C++ Question about default constructor

Question

What does it mean to call a class like this:

class Example
{
 public: 
  Example(void);
  ~Example(void);
}

int main(void)
{
 Example ex(); // <<<<<< what is it called to call it like this?

 return(0);
}

Like it appears that it isn't calling the default constructor in that case. Can someone give a reason why that would be bad?

Thanks for all answers.

Exact duplicate http://stackoverflow.com/questions/180172/ctor-with-no-arguments — Paul Tomblin, Jan 14 '09 at 19:50

JaredPar · Accepted Answer · 2009-01-14T19:40:17.050

16

Currently you are trying to call the default constructor like so.

Example ex();

This is not actually calling the default constructor. Instead you are defining a function prototype with return type Example and taking no parameters. In order to call the default constructor, omit the ()'s

Example ex;

edited Jan 14 '09 at 19:40

answered Jan 14 '09 at 19:31

JaredPar

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Not a function pointer (that would be `Example (*ex)()`) but a function prototype. – Konrad Rudolph Jan 14 '09 at 19:33

score 8 · Answer 2 · answered Jan 14 '09 at 19:32

8

This declares a function prototype for a function named ex, returning an Example! You are not declaring and initializing a variable here.

answered Jan 14 '09 at 19:32

Konrad Rudolph

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score 4 · Answer 3 · edited May 23 '17 at 12:19

4

Does it even compile? Anyway, see this related topic.

edited May 23 '17 at 12:19

Community

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answered Jan 14 '09 at 19:30

Nemanja Trifunovic

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Related? I'd call it a duplicate, myself. – Paul Tomblin Jan 14 '09 at 19:50

score 2 · Answer 4 · answered Jan 15 '09 at 00:10

As has been noted Example ex(); declares a function prototype. Not what anyone would expect. This C++ wart will be fixed by the new C++0x standard. In the future the preferred syntax will be Example ex{};. The new uniform construction has many other nice features, see more here.

C++ Question about default constructor

4 Answers4