static struct in anonymous namespace

Question

that this snippet of code actually do?

#include <iostream>
#include <cstdio>
#include <cstdlib>
using namespace std;

void test();

namespace {
    static struct StaticStruct {
        StaticStruct() { 
            test(); 
        }
    } TheStaticSupport;
}


int main(void) {



    return 0;
}


void test() {
    printf("testing function\n");
}

why does the test function actually get called? and why use the "anonymous" namespace? I found this piece of code in an open source project...

`void teste();` ... `test();`. I don't think so. – James McNellis May 20 '12 at 00:16 — James McNellis, May 20 '12 at 00:16

score 4 · Accepted Answer · answered May 20 '12 at 00:17

This:

static struct StaticStruct {
    StaticStruct() { 
        test(); 
    }
} TheStaticSupport;

Is equivalent to this:

struct StaticStruct {
    StaticStruct() { 
        test(); 
    }
};

static StaticStruct TheStaticSupport;

It defines a type named StaticStruct and an instance of the type named TheStaticSupport with internal linkage (though, since it is declared in an unnamed namespace, the static is redundant).

The constructor for TheStaticSupport is called before main() is entered, to construct the object. This calls the test() function.

score 1 · Answer 2 · answered May 20 '12 at 00:21

1

The anonymous namespace gives the contained objects internal linkage, as their fully qualified name can never be known to anyone outside the translation unit. It's the sophisticated man's version of the old static in C.

Note that you do declare a global object of type StaticStruct, and its constructor (which runs before main() is called) calls test().

answered May 20 '12 at 00:21

Kerrek SB

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Technically, entities in an unnamed namespace may have external linkage. – James McNellis May 20 '12 at 00:23

static struct in anonymous namespace

2 Answers2