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aminura

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Type casting

Oct 19th, 2005

In the following code I am giving an int value to class using type casting--

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#include<iostream>
using namespace std;
class A
{
	int a;
	float b;
public:
	A()
	{
		a=0;
		b=0;
	}
	A(int m)
	{
		a=m;
		cout<<"\n values are"<<a<<endl<<b;
	}
};
void main()
{
	A a1;
	a1=1;
 
}
#include<iostream>
using namespace std;
class A
{
	int a;
	float b;
public:
	A()
	{
		a=0;
		b=0;
	}
	A(int m)
	{
		a=m;
		cout<<"\n values are"<<a<<endl<<b;
	}
};
void main()
{
	A a1;
	a1=1;
	
}

The members of this class are being initialised with zero but after assigning 1 to the object a1.. the value of b changes to a garbage value. Why does it do this way?

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Re: Type casting

Oct 19th, 2005

Just do this

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A(int m)
	{
		a=m;
		b=m;
		cout<<"\n values are"<<a<<endl<<b;
	}
A(int m)
	{
		a=m;
		b=m;
		cout<<"\n values are"<<a<<endl<<b;
	}

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aminura

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Re: Type casting

Oct 19th, 2005

I don't want to assign this value to b. My question is that because of default constructor b is initialized to 0 but after typecasting why doesn't it retain it's value zero..it changes to garbage value.

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Re: Type casting

Oct 19th, 2005

becoz the initial value given by default constructor(i.e 0,0) is given to object a1....but when u write a statement

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a1=1//implicit type conversion from int to user-defined type

a temporary(unnamed) object is formed which calls ur parameterized contructor which initilizes only a and does not initilizes b...that is why b is having garbage value....

Note: Read about explicit keyword

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Re: Type casting

Oct 19th, 2005

the code you posted should not even compile because class A is missing the overloaded = operator. in main(), the line "a = 1" does NOT call the default constructor as you might think, but will call the = operator. And since you did not define one the compiler should have complained (screamed loudly) to you.

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Re: Type casting

Oct 19th, 2005

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Originally Posted by Ancient Dragon

the code you posted should not even compile because class A is missing the overloaded = operator. in main(), the line "a = 1" does NOT call the default constructor as you might think, but will call the = operator. And since you did not define one the compiler should have complained (screamed loudly) to you.

It will compile perfectly fine...but won't give the output...which was expected...

When compiler finds that there is wrong type on the right hand side of = it would look for a conversion function which can convert an int to A type. The argument constructor can meet this requirement. hence the compiler would decide to call it. This is an implicit conversion, one that you may not have intended to make possible......as far as = operator is concerned...compiler will supply that for you...don't worry about that

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Re: Type casting

Oct 19th, 2005

•

Originally Posted by sunnypalsingh

becoz the initial value given by default constructor(i.e 0,0) is given to object a1....but when u write a statement
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a1=1//implicit type conversion from int to user-defined type
a1=1//implicit type conversion from int to user-defined type
a temporary(unnamed) object is formed which calls ur parameterized contructor which initilizes only a and does not initilizes b...that is why b is having garbage value....

Note: Read about explicit keyword

So is it like that the temporary object formed during the calling of parameterised constructor returns the value to the actual object?

 Help with Code Tags 
 C++ Syntax (Toggle Plain Text) 
#include<iostream>
using namespace std;
class A
{
public:
	int a;
	float b;
public:
	A()
	{
		a=0;
		b=0;
	}
	A(int m)
	{
		a=m;
		cout<<"\n values are"<<a<<endl<<b;
	}
};
void main()
{
	A a1;
 
	a1=1;
	cout<<a1.b;
}
#include<iostream>
using namespace std;
class A
{
public:
	int a;
	float b;
public:
	A()
	{
		a=0;
		b=0;
	}
	A(int m)
	{
		a=m;
		cout<<"\n values are"<<a<<endl<<b;
	}
};
void main()
{
	A a1;
	
	a1=1;
	cout<<a1.b;
}

When I make the members public, the output of a1.b is also a garbage value. And if the constructor is returning the value to the object then isn't it contradicting the concept I asked about in one of my previous threads..

See this thread
http://daniweb.com/techtalkforums/thread32689.html

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Re: Type casting

Oct 19th, 2005

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Originally Posted by aminura

So is it like that the temporary object formed during the calling of parameterised constructor returns the value to the actual object?

Constructors don't return any value(not allowed in C++)....i am saying when u write

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a1=1

when compiler finds 1(int type which is not same as user defined type)...it will perform implicit type casting...and by doing that it calls parameterized constructor....which initilizes ur nonamed object....and since u haven't supplied ur own overloaded =operator...compiler will supply it for you

Note:I didn't see the link of the thread u provided

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Re: Type casting

Oct 19th, 2005

what was the doubt(contradiction) you had in the thread you referred in the last post of yours

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Re: Type casting

#10

Oct 19th, 2005

you answered that ...regarding returning value.

Now I am little confused...since constructors don't return any value how does type casting using parameterised constructors work. Like in the program a1=1; calls the constructor and value of a and b changes accordingly in the actual object..how does this happen?

Last edited by aminura; Oct 19th, 2005 at 3:02 pm. Reason: one more question