Does the compiler optimise structs of size 0?

Question

If I have a struct defined such as the following:

struct blank {
    int : 0;
};

Will the compiler optimise this away to nothing at runtime?

I ask because of this rather popular SO question. I would like to employ similar compile-time checks in my own C/C++ code. I'm trying to make the program as optimal as possible, so I don't want these empty structs hanging around at runtime if I'm only going to use them for compile-time checking.

On a side note, is there a C++-idiomatic way of achieving the same outcome as in the link?

What compile-time checking does C allow? This seems to be a C++-only question, retagging — Potatoswatter, Feb 22 '13 at 00:37
@Potatoswatter: C __does__ allow compile time checking. For instance, one can implement a static assert in C. — , Feb 22 '13 at 00:38
Did you profile and discover that your program spends a large amount of time constructing instances of empty structs? If not, *why do you care?* — Jon Purdy, Feb 22 '13 at 00:39
basically if you don't want to have them in the binary. make sure they are unused. which in turn means, even if the size is > 0 it would work — Alexander Oh, Feb 22 '13 at 00:43
Hint: your `blank` struct does not have a size of zero. No type ever does. — GManNickG, Feb 22 '13 at 01:16

score 5 · Accepted Answer · answered Feb 22 '13 at 00:41

C++ only allows that optimization when the object implements a base class. (Hence it's called the empty base optimization or EBO.) For an object which stands alone, or as a member of another struct, it must be at least one byte large even if that byte is just padding.

As for constructs like the one in the linked question, don't worry. There is no object there, only a type. Even if there were an object, it would be stripped by the linker. Unreferenced functions and global objects don't go into the executable unless you specifically ask to export everything. Even in that case, it would still probably need to have a name to be kept.

As others have mentioned, the C++11 way of doing that is static_assert. It requires an error message:

static_assert( condition, "Error: condition was false." );

Fred Foo · Answer 2 · 2013-02-22T00:41:10.753

2

Depends. If, by the as-if rule, all instances of this struct can be completely eliminated, then the compiler may perform that optimization. Whether your compiler does that depends on which compiler it is and possibly on the optimizer settings.

In practice, passing the address of a struct instance across module boundaries is a pretty sure way of materializing the instance in RAM with a size of at least one byte (unless a whole-program optimizer catches it). So is using this struct as the type of a member in another struct or class.

edited Feb 22 '13 at 00:41

answered Feb 22 '13 at 00:34

Fred Foo

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@LokiAstari: yes, but if you never take the address, the optimizer might remove the instance -- after all, you can't do anything else with it, so effectively it's dead code :) – Fred Foo Feb 22 '13 at 00:42

score 2 · Answer 3 · answered Feb 22 '13 at 00:40

2

The empty struct is guaranteed to have its own address, but that's about it: you wouldn't spend time building it (after all, it's empty; there's nothing to build).

The C++ idiomatic way of doing the same thing is using static_assert (available in compilers compliant with the C++11 standard).

answered Feb 22 '13 at 00:40

Sergey Kalinichenko

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http://stackoverflow.com/questions/621616/c-what-is-the-size-of-an-object-of-an-empty-class here is a justification. – Alexander Oh Feb 22 '13 at 00:41

Does the compiler optimise structs of size 0?

3 Answers3