Type qualifiers (const) and complex declarations

Question

Say we have such declaration :

int **const*** k;

Then, it could be well translated literally (according to cdecl.org) to

declare k as pointer to pointer to pointer to const pointer to pointer to int

However, still, I'm not sure to understand what it does not permit? Which operation does it restrict? Could we still do

(***k)++

In other words, what is the effect of adding const there?

And... Same question for

int *const**const* k;

Which difference would it make?

Thanks!

It's good for a question, but don't do weird things in real life, everyone will hate you. — Maroun, Jan 08 '14 at 15:27
There's a quote by someone: "The C syntax is a boon to teachers of trivia and people who like to make fun of the language, but not actually a problem". — Kerrek SB, Jan 08 '14 at 15:30
I was waiting for someone to comment! Actually, I'm making a kind of C interpreter. Hence why I must be able to reproduct accurately even the most unusual types! Else, I wouldn't even bother asking. — Yannick, Jan 08 '14 at 15:30

haccks · Accepted Answer · 2014-01-09T11:14:15.750

4

still, I'm not sure to understand what it does not permit?

It does not permit the modification that ***k points to.

Could we still do

(***k)++

No. You can't. This is because ***K is a pointer to pointer to pointer to const pointer and you can't modify it. But yes, modification to K, *k, **k is valid.

For the sake of convenience, you can understand this as follows:

int *const k;     // k is a const pointer to integer. No modification to k.  
int *const *k;    // *k is a const pointer to integer. No modification to *k.  
int *const **k;   // **k is a const pointer to integer. No modification to **k.  
int *const ***k;  // ***k is a const pointer to integer. No modification to ***k.  
int **const ***k; // ***k is a const pointer to integer. No modification to ***k.

edited Jan 09 '14 at 11:14

answered Jan 08 '14 at 15:29

haccks

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Aren't `const` protecting `int **`? – Maroun Jan 08 '14 at 15:40
No. In these cases they are not. – haccks Jan 08 '14 at 15:41
I thought `const` protects its left side, unless there is nothing to its left, then and only then, it protects its right side.. Why this doesn't apply here? – Maroun Jan 08 '14 at 15:42
What do you think, what is the difference between `int const *p` and `int *const p? – haccks Jan 08 '14 at 15:43
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Now it's clear.. Two years without C++ is hard.. I'm refreshing my knowledge nowadays, +1 nice explanation my friend. – Maroun Jan 08 '14 at 15:44
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Do you remember? Once you told me to learn about pointers and I replied to you that "I am a bit afraid of this pointer thing" :) – haccks Jan 08 '14 at 15:46
I think I'm starting to get the point. So with "int \*const**const* k", we could do (**k)++, (****k)++, but not (*k)++ and (***k)++ right? – Yannick Jan 08 '14 at 15:47
@haccks Now I do remember... :)) – Maroun Jan 08 '14 at 15:49
@ᴍarounᴍaroun; Oh! That's nice. Good memory. I think its about 4 to 5 months ago. – haccks Jan 08 '14 at 15:49
@haccks I'm looking for that question now :D – Maroun Jan 08 '14 at 15:51
Me too :D. Lets find that question :) – haccks Jan 08 '14 at 15:54
@Yannick; You got it now. Congratulations! :) – haccks Jan 08 '14 at 15:55
Okie. Just a quick and dirty question too. For example, char ((* const k)[5][6]). Basically, I could modify here the content of the array to which 'k' points to, but I could not increment k, is that right? In other words, const only applies to the element on his left, righty? Thanks, a lot! – Yannick Jan 08 '14 at 15:57
@Yannick; Yes. You are going right. Now my tern, What about `const int *p` ? :) – haccks Jan 08 '14 at 16:09
@Yannick; Are you there? Answer my question. If not then I will explain you (do not hesitate to clear your questions). – haccks Jan 08 '14 at 16:16
Well, I would say you cannot modify on what p points to? in other words, 'ints' are not modifiable but p is? – Yannick Jan 08 '14 at 16:26
@Yannick; Excellent. Now you understand it very well. – haccks Jan 08 '14 at 16:27
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Well, chosen answer for sure :) – Yannick Jan 08 '14 at 16:29

Type qualifiers (const) and complex declarations

1 Answers1