Setting a pointer variable to multiple values

Question

I am working on code which uses a custom linked list class. The list class has the function:

void linkedList::expire(Interval *interval, int64 currentDt)
{
    node *t = head, *d;
    while ( t != NULL )
    {
        if ( t->addedDt < currentDt - ( interval->time + (((long long int)interval->month)*30*24*3600*1000000) ) )
        {
            // this node is older than the expiration and must be deleted
            d = t;
            t = t->next;

            if ( head == d )
                 head = t;

            if ( current == d )
                 current = t;

            if ( tail == d )
                 tail = NULL;

             nodes--;
             //printf("Expired %d: %s\n", d->key, d->value);
             delete d;
         }
         else
         {
            t = t->next;
         }
     }
}

What I don't understand is the first line of code in the function:

node *t = head, *d;

How is it that this code compiles? How can you assign two values to a single variable, or is this some shorthand shortcut? head is a member variable of type *node, but d isn't found anywhere else.

score 4 · Accepted Answer · edited May 23 '17 at 12:23

4

These are two definitions, not the comma operator¹. They are equivalent to

node* t = head;
node* d;

¹ the comma operator has the lowest precedence of all operators in C++, so invoking it requires parantheses:

node* t = (head, *d);

This would work properly if d was of type node**.

edited May 23 '17 at 12:23

Community

1
1

answered Oct 21 '15 at 13:06

cadaniluk

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score 0 · Answer 2 · answered Oct 21 '15 at 13:55

Generally in c++ you can list multiple definitions separating them with a comma:

int a,b,c,d;

will define 4 integers. The danger is that pointers are not handled in a way that might be obvious:

int* a,b,c,d;

Will declare a to be a pointer to an int, and the remaining will just be ints. Hence the not-uncommon practice of declaring pointers in the style:

int *a, *b;

which declares two integer pointers.

Setting a pointer variable to multiple values

2 Answers2