Initializing struct in C

Question

OK, this is the definition of the struct:

typedef struct { int first; int last; int count; char * Array [50]; } queue;

and I use another function to initialize it

void initialize(queue * ptr){ ptr=malloc(sizeof(queue)); ptr->first=0; ptr->last=0; ptr->count=0; }

Then I use printf to print out first, last and count. All three should be zero. However, what I actually get is, count is 0 as I expected, but first&last are two very large strange numbers and they change every time I run the program. Can anybody tell me what's wrong here? Thank you.

How are you calling your function? Can you post the code for that too? — Mark Byers
– Mark Byers, Commented Apr 27, 2012 at 22:44

Mark Byers · Accepted Answer · 2012-04-27 22:42:49Z

You are passing your pointer by value. The function changes a copy of the argument it receives, but the caller's pointer is not modified and is probably unintialized.

You need to change your function to take a queue** and pass the address of the pointer you want to initialize.

Alternatively you could return a pointer instead of passing it in as an argument. This is a simpler approach.

justin · Accepted Answer · 2012-04-27 22:46:19Z

Given:

void initialize(queue * ptr);

Pass it like this:

queue q; // caller allocates a queue initialize(&q); // now q is initialized

Also, it's allocated by the caller -- don't malloc it.

// bad void initialize_bad(queue * ptr){ ptr=malloc(sizeof(queue)); << nope. already created by the caller. this is a leak ptr->first=0; ptr->last=0; ptr->count=0; } // good void initialize_good(queue * ptr){ ptr->first=0; ptr->last=0; ptr->count=0; // ptr->Array= ???; }

If you prefer to malloc it, then consider returning a new allocation by using this approach:

queue* NewQueue() { // calloc actually works in this case: queue* ptr = (queue*)calloc(1, sizeof(queue)); // init here return ptr; }

Ultimately, what is 'wrong' is that your implementation passes a pointer by value, immediately reassigns the pointer to a new malloc'ed allocation, initializes the malloc'ed region as desired, without ever modifying the argument, and introducing a leak.

Heath Hunnicutt · Accepted Answer · 2012-04-27 22:56:21Z

Here is the smallest alteration to your program which should correct your problem:

void initialize(queue * * pptr) { queue * ptr; ptr=malloc(sizeof(queue)); if (ptr) { ptr->first=0; ptr->last=0; ptr->count=0; } /* The assignment on the next line assigns the newly allocated pointer into memory which the caller can access -- because the caller gave us the address of (i.e. pointer to) such memory in the parameter pptr. */ *pptr = ptr; }

The most important change is to pass a queue ** to your initialize function -- otherwise you are changing a copy of the queue * supplied as the actual parameter when you call initialize(). By passing a pointer to the pointer, you can access the original variable which stores the pointer in your caller.

I couldn't resist and also added a check for NULL returned from malloc(). That doesn't address your problem, but I couldn't bring myself to post code that didn't do it.

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