[SOLVED] what's this? (unsigned int) & 'something'

lesca · 10-30-2010, 08:19 PM

Hello everyone

When I 'call' offsetof(3) from <stddef.h>, I find 'offsetof' is actually a macro.

I expand it with Eclipse (with CDT), the expanded form is:

Code:

((size_t) &((Node *)0)->next)

Wherein, the Node is a type I defined:

Code:

typedef struct NODE
{
	int value;
	struct NODE * next;
} Node ;

And I know size_t is actually unsigned int in my computer.
So the fully expanded form is:

Code:

( (unsigned int) & ((Node *)0)->next )

We offten evaluate values like " 0x000f & var ", so that we can get the designated bits we want.
But here, the AND opreator seems doesn't work in that way.
And I find my questions:
1. How does it work?
2. On the left side of AND, that is a built-in type. On the right side, what is it?

Thanks!

paulsm4 · 10-30-2010, 10:33 PM

Hi -

The "&" operator has two completely different meanings in C/C++:

a) boolean "and" operator
b) "address of" operator

This is an example of "address of":

Code:

((size_t) &((Node *)0)->next)

Here's a good link:
http://www.cprogramming.com/tutorial/lesson6.html

lesca · 10-30-2010, 11:48 PM

Aha! Thanks!

I know pointer, I just couldn't associate this expression with that "address of" operator.

"&((Node *)0)->next" gets the address of next from the beginning of the memory, and then cast the value of address to size_t.

I figure it out!

Thanks very much for your tips!

TobiSGD · 10-31-2010, 12:17 AM

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