Memory size of a method

[mojumbo]

I am passing around a structure, say:

tsBob
{
int a;
int b;
int bobMethod(bool lb);

};

I would like to make sure the structure is aligned along a 64-bit
boundary (it's being thrown around the system). I assumed it was the
size of a pointer (32-bits) but when I googled & wrote a little driver
(See below) it was only 1 byte. Since I can't believe this does
anyone have an answer?

int main(int argc, char* argv[])
{
tsBob aBob;

cout << "Size of bobMethod is " << sizeof(aBob.bob Method) << endl;

return 0;

}

[Matthias Buelow]

Re: Memory size of a method

mojumbo wrote:

I am passing around a structure, say:
>
tsBob
{
int a;
int b;
int bobMethod(bool lb);
>
};
>
I would like to make sure the structure is aligned along a 64-bit
boundary (it's being thrown around the system).

Function members do not contribute to a structure/class's size.
If you call aBob->bobMethod(true ), the compiler essentially
underhandedly rewrites the expression into something similar to
bobMethod(aBob, true) in an unambiguous way.

I assumed it was the
size of a pointer (32-bits)

A function is not a function pointer.

[mojumbo]

Re: Memory size of a method

Thanks

[Richard Herring]

Re: Memory size of a method

In message
<b32bd960-ba0a-4e73-8413-babc44091c46@t5 4g2000hsg.googl egroups.com>,
mojumbo <jnbbender@gmai l.comwrites

>I am passing around a structure, say:
>
>tsBob
>{
int a;
int b;
int bobMethod(bool lb);
>
>};
>
>I would like to make sure the structure is aligned along a 64-bit
>boundary (it's being thrown around the system). I assumed it was the
>size of a pointer (32-bits)

Why do you assume a member function needs to occupy any memory at all
within individual instances of the class?

--
Richard Herring

[James Kanze]

Re: Memory size of a method

On Sep 19, 2:28 pm, mojumbo <jnbben...@gmai l.comwrote:

I am passing around a structure, say:

tsBob
{
int a;
int b;
int bobMethod(bool lb);
};

I would like to make sure the structure is aligned along a 64-bit
boundary (it's being thrown around the system).

Generally speaking, it's the compiler's job to ensure proper
alignment, not yours.

I assumed it was the size of a pointer (32-bits) but when I
googled & wrote a little driver (See below) it was only 1
byte. Since I can't believe this does anyone have an answer?

int main(int argc, char* argv[])
{
tsBob aBob;

cout << "Size of bobMethod is " << sizeof(aBob.bob Method) << endl;
return 0;
}

This shouldn't compile; in C++, functions don't have any size
(or at least, you can't use sizeof on them). And of course,
something like 'aBob.bobMethod ' is only a legal expression if it
is an operand to a function call operator, i.e.
aBob.bobMethod( ).

--
James Kanze (GABI Software) email:james.kan ze@gmail.com
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