What does it mean when the function in global namespace is declared as static C++?

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What is a “static” function?

I have seen a function in a global namespace that is declared like this:

static int function_name(int a, double* p, int c, float * u)
{
     //do something with these arguments
}

What the static keyword means here?

EDIT: Now when I know what is for static, please explain what advantage gives the restriction of a function to be visible in a file only where it is declared? I mean why I should restrict my function visibility, what it gives to me?

It's not the return type which is static, it's rather the function which is declared static! — Nawaz, Mar 14 '11 at 15:35
Please stop voting for closing this question. Let's get the answer of my edit part as well. — Narek, Mar 14 '11 at 15:42
Fix the title, then. Or let it be closed and ask a new question about the benefits of file-static. — Lightness Races in Orbit, Mar 14 '11 at 16:28

score 6 · Answer 1 · edited May 23 '17 at 10:29

6

The return value is not static int. The function is a static function returning an int.

See what is a static funciton

edited May 23 '17 at 10:29

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answered Mar 14 '11 at 15:23

Himadri Choudhury

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But it is not a class member function, then what is it for the "static"? – Narek Mar 14 '11 at 15:24
In that case it means that the function has internal linkage, which means that it cannot be referred to from other translational units. Both uses are mentioned in the link that I gave. – Himadri Choudhury Mar 14 '11 at 15:26
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What @DasBoot said is correct, but I'd like to explain a bit. "Translational unit" normally means: the same file. So if a function is declared static you can only call it within the same *.c file, not from other *.c files. The point is to avoid name collisions and also to make the function "private". – DarkDust Mar 14 '11 at 15:31

score 5 · Accepted Answer · answered Mar 14 '11 at 15:53

5

You would use a static function with the scope in the compilation unit when you really want the function with that name to be known only inside that compilation unit. A class or function that is not in that scope cannot accidentally call the function. (I'd have put this in a comment but I don't have privileges for that yet)

answered Mar 14 '11 at 15:53

titania424

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One example I can think of is if you need to pass a callback function to some other API. – Dre Mar 14 '11 at 16:00
It's hard to come up with a specific example, because ideally, in C++, if you really wanted a utility function that was only seen in a certain scope, you'd put it in a class. I can see why you are wondering "why would I use this". I think it is more that it can be done, and it is because C++ is based on C, vs. some purity of the language and the ability to do this having some great meaning. Java changed that and said "put your functions in a class" so you can't do this. But in C++ it is there because of C. – titania424 Mar 14 '11 at 16:01

Gareth McCaughan · Answer 3 · 2011-03-14T15:36:47.860

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It isn't the return value that's static, it's the function. It means that the function is visible within that compilation unit (= file, approximately) but not elsewhere.

edited Mar 14 '11 at 15:36

answered Mar 14 '11 at 15:23

Gareth McCaughan

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to be completely correct, a compilation unit != file, though that is most often the scope and is, in fact, the smallest scope. A case where a compilation unit is more than a file is when one file #include's another file. – Chris Cleeland Mar 14 '11 at 15:26
It's because "compilation unit" isn't strictly the same as "file" that I said "compilation unit" rather than just "file". It's because most of the time you can treat them as the same that I said "(= file)" rather than just "compilation unit". But yes, I should be a little more explicit about the fact that the two aren't quite equivalent; I'll add a note to my answer. (Apologies for the fact that it will make nonsense of your comment to anyone who doesn't bother reading this!) – Gareth McCaughan Mar 14 '11 at 15:36

score 1 · Answer 4 · edited May 23 '17 at 12:02

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As already mentioned, static refers to the whole function, not its return type. I want to add that the use of the static keyword in this context is deprecated in C++. Anonymous namespaces are the better way to go.

namespace 
{
   int function_name(int a, double* p, int c, float * u)
}

edited May 23 '17 at 12:02

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answered Mar 14 '11 at 15:25

Armen Tsirunyan

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This specific deprecation has been un-deprecated in C++0x, by the way. – fredoverflow Mar 14 '11 at 15:28
It is actually un-deprecated again in C++0x, so it is not obviously too bad to use it. :-) – Bo Persson Mar 14 '11 at 15:32
But why to use it? What it gives me? – Narek Mar 14 '11 at 15:36
@Narek: This is a way to specify that your function has internal linkage, that is, it cannot be seen or called from other translation units (cpp files). – Armen Tsirunyan Mar 14 '11 at 15:42
@Narek: One advantage is less namespace pollution, and thus less risk of name clashes. – fredoverflow Mar 14 '11 at 15:54
No, for that case I have a better solution with namespaces, indeed. So I will never use static for that reason. Another reason is necessary for its use! – Narek Mar 14 '11 at 15:57
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@Narek: Are you asking about a useful application of `static` in C++, or are you wondering why it exists? Hint: C does not have namespaces. – fredoverflow Mar 14 '11 at 16:35

What does it mean when the function in global namespace is declared as static C++?

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