On Wed, Mar 16, 2016 at 11:28:30AM +0100, Harald Welte wrote:

Why does gsm48_hdr_pdisc() not take a 'const struct gsm48_hdr *' argument in the first place? I don't think it modifies the contents of the structure...

Yes, Holger has applied the patch with exactly that change.

Yet this has me confused. A while back I concluded that C doesn't allow passing non-const instances to functions that expect a const arg. That was discussed on this list in January:

http://lists.osmocom.org/pipermail/openbsc/2016-January/001051.html http://lists.osmocom.org/pipermail/openbsc/2016-January/001055.html

How is this different from the conclusion that Jacob confirmed two months ago? Is it int (primitives) vs. struct??

Thanks!

~Neels

On Thu, Mar 17, 2016 at 08:06:32AM +0100, Harald Welte wrote:

This is standard practise, see the definition of memcpy:

void *memcpy(void *dest, const void *src, size_t n);

I'm very familiar with that and want to use this everywhere and always, which is why my conclusion that it apparently didn't work was quite a disappointment. Glad to see that it does work after all...

What bugs me is that I don't understand the apparent corner case where it doesn't work.

...

http://lists.osmocom.org/pipermail/openbsc/2016-January/001051.html http://lists.osmocom.org/pipermail/openbsc/2016-January/001055.html

How is this different from the conclusion that Jacob confirmed two months ago? Is it int (primitives) vs. struct??

from a quick look, the above discussion was about passing the address of a pointer, and of course if you pass that into a function, you expect the function to be able to modify the pointer stored at that address?

First, let's clarify:

const int * const x; the pointer x can't be modified, neither can the int at *x be modified.

const int *x; pointer x can be modified, but the int at *x can't be modified.

const int **x; x can be modified, *x can be modified, but the int at **x can't be modified.

right?

Let's look at a simple example that works without warning: I only want to modify a pointer, not the ints.

const int *func(const int *yy) { return yy + 1; }

int main(void) { int y[2] = {23, 42}; int *yy = y; return *func(yy); // returns 42 }

This works perfectly. I pass a pointer to a mutable int, though the function expects a pointer to a const int. No compiler warning.

And the failing example again; it does the exact same thing, only it returns the modified pointer in an output-argument, using a pointer to a pointer to a const int:

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

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

return *yy; // returns 42 }

And here I get the warning:

cc check.c -o check check.c: In function ‘main’: check.c:11:14: warning: passing argument 1 of ‘func’ from incompatible pointer type func(yyy); ^ check.c:1:6: note: expected ‘const int **’ but argument is of type ‘int **’ void func(const int **yyy) ^

I'm quite happy that it seems to be only a corner case, and I'll be consting aggressively and happily ever after :)

Yet it's a stupid corner case. Would be great if someone had an explanation to this odd gcc behavior.

Maybe it's rather a question for another mailing list, too...

~Neels

On Thu, Mar 17, 2016 at 04:25:48PM +0100, Harald Welte wrote:

Hi Neels,

On Thu, Mar 17, 2016 at 04:02:07PM +0100, Neels Hofmeyr wrote:

...

What bugs me is that I don't understand the apparent corner case where it doesn't work.

I'm sorry, I cannot help you there. My knowledge about the corner cases is limited, and I'm happy to accept it as it hasn't been an issue yet. we rarely use pointers to pointers in the osmocom code, luckily ;)

Call me ignorant, if you'd like :)

Heh, good answer :)

At least I know now that it's a corner case.

~Neels