
    	{
    		int exp;
    		float koef;
    		struct node *next;
    	} monomial;
    
    monomial *p[5];
    Some of the pointers are NULL some aren't and have value. I would like to have an array q which has the exact same values as p does, but not in a way that q[i] points to the same memory location as p[i]. I want q[i]->koef and q[i]->exp to have the same value as p[i]->koef and p[i]->exp but not to be in the same memory space.

    I hope you understand me

  2. #2
    C++まいる!Cをこわせ! Elysia's Avatar
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    I suppose you are talking about a deep copy.
    So allocate memory to q using malloc, then use assignments in a loop (or memcpy).

    q[i] = malloc(...);
    memcpy(q, p, size_of_p);
    Quote Originally Posted by Adak View Post
    io.h certainly IS included in some modern compilers. It is no longer part of the standard for C, but it is nevertheless, included in the very latest Pelles C versions.
    Quote Originally Posted by Salem View Post
    You mean it's included as a crutch to help ancient programmers limp along without them having to relearn too much.

    Outside of your DOS world, your header file is meaningless.

  3. #3
    Woof, woof! zacs7's Avatar
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    Just chuck it all in a loop.

    I'd recommend allocating as you go, otherwise you'll have to keep track of where you started from if you want to free the entire array in one go.

  4. #4
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    thanks. works like a charm. put it in a loop

    Code:
    for (i=0; i<=n; i++) 
    	{
    		q[i] = malloc(sizeof(monomial));
    		q[i] = p[i];
    	}

  5. #5
    C++まいる!Cをこわせ! Elysia's Avatar
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    This is wrong. Remember that q and p are arrays of pointers. You're just overwriting q's newly acquired memory with the memory address of p.
    It should be
    *q[i] = *p[i];
    (To dereference each pointer in the array and copy its contents.)
    Quote Originally Posted by Adak View Post
    io.h certainly IS included in some modern compilers. It is no longer part of the standard for C, but it is nevertheless, included in the very latest Pelles C versions.
    Quote Originally Posted by Salem View Post
    You mean it's included as a crutch to help ancient programmers limp along without them having to relearn too much.

    Outside of your DOS world, your header file is meaningless.

  6. #6
    Woof, woof! zacs7's Avatar
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    > q[i] = p[i];

    I wouldn't be doing that. Copy each element manually, plus g[i]->next is the same as p[i]->next (not q[i]->next :-D

  7. #7
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    oops. thanks.

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