What is the meaning of prepended double colon “::”

问题:

I found this line of a code in a class which I have to modify:

::Configuration * tmpCo = m_configurationDB;//pointer to current db

and I don\'t know what exactly means the double colon prepended to the class name. Without that I would read: declaration of tmpCo as a pointer to an object of the class Configuration... but the prepended double colon confuses me.

I also found:

typedef ::config::set ConfigSet;

回答1:

This ensures that resolution occurs from the global namespace, instead of starting at the namespace you\'re currently in. For instance, if you had two different classes called Configuration as such:

class Configuration; // class 1, in global namespace
namespace MyApp
{
    class Configuration; // class 2, different from class 1
    function blah()
    {
        // resolves to MyApp::Configuration, class 2
        Configuration::doStuff(...) 
        // resolves to top-level Configuration, class 1
        ::Configuration::doStuff(...)
    }
}

Basically, it allows you to traverse up to the global namespace since your name might get clobbered by a new definition inside another namespace, in this case MyApp.

回答2:

The :: operator is called the scope-resolution operator and does just that, it resolves scope. So, by prefixing a type-name with this, it tells your compiler to look in the global namespace for the type.

Example:

int count = 0;

int main(void) {
  int count = 0;
  ::count = 1;  // set global count to 1
  count = 2;    // set local count to 2
  return 0;
}

回答3:

Lots of reasonable answers already. I\'ll chip in with an analogy that may help some readers. :: works a lot like the filesystem directory separator \'/\', when searching your path for a program you\'d like to run. Consider:

/path/to/executable

This is very explicit - only an executable at that exact location in the filesystem tree can match this specification, irrespective of the PATH in effect. Similarly...

::std::cout

...is equally explicit in the C++ namespace \"tree\".

Contrasting with such absolute paths, you can configure good UNIX shells (e.g. zsh) to resolve relative paths under any element in your PATH environment variable, so if PATH=/usr/bin:/usr/local/bin, then...

X11/xterm

...would happily run /usr/bin/X11/xterm if found, else /usr/local/bin/X11/xterm. Similarly, say you were in a namespace called X, and had a \"using namespace Y\" in effect, then...

std::cout

...could be found in any of ::X::std::cout, ::std::cout, ::Y::std::cout, and possibly other places due to argument-dependent lookup (ADL, aka Koenig lookup). So, only ::std::cout is really explicit about exactly which object you mean, but luckily nobody in their right mind would ever create their own class/struct or namespace called \"std\", nor anything called \"cout\", so in practice using only std::cout is fine.

(A noteworthy difference is that shells tend to use the first match using the ordering in PATH, whereas C++ gives a compiler error when you\'ve been ambiguous.)

General discussion on namespaces and explicitness of symbols

Using absolute ::abc::def::... \"paths\" can sometimes be useful to isolate you from any other namespaces you\'re using, part of but don\'t really have control over the content of, or even other libraries that your library\'s client code also uses. On the other hand, it also couples you more tightly to the existing \"absolute\" location of the symbol, and you miss the advantages of implicit matching in namespaces: less coupling, easier mobility of code between namespaces, and more concise, readable source code.

As with many things, it\'s a balancing act. The C++ Standard puts lots of identifiers under std:: that are less \"unique\" than cout, that programmers might use for something completely different in their code (e.g. merge, includes, fill, generate, exchange, queue, toupper, max). Two unrelated non-Standard libraries have a far higher chance of using the same identifiers as the authors are generally un- or less-aware of each other. And libraries - including the C++ Standard library - change their symbols over time. All this potentially creates ambiguity when recompiling old code, particularly when there\'s been heavy use of using namespaces: the worst thing you can do in this space is allow using namespaces in headers to escape the headers\' scopes, such that an arbitrarily large amount of direct and indirect client code is unable to make their own decisions about which namespaces to use and how to manage ambiguities.

So, a leading :: is one tool in the C++ programmer\'s toolbox to actively disambiguate a known clash, and/or eliminate the possibility of future ambiguity....

回答4:

:: is the scope resolution operator. It\'s used to specify the scope of something.

For example, :: alone is the global scope, outside all other namespaces.

some::thing can be interpreted in any of the following ways:

• some is a namespace (in the global scope, or an outer scope than the current one) and thing is a type, a function, an object or a nested namespace;
• some is a class available in the current scope and thing is a member object, function or type of the some class;
• in a class member function, some can be a base type of the current type (or the current type itself) and thing is then one member of this class, a type, function or object.

You can also have nested scope, as in some::thing::bad. Here each name could be a type, an object or a namespace. In addition, the last one, bad, could also be a function. The others could not, since functions can\'t expose anything within their internal scope.

So, back to your example, ::thing can be only something in the global scope: a type, a function, an object or a namespace.

The way you use it suggests (used in a pointer declaration) that it\'s a type in the global scope.

I hope this answer is complete and correct enough to help you understand scope resolution.

回答5:

:: is used to link something ( a variable, a function, a class, a typedef etc...) to a namespace, or to a class.

if there is no left hand side before ::, then it underlines the fact you are using the global namespace.

e.g.:

::doMyGlobalFunction();

回答6:

its called scope resolution operator, A hidden global name can be referred to using the scope resolution operator ::
For example;

int x;
void f2()
{
   int x = 1; // hide global x
   ::x = 2; // assign to global x
   x = 2; // assign to local x
   // ...
}

回答7:

(This answer is mostly for googlers, because OP has solved his problem already.) The meaning of prepended :: - scope resulution operator - has been described in other answers, but I\'d like to add why people are using it.

The meaning is \"take name from global namespace, not anything else\". But why would this need to be spelled explicitly?

Use case - namespace clash

When you have the same name in global namespace and in local/nested namespace, the local one will be used. So if you want the global one, prepend it with ::. This case was described in @Wyatt Anderson\'s answer, plese see his example.

Use case - emphasise non-member function

When you are writing a member function (a method), calls to other member function and calls to non-member (free) functions look alike:

class A {
   void DoSomething() {
      m_counter=0;
      ...
      Twist(data); 
      ...
      Bend(data);
      ...
      if(m_counter>0) exit(0);
   }
   int m_couner;
   ...
}

But it might happen that Twist is a sister member function of class A, and Bend is a free function. That is, Twist can use and modify m_couner and Bend cannot. So if you want to ensure that m_counter remains 0, you have to check Twist, but you don\'t need to check Bend.

So to make this stand out more clearly, one can either write this->Twist to show the reader that Twist is a member function or write ::Bend to show that Bend is free. Or both. This is very useful when you are doing or planning a refactoring.

回答8:

:: is a operator of defining the namespace.

For example, if you want to use cout without mentioning using namespace std; in your code you write this:

std::cout << \"test\";

When no namespace is mentioned, that it is said that class belongs to global namespace.