What does return X() mean in this C++ program, as X is not an object?

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In the below program, i want to know what function f5 is trying to do. What does
return X() mean, since X() is not a object.


class X
{
int i;
public:
X(int ii = 0);
void modify();
};

X::X(int ii)
{
i = ii;
}

void X::modify()
{
i++;
}

X f5()
{
return X();
}
 

Re: For C++ expert

can u post full program ......... i think its only a part of the program..
 

Re: For C++ expert

The return X() in the f5 function creates a temporary object of class X with the default constructor (ii=0) and copies it to the variable where you assigned the result. E.g.:

X myVar;

myVar = f5();

myVar.i will be equal to 0. (I don't remember if the compiler generates the default copy constructor)

Best regards,
Mauro H. Leggieri
**broken link removed**
 

Re: For C++ expert


The compiler generate the default copy constructor, destructer and conscructor without parameters
 

Re: For C++ expert

I dont completly understand your question. f5() func simply return X object with default construction and the func Modify() was never called. I think that's all.
 

For C++ expert

here is it returning a result of function X()
 

Re: For C++ expert

Hi,

X f5()
{
return X();
}
Click to expand...

The function f5 creates temporary object of class X by calling the constructor you have declared X::X(int ii) with ii value with 0 and returns the same to the caller. Remember you have defined a single argument constructor with default value. If you dont supply a value while calling a constructor, it will call this single argument constructor with default value you specified ie 0 in this case.

If there is no copy-constructor, constructor, destructor then compiler will create.

Kars
 

For C++ expert

function f5 returns an X object by calling the constructor without any arguments.
 

Re: For C++ expert

f5() func simply return X object with default construction and the func
 

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