C++11-style [[unused]] attribute in gcc?

Question

Under gcc/g++ 4.9 I can write:

int x __attribute__((unused)) = f();

to indicate that x is intentionally unused.

Is it possible to d

Answer 1

Yes, use [[gnu::unused]]

Like already said unused isn't part of the standard attributes specified by the standard.

The standard allows implementation defined attributes too like the __attribute__ and __declspec ones to be used with the new syntax. If a compiler doesn't recognize an attribute (a gcc attribute when compiling on MSVC as example) it'll simply be ignored. (probably with a warning)

For gcc you can use the gnu prefix and the C++11 attribute syntax: [[gnu::unused]] instead of __attribute__((unused)) the same should apply for the other gcc attributes too.

example without gnu prefix

example with gnu prefix

Answer 2

The thing you are referring to is known as attribute specifiers. It is an attempt to standardize various, platform dependent, specifiers:

• __attribute__ in case of GCC / ICC (Linux)
• __declspec on MSVC / ICC (Windows)

As you can see in attached doc link, the only specifiers supported in C++11 are:

• [[noreturn]]
• [[carries_dependency]]

and in C++14:

• [[deprecated]] (also supported as: [[deprecated("reason")]])

So the answer is: no, it's not possible, using only C++11 features.

---

If you are not interested only in portable solutions, there might be a way. C++ standard does not limit this list:

Only the following attributes are defined by the C++ standard. All other attributes are implementation-specific.

Various compilers can support some non-standard specifiers. For example, you can read this page in order to find out, that Clang supports:

• [[gnu::unused]]

Perhaps your version of GCC also supports this specifier. This page contains a bug report referring to generalized attributes support. [[gnu::unused]] is also mentioned.

Answer 3

There is [[maybe_unused]] attribute in C++17. It's implemented in GCC 7, see C++ Standards Support in GCC .

Example from P0212R1 proposal:

[[maybe_unused]] void f([[maybe_unused]] bool thing1,
                        [[maybe_unused]] bool thing2) {
    [[maybe_unused]] bool b = thing1 && thing2;
    assert(b);
}