Why is it not possible to implicitly convert char to const char

Question

I have a code which does more or less:

bar(const char**)
{
// stuff
}

foo(char** arr)
{
  bar(arr);
}

The compiler notifies me that i am doing an invalid conversion from ‘char**’ to ‘const char**’. While I know what it means, I fail to understand why is it not allowed. It is perfectly ok to pass char* where const char* is needed.

I mean, in the function signature of `bar()` you have `const char***`, there's 3 asterisks. Didn't you mean to have two perhaps? — Andy Prowl, Mar 29 '13 at 13:10
It's nothing to do with the `const`, it's the extra `*` as Andy Prowl suggested. — Jonathan Potter, Mar 29 '13 at 13:11
One of the many duplicates. See [comp.lang.c FAQ list · Question 11.10 - Why can't I pass a char ** to a function which expects a const char **?](http://c-faq.com/ansi/constmismatch.html). — Alexey Frunze, Mar 29 '13 at 13:15
While I agree that the problem is the same as in the submitted duplicate, I would suggest leaving the question and posting an answer. I did not manage to find the resolution to my problem in google and I believe that the question can help other people — Dariusz, Mar 29 '13 at 13:19
@PeteBecker you edited the question - but the original was better. Not only can't you implicitly convert char** to const char**, you can't do it explicitly in C++. Please revert. — Dariusz, Mar 29 '13 at 14:16
@Dariusz - you can force the conversion with a `const_cast`. — Pete Becker, Mar 29 '13 at 14:25
@Dariusz: well you can do it explicitly (e.g. `const_cast`), and Pete knows a bit about what you can do, being the one who penned the current standard. u now. so -- i was too tired to notice when i edited the title, but Pete fixed it. — Cheers and hth. - Alf, Mar 29 '13 at 14:25

Cheers and hth. - Alf · Accepted Answer · 2013-03-29T13:26:07.387

1

because if it was allowed then you could inadvertently change something that was declared const.

here's a concrete abuse example, if the rules allowed this:

char const* s = "a literal, very const";

bar(const char** pp )
{
    *pp = s;
}

foo(char** arr)
{
  bar(arr);
  char* unconsted_s = *arr;
  unconsted_s[0] = 'X';
}

This is also a FAQ. It’s often a good idea to check the FAQ (or just google) before asking.

edited Mar 29 '13 at 13:26

answered Mar 29 '13 at 13:19

Cheers and hth. - Alf

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score 0 · Answer 2 · answered Mar 29 '13 at 13:26

It can lead to similar scenario if its allowed.

void foo(int** arr)
{
    /*Can freely modify value at location pointed by a*/
   **arr = 6;
}


int main()
{
 /*Pointer to constant*/
 int i = 5;

 const int* a =&a;

 /*Not a pointer to constant*/
 int** aa= &a;

 foo(aa);

return 0;
}

Why is it not possible to implicitly convert char** to const char **

2 Answers2

Why is it not possible to implicitly convert char to const char