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Interview Question


Country: India



Comment hidden because of low score. Click to expand.
5
of 5 vote

Output is:
der 10 func
der 1 func

The explanation goes as follows:

The default argument is a compile-time feature i.e. the substitution of default arguments is done at compile-time.
For this reason, obviously, there's no way the default argument mechanism for member functions can depend on the dynamic (i.e. run-time) type of the object. It always depends on static (i.e. compile-time) type of the object.

The call you wrote in your code sample[bb->func(10)] is immediately interpreted by the compiler as bb->func(10,10) regardless of anything else.

- Aashish July 14, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 votes

shondik, can you elaborate on that? Since I am not sure I understand the slightly modified version below.

class base{
public:
virtual void func(int i,int j=10)
{
cout<<"base func"<<endl;
}
};


class der:public base
{
public:
void func(int i,int j=5)
{
cout<<"der "<<j<<" func"<<endl;
}
};


int main()
{
der dd;
base *bb = new der();
bb->func(10);
dd.func(10,1);
dd.func(10,5);
}

- Oliver March 16, 2013 | Flag
Comment hidden because of low score. Click to expand.
2
of 2 vote

der 10 func
der 1 func

- Harsha July 13, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 2 vote

first out put : base func

second output : der 1 func

- sam July 14, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 votes

Please check the output before writing anything.

- Aashish July 14, 2012 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 vote

Is there any way one can call the "base func" to get printed?

- jaigupta July 16, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 votes

Sure - make the base func non-virtual

- Tom July 21, 2012 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 vote

der 10 func
der 1 func

- deveshucer August 06, 2012 | Flag Reply


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