C pointers and || operators

I'm just wondering whether this is "good" code for a C89 program.

obj_ptr = (obj*) (ptr1 || ptr2);

Essentially what it does (atleast in GCC on my computer) is set obj_ptr as ptr1 if ptr1 != NULL and ptr2 otherwise.

I've looked around and I can't see whether this i开发者_JAVA技巧s proper, but judging by the fact that the || operator has to convert the pointers to integers and then I have to cast them back is a hint of bad style.

If this is bad style or unportable, and is there a better and (hopefully) equally as terse solution?

EDIT: My primary concern whether the code I have written is portable and doesn't rely on undefined behavior.

I may have found a better way which is portable and which I think is "good style" (unless you don't like assignment in if statements).

if(!(obj_ptr = ptr1))
    obj_ptr = ptr2;

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No, what it does is set obj_ptr to 1 if either ptr1 is not NULL or ptr2 is not NULL, and 0 otherwise. You need to use the ternary operator:

obj_ptr = ptr1 ? ptr1 : ptr2;

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Well, it would definitely be invalid in C++ (where both operands are promoted to bool). I admit I am not sure about C.

[Update] OK, found it, C99 spec section 6.5.14:

Semantics

The || operator shall yield 1 if either of its operands compare unequal to 0; otherwise, it yields 0. The result has type int.

So || always evaluates either to 0 or to 1.

The usual way to formulate that expression is:

obj_ptr = (ptr1 ? ptr1 : ptr2);

If you actually need the (obj *) cast, there is a good chance you are doing something wrong.

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if you don't like writing ptr1 twice, you can use a macro:

#define or(a, b) (a ? a : b)

obj_ptr = or(ptr1, prt2);