"duplicating" an array of pointers

budala · 08-13-2009, 06:01 AM

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

Elysia · 08-13-2009, 06:13 AM

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);

zacs7 · 08-13-2009, 06:24 AM

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.

budala · 08-13-2009, 06:46 AM

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];
	}

Elysia · 08-13-2009, 06:49 AM

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.)

zacs7 · 08-13-2009, 06:50 AM

> 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

budala · 08-13-2009, 08:16 AM

oops. thanks.

08-13-2009, 06:01 AM	#1
budala Registered User Join Date: Apr 2009 Posts: 48	"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
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08-13-2009, 06:24 AM	#3
zacs7 Woof, woof! Join Date: Mar 2007 Location: Australia Posts: 3,139	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. __________________ "I.T. gets the chicky-babes" - M. Kelly bakefile \| vim
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08-13-2009, 06:46 AM	#4
budala Registered User Join Date: Apr 2009 Posts: 48	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]; }
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08-13-2009, 06:50 AM	#6
zacs7 Woof, woof! Join Date: Mar 2007 Location: Australia Posts: 3,139	> 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 __________________ "I.T. gets the chicky-babes" - M. Kelly bakefile \| vim
zacs7 is offline

08-13-2009, 08:16 AM	#7
budala Registered User Join Date: Apr 2009 Posts: 48	oops. thanks.
budala is offline

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