Function overloading not working

Question

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.

Overloading doesn't magically occur based on what you do with the result down the line. — Lightness Races in Orbit
– Lightness Races in Orbit, Commented Jan 21, 2014 at 17:14
@StoryTeller You can overload on const though AFAIK. I think the problem here is the choice between the two would probably only be affected by the constness of test, or if it was used as a parameter to a function requiring a const parameter (I'm not sure of the latter though). — Borgleader
– Borgleader, Commented Jan 21, 2014 at 17:14
@Borgleader, you can overload on the constness of this. But the OP was clearly trying to overload on the return value. — StoryTeller - Unslander Monica
– StoryTeller - Unslander Monica, Commented Jan 21, 2014 at 17:17

juanchopanza · Accepted Answer · 2014-01-21 17:12:14Z

10

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();

answered Jan 21, 2014 at 17:12

juanchopanza

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Luchian Grigore · Accepted Answer · 2014-01-21 17:12:18Z

4

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

const test t;
t.data();

answered Jan 21, 2014 at 17:12

Luchian Grigore

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Mike · Accepted Answer · 2014-01-21 17:41:28Z

0

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

answered Jan 21, 2014 at 17:41

Mike

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