Re: sizeof operator implementation
- From: Skarmander <invalid@xxxxxxxxxxxxxx>
- Date: Fri, 08 Dec 2006 01:01:47 +0100
Stephen Sprunk wrote:
"Chris Dollin" <chris.dollin@xxxxxx> wrote in message news:el9dua$5ob$1@xxxxxxxxxxxxxxxxxxxxxLet's ignore the fact that you'd better put parentheses around "val" in the expansion to avoid some of the sillier pitfalls. Even so, "sizeof_op1(char)" is a syntax error and "sizeof_op1('0')" a constraint violation. If "f" is declared asvijay wrote:...if you see assembly then sizeof operator has sizeof(type) or
sizeof(variable) at compile time. How does C compiler gets value at
compiler time.?
the implementation as macro is as below
#define sizeof_op1(val) (&val +1 ) - &val // for variable ex
sizeof_op1(n)
This is homework, isn't it. Oh well.
When this "works", it will get the answer `1`. It won't work for
things that don't have addresses or, more accurately, things that
aren't legal operands of &.
Conclusion: not useful. And you didn't test it.
Wouldn't this work (for some definition of "work") if you did:
#define sizeof_op1(val) ((char*)(&val+1)-(char*)(&val))
In fact, I can't see any reason that'd even invoke UB.
void f(void);
then "sizeof_op1(&f)" and "sizeof_op1(f)" are both constraint violations (of course, so is "sizeof f"). If "a" is declared as
int a[4];
then "sizeof_op1(a[4])" yields UB.
In C99, an expression not involving sizeof can never return correct results for variable-sized arrays.
Even if we're very generous and restrict use of the macro to valid objects of scalar type only, it will still yield UB on objects declared with register class.
The sizeof operator has none of the above restrictions. But of course, you did say "some definition of 'work'". So yes, for some definitions, I suppose it works. But that's not really the point.
Casting 10 to a pointer type does *not* invoke UB, it invokes implementation-defined behavior. Conversions of pointers to and from integers are perfectly legal. What will most likely invoke UB is adding 1 to the result, since it will probably not point to an object.#define sizeof_op2(type) ((type*) (10) + 1) - (type*) (10) // for type
as sizeof_op2(int)
This will /also/ get the answer `1` if it works, and that will
be by coincidence I think, since casting `10` to any pointer type
is at best iffy.
You didn't test this, either.
The same trick as above (casting to char* before subtracting) should "work" for this as well. However, it definitely invokes UB due to casting 10 to a pointer type; might as well cast 0 and make what you're doing obvious.
However, "casting 0" and then adding 1 to the result will *definitely* invoke UB, since you're adding to a null pointer, which by definition cannot point to any object.
The real answer to this is there's no portable or even unportable way to implement sizeof() in user code. That's why it's a core language feature instead of being left to users to roll their own.
Indeed. So it's not a good reason to suggest approaches to it anyway. There be dragons.
S.
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