Overriding '+' operator

Hello, everyone.

I was wondering is it possible to override the addition operator so that it will take two constants of type char*.

I tried writing the following code but it only gave me an error message stating "'operator +' must have at least one formal parameter of class type".

Code:

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char* operator+(const char* left, const char* right)
{
        char* c, *d;
        d = new char[strlen(left) + 1];
        strcpy(d, left);
        c = new char[strlen(left) + strlen(right) + 1];
        c = strcat(d, right);
        return c;
}

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So, is it fundamentally impossible to override this operator outside a user defined class, or is there a way to do it?

If so, how?

Thanks.

Why do you want to overload it with two const char *s? That already exists, and that's what the compiler's complaining about.

Oops, didn't explain enough in my post.

I just wanted to try and create something like the string handlers you find in Visual Basic.

In visual basic you can do this:

Code:

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Dim S as String
S = "Hello" + "World!"
S = "Hello" & "World!"

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Either '+' or '&' would function as concatenation operators if they were passed strings.

I know you can't do that with character pointers in c++. But can you override them to act like the Visual Basic concatenators if the operands were character pointer constants?

Quote:

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I was wondering is it possible to override the addition operator so that it will take two constants of type char*.

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No, that is why you are getting this error:

Quote:

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I tried writing the following code but it only gave me an error message stating "'operator +' must have at least one formal parameter of class type

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You can only overload operators for members of a user defined class:
http://geneura.ugr.es/~jmerelo/c++-f...html#faq-26.10

Just use the C++ string class and its operator+

Code:

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std::string S;
S = "Hello" + std::string("World!");
// or
S = std::string("Hello") + "World!";
// or the best option for small additions:
S = "Hello";
S += "World";

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Thanks. Now I know better.

Quote:

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If C++ let you redefine the meaning of operators on built-in types, you wouldn't ever know what 1 + 1 is: it would depend on which headers got included and whether one of those headers redefined addition to mean, for example, subtraction.

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...or

Code:

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string char1 = "o";
string char2 = "k";
cout<<char1 + char2<<endl;

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