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I use an extern C function that returns dynamically allocated const char*.
I want to use unique_ptr<const char, decltype(std::free)> to manage it.
But there is no std::free(const void*) overload so I get invalid conversion from 'const void*' to 'void*' and have to use const_cast<char*>().
Is this just a Standard Library imperfection or there is something else behind it?
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That is it is a constant pointer that points to a const object of type GLvoid . This parameter declaration const GLvoid * const * indices. means that using pointer indices you may not change the pointer (or pointers if this pointer points to the first element of an array of pointers) it points to.
void* is normally used as a way to pass non-specific pointers around (e.g. with memcpy() ). If you want to pass a const char* to such a function then you cannot use void* or you lose the fact that the thing it points to is constant and cannot be altered.
const void is a type which you can form a pointer to. It's similar to a normal void pointer, but conversions work differently. For example, a const int* cannot be implicitly converted to a void* , but it can be implicitly converted to a const void* .
std::free is inherited from the C standard library. C does not have overloading, so a const overload could not have been inherited.
While the C++ standard library has extended the inherited C library with some useful overloads, a const overload has not been added for free.
Either such an overload has never been considered, or it has not been considered necessary enough to warrant its addition to the standard. I have not come across any openly available proposal for such an addition -- although I haven't read all proposals there have ever been, nor have I attended any committee meetings or workshops, so I cannot deny the possibility of the existence of such a document.
To work around the lack of such an overload, you can indeed use const_cast. In this context, it is completely safe.
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