Hi,

my attention was directed at the gprs_shift_v_fixed() function:

int gprs_shift_v_fixed(uint8_t **data, size_t *data_len, size_t len, uint8_t **value)

It advances the *data pointer and returns a *value pointer. So it could technically be consting much more aggressively. However, the compiler warnings confuse me.

I've created this test code to illustrate; the ints should be treated read-only by func():

void func(const int **yyy) { (*yyy) ++; }

int main(void) { int y[2] = {23, 42}; // mutable "by coincidence" int *yy = y; func(&yy);

return *yy; // returns 42 }

The idea is that yy is a pointer to a mutable int. func() expects a const int, but what harm could be done by passing a mutable-int pointer as an immutable-int pointer. Apparently this harm:

check.c: In function ‘main’: check.c:10:14: warning: passing argument 1 of ‘func’ from incompatible pointer type func(&yy); ^ check.c:1:6: note: expected ‘const int **’ but argument is of type ‘int **’ void func(const int **yyy)

Changing yy to a pointer-to-const-int clears the warning:

const int *yy = y; func(&yy);

I'm expecting the const to apply to the int pointed at, so I should be able to modify the pointer while the compiler shouldn't waste a thought at whether the int pointed at is const or not. func() is stricter, not less strict than y[].

To further illustrate, this would make the pointer address itself immutable, which is not what I want:

int * const yy = y;

I just want to declare the underlying int data to be unchangeable by func(), not caring whether the caller passes a const int or a mutable int buffer.

Is the compiler merely nitpicking or am I getting something wrong fundamentally?

Thanks!

~Neels