Answer / avi

Data members of a class are initialized in the order they
are declared in class. in case class is derived from some
base class then the base class will be initialized first
followed by the derived class.

Answer / sanjay makwana, puna

In order of class declared memeber variables.

Answer / nilay

all the static memebers are created at the time of class
definition and rest all non static memebers are created
when object is created of that class. member functions are
also create in memory when class is defined. (However only
single copy of static data members and functions is created)

Answer / vinay bondade

It is same as the order of declaration in the class.

Answer / kasi

1) Brief details on initialization:

To perform true initialization (not assignments) for a
class data member, C++ provides extened syntax for
constructor function.

Syntax:
class A
{
int ssn;
double sal;
public:
A(int j,double k):sal(k),ssn(j) //initialization
{
//assignement
}
}

Therefore constructor function has 2 parts they are
- initialization
- Assignment
Note - initialization will be excuted first and then
assignment

2) When is the initialization list mandatory?
- If the instance variable is constant
- If the class contains reference variable

3) coming to the posted question: order of member
initialization will be done in 2 ways by the compiler.

A) Declaration order
B) The order of members in the intialization list.

For the above example ssn is initialized first then sal.

Answer / sachin magdum

Kasi, I am not agreed with you, in your case sal will be initialized before ssn.

Debug this program and see,


class A
{
int i ;
public :
A (int _i)
{
i = _i;
}
};

class B
{
A a;
A b;
A c;

public :
B(int _a, int _b, int _c) : a(_a), c(_c), b(_b)
{}
};

int main(int argc, char* argv[])
{
B b(1,2,3);

return 0;
}

Here the initialization order is first 'a' then 'c' and then 'b' and not a, b, c.

However, if you have base class then the case is little different


class A
{
int i ;
public :
A (int _i)
{
i = _i;
}
};

class B : public A
{
A a;
A b;
A c;

public :
B(int _a, int _b, int _c, int _base) : a(_a), c(_c), b(_b), A(_base)
{}
};

int main(int argc, char* argv[])
{
B b(1,2,3,4);

return 0;
}

In this case first 'A' that is base class of B, then 'a' then 'c' then 'b' will be initialized.

If you have multiple base classes, then they will be initialized in the order of their declaration, and not in the order how they are initialized.

class A
{
int i ;
public :
A (int _i)
{
i = _i;
}
};

class C
{
int i ;
public :
C (int _i)
{
i = _i;
}
};


class B : public A, public C
{
A a;
A b;
A c;

public :
B(int _a, int _b, int _c, int _baseA, int _baseC) : a(_a), C (_baseC), c(_c), b(_b), A(_baseA)
{}
};

int main(int argc, char* argv[])
{
B b(1,2,3,4,5);

return 0;
}

In this case first 'A' then 'C' then 'a' then 'c' and then 'b' will be initialized.

Answer / g. maheshwari

The member and base classes are initialized in the order they appear in the class definition, not the order they appear in initialization list.

Answer / karthik

there is no need for ordering a initialization of data.It
depends upon the program

Answer / mithali

I think initialization doesn't have any order.we r just
initializing the variables to some value.

Answer / rejeesh r

Class is a connection of objects.

Is c++ double?

0 Answers  

string somestring ; Which of the following choices will convert a standard C++ string object "somestring" to a C string? a) Copy.somestring () ; b) somestring.c_str () c) &somestring [1] d) std::cstring (somestring) e) (char *) somestring

1 Answers   Quark,

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2 Answers   Quark,

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0 Answers   Genpact,

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