Volatile Struct Semantics

c, struct, volatile

Solution

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

First off, `volatile uint8_t *foo;` tells 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;

Problem

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.

Original source