Error Can not convert char to const char

Question

I have following code:-

static char* ListOfStr[] = { "str1", "str2", "str3" };
void Foo(const char** listOfStr)
{
// do something
}

When I call Foo like;

Foo(ListOfStr);

I get Error Can not convert char** to const char** (C2664 - vc++)

I know how to solve the problem using casting or other way around like defining const array at first place.

But isn't it safe to use char** as const char** than why it gives error ? I supposed there should be auto convertion like std::string to const std::string when passing to function. Only the reverse of this cont char** to char** must give the Error without cast.

This is because in modern C++ a `"literal string"` is a `const char *` and not a `char *`. — Sam Varshavchik, Dec 04 '21 at 15:52
@SamVarshavchik The error message seems to indicate *the opposite* problem. — Peter - Reinstate Monica, Dec 04 '21 at 16:04
If you don't get an error on the first line, then this is C. Why have you tagged the question with C++? — 273K, Dec 04 '21 at 16:05
@S.M. Probably every compiler out there permits this legacy initialization with some option; VC++ will need "conformance mode off". This is not the issue here. The issue is that with double pointers certain conversions *towards const* are prohibited. — Peter - Reinstate Monica, Dec 04 '21 at 16:14
@Peter-ReinstateMonica yes, you are right I missed it, I blame Sam ;) — anastaciu, Dec 04 '21 at 16:15
@Peter OP did not indicate this, thus we can assume the default compiler settings used. — 273K, Dec 04 '21 at 16:17
Take a look here https://stackoverflow.com/q/2220916/6865932 — anastaciu, Dec 04 '21 at 16:19
You can declare `Foo` as `Foo(const char* const* listOfStr)`, which avoids the trap in the duplicate, since the passed pointer is no longer mutable so you can't use it to swap in a const for a non-const. — Raymond Chen, Dec 04 '21 at 16:48
@RaymondChen Thanks your way of declaring the `Foo` Also explains why it is forbidden to convert `char**` to `const char**`. I get my answer. — vkchavda, Dec 04 '21 at 16:58
@RaymondChen Nitpick: The "passed pointer", listOfStr, is still not const in `const char* const* listOfStr`; that would only be `const char* const* const listOfStr`! (The second const makes the intermediate pointer level const.) — Peter - Reinstate Monica, Dec 04 '21 at 20:14

score 2 · Accepted Answer · answered Dec 04 '21 at 16:34

2

The issue is covered in the C++ FAQ, thanks to Steve Summit: https://isocpp.org/wiki/faq/const-correctness#constptrptr-conversion

I have looked at it for 10 minutes and still don't fully understand it, but the problem is apparently that you could create non-const aliases and thus modify an originally const object if this were allowed.

answered Dec 04 '21 at 16:34

Peter - Reinstate Monica

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At the end, if every step was legal, q would point to p and p would point to x, with q and x being const, but not p, which is a problem. q was basically used in order to make p point to x within `*q = &x; ` (with *q being p at that moment). Thus, one step must be forbidden. All lines except the one in the middle are something done commonly. I think if I hadn't read your link, I'd never guessed that this harmless looking line is actually dangerous in a hundred years. – Aziuth Dec 04 '21 at 20:16
@Aziuth q is not const either... it points to pointers to const. – Peter - Reinstate Monica Dec 04 '21 at 20:24

Error Can not convert char** to const char**

1 Answers1

Error Can not convert char to const char