Using C array to store structs and then trying to read the array using pointer to the first element

Question

So I'm using an array to store sructs (struct is Path). Defined as such:

Path **array = malloc(items * sizeof(Path *));

Then each element is alloc'd:

for (i=0;i<items;i++) 
{
    array[i] = malloc(sizeof(Path *));
}

Later in the program variables are updated as such:

Path *path = array[id];
path->next_hop=-1;
path->weight=INT_MAX;

I then have a separate function to print the values of the array, taking the pointer to the first item in the array (array[0]) and the size of array as parameters, declared as such:

void PrintTable(Path *paths, int n)
{
    if(paths == NULL || n <= 0)
        return;
    printf("%s %8s %8s\n", "Net ID", "Weight", "Next Hop");
    for(int i = 0; i < n; i++)
    {
        Path *p = paths + i;
        printf("%d %8.2f %8d\n", vertices[i],p->weight, p->next_hop);
    }

}

When I was debugging I noticed that the stuct pointers seem to be in memory addresses with locations 0x000000000020 apart while in the loop the memory addresses being read are 0x000000000010 apart. Therefore by making this change:

Path *p = paths + i;

to

Path *p = paths + (2 * i);

It seems to work. Why is this? And how do I fix it?

Shouldn't Path *paths be Path *paths[]?

Barmar
– Barmar

2023-01-11 19:53:17 +00:00
Commented Jan 11, 2023 at 19:53 — Barmar
– Barmar, Commented Jan 11, 2023 at 19:53

Barmar · Accepted Answer · 2023-01-11 19:56:14Z

2

You have an array of pointers to Path, not an array of Path. So the argument to PrintTable should be Path **paths or Path *paths[], not Path *paths.

Then you can simply index it rather than using pointer arithmetic in the loop.

void PrintTable(Path *paths[], int n)
{
    if(paths == NULL || n <= 0)
        return;
    printf("%s %8s %8s\n", "Net ID", "Weight", "Next Hop");
    for(int i = 0; i < n; i++)
    {
        printf("%d %8.2f %8d\n", vertices[i], paths[i]->weight, paths[i]->next_hop);
    }

}

answered Jan 11, 2023 at 19:56

Barmar

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6 Comments

Eric Postpischil Over a year ago

Do not overlook array[i] = malloc(sizeof(Path *));, where the type of array[i] is Path *, so the size should be of *array[i] or Path.

Nat569On Over a year ago

Thanks but is there a way to do it without modifying the PrintTable function? (I am not allowed to change it)

Barmar Over a year ago

@Nat569On You could change your array to be Path *array instead of Path **array.

Nat569On Over a year ago

@Barmar but isn't that for a single struct pointer, not an array of struct pointers?

Barmar Over a year ago

Why use an array of struct pointers instead of an array of structs? That's what PrintTable expects. If you don't want to modify the PrintTable function, you have to make the data match what it expects.

|

pmacfarlane · Accepted Answer · 2023-01-11 20:03:40Z

0

You seem to be doing two contradictory things. You have this:

Path **array = malloc(items * sizeof(Path *));

which does not declare an array of Path structures. It declares an array of pointers to Path structures.

You then have this:

for (i=0;i<items;i++) 
{
    array[i] = malloc(Path *); // BUG
}

in an attempt to allocate all the structure pointers you need for your array. malloc(Path *) is nonsense, and I'm amazed your compiler let you do that. You probably want:

    array[i] = malloc(sizeof(Path));

You then have things like this, which are reasonable for an array of pointers to Paths:

Path *path = array[id];
path->next_hop=-1;
path->weight=INT_MAX;

But your PrintTable() function treats your array as an array of Paths, not an array of pointer to Paths.

@Barmar has just posted the fix for your PrintTable() function, which I won't repeat here.

Your compiler probably warned you about some of these issues.

answered Jan 11, 2023 at 20:03

pmacfarlane

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1 Comment

Nat569On Over a year ago

Yes that was a mistake. I have sizeof(Path *) in my code.

user17592432 · Accepted Answer · 2023-01-11 20:46:52Z

0

One of the problems is the tendency to "know" what you're doing and tell the compiler that you want something that doesn't make sense...

array[i] = malloc(sizeof(Path *)); // space for a pointer where a struct is needed

Get used to a better syntax:

array[i] = malloc( sizeof *array[0] );

Allocate enough space to store an instance of what the pointer will be pointing at.

(Much less maintenance and fewer bugs caused by naming an incorrect datatype.)

answered Jan 11, 2023 at 20:46

user17592432

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