CODEWITHSUNDEEP
c++
createproblem
createproblem

Reputation: 1612

Function overloading not working

I've a problem with c++ function overloading. Here is an example class.

class test
{
public:
  const char* data() const
  {
    std::cout << "const char* data() const" << std::endl;
    return data_;
  }

  char* data()
  {
    std::cout << "char* data()" << std::endl;
    return data_;
  }
private:
  char data_[512];
};

In my example I've two function calls.

test t;
const char *t1 = t.data();
char* t2 = t.data();

And my output is char* data() twice. Can someone explain me whats going on? Why is const char* data() const never been called?

Thanks for help.

Upvotes: 4

Views: 1118

Answers (3)

Mike
Mike

Reputation: 126

test t;
t.data();
static_cast<const test>(t).data();

this will give you the desired results, and you wouldn't need to store it as a temporary as shown in the other answer.

http://ideone.com/XDdacr

Upvotes: 0

Luchian Grigore
Luchian Grigore

Reputation: 258548

The const version will only be called on a const object.

const test t;
t.data();

Upvotes: 4

juanchopanza
juanchopanza

Reputation: 227370

Because t is not const, you get the non-const overload of the method. Note that the constness of the return type does not participate in overload resolution, and you can convert char* to const char*.

If you were to try this

const test t;
const char *t1 = t.data();

you would get the const overload, and this wouldn't compile:

char* t2 = t.data();

Upvotes: 10

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