sizeof operator

[dharmesh Gupta]

Hi All,

the following program gives out put as

4 1

but i am unable to understand it , coulf anyone help me out

Thanks
Dharmesh
//////////////////////////////////////////
#include <stdio.h>

class x
{
public:
virtual fn1();
};

class y
{
public:
fn1();
};

void main()
{
int a,b;
a = sizeof(x);
b = sizeof(y);
printf("%d\t %d ",a,b);
}
/////////////////////////////////////////////////////////

 

[Peter van Merkerk]

Hi All,

the following program gives out put as

4 1

but i am unable to understand it , coulf anyone help me out

Thanks
Dharmesh
//////////////////////////////////////////
#include <stdio.h>

class x
{
public:
virtual fn1();
};

class y
{
public:
fn1();
};

void main()
{
int a,b;
a = sizeof(x);
b = sizeof(y);
printf("%d\t %d ",a,b);
}
/////////////////////////////////////////////////////////

Class x has a virtual function; as a result instances of class x will have
a (hidden) vtable pointer to implement the virtual function call mechanism
with many compilers. A compiler may choose to implement the virtual
function call mechanism differently, but the vtable mechanism is used by
many C++ compilers. On many platforms (including yours) the size of the
vtable pointer is 4 bytes.

Class y doesn't have virtual functions and doesn't have member functions.
So instances of class y don't need to occupy memory. However the C++
standard dictates that the sizeof() operator should yield a value greater
than 0 for most derived classes; this explains why sizeof(y) produces 1
instead of 0.

 

[Ralf]

In the first case your type has an implicit member. The vtable pointer which
has the size of 4 bytes.
In the second case there is no member inside the class. The definition in
C++ is that a type has the
minimum size of one byte.


Ralf

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