extern "C" nests suspiciously

[Phlip]

Language Lawyers:

Compare this:

extern "C" int Maine()
{
...
bool runTests();
return !runTests();
}

to this:

bool runTests();

extern "C" int Maine()
{
...
return !runTests();
}

The first does not link because the true runTests() is not extern "C". But I
didn't specify extern "C" directly on bool runTests().

Should the extern "C" have been "inherited" from the enclosing function like
that?

Just curious. (But always glad to find yet another bug in Visual C++ 6!)

 

[Victor Bazarov]

Language Lawyers:

Compare this:

extern "C" int Maine()
{
...
bool runTests();
return !runTests();
}

to this:

bool runTests();

extern "C" int Maine()
{
...
return !runTests();
}

The first does not link because the true runTests() is not extern "C". But I
didn't specify extern "C" directly on bool runTests().

Should the extern "C" have been "inherited" from the enclosing function like
that?

According to the Standard, language linkage specifications do nest. Every
declaration of a function inside another function will have the same
language linkage specification as the outside function. You can test your
compiler with
------------------------
extern "C++" void foo();

extern "C" void bar() {
void foo();
return foo(); // should give an error
}
------------------------
If your compiler throws a fit, you got a decent one (WRT language linkage
specifications, anyway).

Just curious. (But always glad to find yet another bug in Visual C++ 6!)

V