Why Compiler provided Default Constructor Hides when parameterized constructor is Provided

Question

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hi Can any one help me to solve this Problem

C#

class sample
{
   int a,b;
public:
   sample(int x, int y)
   {
      a = x;
      b = y;
      cout<<"a="<<a<<endl<<"b="<<b<<endl;
   }
};

void main()
{
   sample obj1;
}

In the Above Program the Error it is showing is "No Appropriate Default Constructor is provided"

But The Compiler internally provides One Default Constructor know.
Why doesn't the compiler use the Compiler provided Default Constructor?

thanks
sarath

Posted 16-Feb-11 16:07pm

sarathbabununna

Updated 16-Feb-11 16:21pm

HimanshuJoshi

v2

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2 solutions

Solution 1

You have provided your own constructor which overrides the default constructor. And when you are creating a object, you are not passing any arguments which are required by your constructor.

Try changing your line

sample obj1;

to

int x = "some value you want in x";
int y = "some value you want in y";
sample obj1(x, y);

Replace some values you want in x/y to something meaningful.

Posted 16-Feb-11 16:20pm

HimanshuJoshi

Comments

Sergey Alexandrovich Kryukov 16-Feb-11 22:31pm

It does not "override" it (C++ constructors are not virtual functions).
My 4.
--SA

Emilio Garavaglia 17-Feb-11 8:59am

hummmm ... "overriding" is not necessarily something related to virtual functions: I "override" something (A) with something else (B) every time B "masks" the usage of A.
The "scope" where this happens, for virtual function, is the (runtime) lifetime of an object, for non virtual function is the name scope of a class (that include all derived's), for a template definition are all the not-elsewhere specialized instances.

They are completely different semantics, but always "over-ride" is.

Sergey Alexandrovich Kryukov 17-Feb-11 20:13pm

Sorry, in strict C++ terms it is, will you look at the text book.
A common misconception is "override happens in run-time". No, it's statically defined at compile time: VTable is defined for a class. It's run-time (not defined at the moment of compilation) what function defined in base class will be called, due to late binding, but subtle difference is: this is about a call, not about override.
The novice need to feel the difference in the situation when base-class name is hidden. It looks like "override" because new method is called is derived class is explicitly used, but late binding is lost. Also modifies "new" needs explanation.
--SA

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Sergey Alexandrovich Kryukov · Accepted Answer · 2011-02-16T16:40:00

You're almost right, it's easy to get confused. You misunderstood the syntax rules only slightly.

The C++ rule about default constructor does not say:

"If parameter-less constructor is not provided, default constructor is assumed".

Instead, it says:

"If you provide a constructor that takes a non-void parameter, and you want to allow your class to be created with no parameters, you must also provide a default constructor. The default constructor can be a constructor with default values for all parameters."

Do you see the subtle difference?

So, assuming you want to have the same members, you need to change one of the following, or both:

1) Add parameter-less constructor, something like this:

C++

class sample
{
   int a,b;
public:
   sample() : a(0), b(0) {}
   //...
}

2) Use your available constructor with two parameters:

C++

sample obj1(x, y);

—SA