Warning about assignment from incompatible pointer type when using pointers and arrays?

Question

For the struct

typedef struct sharedData
{
    sem_t *forks;
}sharedData;

I get a warning when I try to do this:

sharedData sd;
sem_t forks[5];
sd.forks = &forks; // Warning: assignment from incompatible pointer type

Am I misunderstanding or missing something?

templatetypedef · Accepted Answer · 2011-02-04 06:19:35Z

12

The problem is that &forks has type

sem_t (*)[5]

That is, a pointer to an array of five sem_ts. The compiler warning is because sd.forks has type sem_t*, and the two pointer types aren't convertible to one another.

To fix this, just change the assignment to

sd.forks = forks;

Because of C's pointer/array interchangeability, this code will work as intended. It's because forks will be treated as &forks[0], which does have type sem_t *.

answered Feb 4, 2011 at 6:19

templatetypedef

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newrev426 · Accepted Answer · 2013-07-06 20:42:41Z

1

The above is a great explanation of but remember that

sd.forks = forks;

is the same as....

sd.forks = &forks[0];

I like the second one for clarity. If you wanted the pointer to point to the third element...

sd.forks = &forks[2];

answered Jul 6, 2013 at 20:42

newrev426

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