Question

I have a simple class Name:

   class Name
{ private:
    char nameStr[30];
  public:
    Name(const char * = "");
    char * getName() const;
    void print() const;
};

I am having two problems, which are just a bit difficult for me wrap my head around. The constructor is supposed to take an argument of const char * and copy that into the nameStr array. However, this is what I have tried and is not working:

Name::Name(const char * setName)
{
    &nameStr = setName;
}

It gives me the error “invalid lvalue in assignment”. If I remove the reference in front of nameStr, it still errors, saying “incompatible types in assignment of ‘const char*’ to ‘char[30]'”. Not quite sure what I’m missing.

The other problem is the accessor method:

char * Name::getName() const
{
    return  &nameStr[0];
}

I have the same problem, it errors “invalid conversion from ‘const char*’ to char*'” and I can’t for the life of me figure out how to make it return a pointer to a const char… I’ve been searching to avail. Found plenty of info on const, but can’t quite figure how to apply it to my situation. Any help would be greatly appreciated!

Answer 1

I would recommend you use an std::string, but first you need to know how pointers work.

Let’s take your first problem: &nameStr = setName;

Here, nameStr is an array of chars, for storing your name. &nameStr refers to the address of this array. You cannot set the address of a variable of any sort, it is illegal. For example, &a = &b (where a and b are ints) is just as illegal. Next you tried, nameStr = setName; and it said incompatible conversion from const char* to char[30]. What you are trying to do is assign one pointer to another. This doesn’t actually copy the data. To copy the data, you need to do this manually, e.g.:

for (int n=0;(n>0?setName[n-1]!=0:true);n++) { nameStr[n] = setName[n]; }
// The above won't work (or might crash) if the ternary operator evaluates both sides. I don't think it does, but correct me if I'm wrong.

Note, the comparion with setName[n-1] instead of setName[n] is so the NULL character gets copied.

For your next problem, cannot convert const char* to char*. Taking the address of a variable will always yeald a constant pointer, because you are not allowed to change the address. Also note that &array[0] is essentially the same as array because you a dereferencing, then re-referencing it, thus making the process unnecessary. You need to state your getName() function as returning const char* and in it, return nameStr.

A word on arrays: Arrays are like other datatypes, except that they contain multiple of a datatype, not one. When referring to an array, its type is of a pointer to its member type. For example, an array of ints would be treated like an int*. The pointer points to it’s first element. This means that using [] on an array will dereference the pointer to any of its elements.

Good luck with C++!