Volatile Struct Semantics

Question

Is it sufficient to declare an instance of a structure-typed variable as volatile (if its fields are accessed in re-entrant code), or must one declare specific fields of the structure as volatile?

Phrased differently, what are the semantic differences (if any) between:

typdef struct {
  uint8_t bar;
} foo_t;

volatile foo_t foo_inst;

and

typedef struct{
  volatile uint8_t bar;
} foo_t;

foo_t foo_inst;

I recognize that declaring a pointer-typed variable as volatile (e.g. volatile uint8_t * foo) merely informs the compiler that the address pointed-to by foo may change, while making no statement about the values pointed to by foo. It is unclear to me whether an analogy holds for structure-typed variables.

Answer 1

In your example, the two are the same. But the issues revolve around pointers.

First off, volatile uint8_t *foo; does tell the compiler the memory being pointed to is volatile. If you want to mark the pointer itself as volatile, you would need to do uint8_t * volatile foo;

And that is where you get to the main differences between marking the struct as volatile vs marking individual fields. If you had:

typedef struct
{
    uint8_t *field;
} foo;

volatile foo f;

That would act like:

typedef struct
{
    uint8_t * volatile field;
} foo;

and not like:

typedef struct
{
    volatile uint8_t *field;
} foo;

Answer 2

if you declare a structure with volatile then all its members will also be volatile