Casting a NULL pointer function argument resulted in violation of Misra Rule 11.3

Question

I define a NULL_PTR as 0U

Then call a function with this NULL_PTR as argument.

read_some_data(2U, (uint8_t *const) NULL_PTR, (uint8_t *const) NULL_PTR);

Called function prototype:

int16_t read_some_data(const uint8_t id,   uint8_t *const data_1, uint8_t *const data_2);

On compilation, Misra raised a rule 11.3 violation error.(A cast should not be performed between a pointer type and an integral type.)

But if I just pass the NULL_PTR as follows, no violation.

read_some_data(2U, NULL_PTR, NULL_PTR);

Which is the better way to do? Suppress Misra 11.3 rule or just pass the NULL_PTR without casting?

Answer 1

I used NULL_PTR in my header file to avoid using IAR internal configuration file yvals.h which defines NULL. But that's not an issue since I may have to use yvals.h later due to other reasons.

Whether using NULL or NULL_PTR, I assume that the general consensus is to pass NULL without casting. My function doesn't have any problem in accepting it. This way, I can avoid suppressing Misra 11.3 rule.

Hope I am proceeding the right way.

Answer 2

What is wrong with the Standard 'NULL'?

Answer 3

Why cast if you can avoid it? Cast always makes code a little bit more dirty and hints that there is something hacky performed.

So I would just pass NULL_PTR without casting. After checking function specification that it can accept NULL_PTR as its second parameter!!!

Answer 4

On compilation, Misra raised a rule 11.3 violation error.

Just to be pedantic, MISRA did not raise a violation error, your compiler raised a MISRA rule violation error.

From what you've posted, I'm not convinced the reported violation is correct...

On the other hand, personally I'd define NULL_PTR as (void *)0u because (strictly speaking) 0 is not a pointer.