:: (scope resolution operator) is used for multiple purposes in C++? - c++

I am learning C++, and want to confirm few thins. I have seen that std::cin has :: operator which means that iostream.h file must have cin related code inside namespace std. Whereas if you include myclass.h and have class NOT inside namespace like this
//myclass.h
class x { void func(){}};
Then you don't need to do this myclass:x.func(); Instead i can do just this x.func();. Correct Becuase it's not inside namespace.
Secondly, :: can be used to define class functions outside of the class like in their cpp file. Correct?
Now my additional question is that are these two functions separate or one? do they just look different to me?


:: (scope resolution operator) is used for multiple purposes in C++?
It's only really serving one purpose - to help find the namespace and/or class/struct/union scope that an identifier's in.
The compiler knows to look first in the tightest scope then work its way back to the global scope, but it also considers any namespaces and identifiers you're "using" (e.g. using namespace xxx; or using ns1::identifier;).
Then you don't need to do this myclass:x.func(); Instead i can do just this x.func();
It has nothing to do with files on disk.
Secondly, :: can be used to define class functions outside of the class like in their cpp file. Correct?
Yes. When you define a member function outside its class, you use the class name so it's not deemed a distinct non-member function in the current scope.
Now my additional question is that are these two functions separate or one? do they just look different to me?
Just the one in the defining-member-function scenario above.


I have seen that std::cin has :: operator which means that iostream.h file must have cin related code inside namespace std.
I realize this isn't a question, but you said iostream.h and I wanted to point out that you should not use iostream.h because it is deprecated, use iostream instead.
Then you don't need to do this myclass:x.func(); Instead i can do just this x.func();. Correct Becuase it's not inside namespace.
This is correct. You could do the same thing if all of your code was inside a namespace rather than a class.
Secondly, :: can be used to define class functions outside of the class like in their cpp file. Correct?
Correct, so long as they have been declared, but not defined, in the class scope this is allowed and usually considered preferable for readability.
Now my additional question is that are these two functions separate or one? do they just look different to me?
If both are in the class scope already, void x::func(){} and void func(){} result in the same thing (one would call both by using x a; a.func(), for example).
EDIT: In response to the comment, you can define a namespace basically the same way you define a class, as:
namespace funcs
{
void innerfunc(){}
void outerfunc();
void funcs::scopedfunc(){}
}
funcs::outerfunc();
As you can infer from std, these are called as free functions, rather than being called from an object like a non-static class function.
int main()
{
funcs::innerfunc();
using namespace funcs;
outerfunc();
}


Then you don't need to do this myclass:x.func(); Instead i can do just this x.func();.
Note that neither of these will compile. You cannot use the dot operator with a class; you can only use it with an object. Specifically, myObject.func(); serves an entirely different purpose than MyClass::func() { }. The first calls the member function func() on the object named myObject. The later is used to define the function func() which belongs to the scope of MyClass.

Related

Why do we have scope resolution operator right after equal sign? [duplicate]

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;

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.

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;
}

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 your current directory or any element in your PATH environment variable, so if PATH=/usr/bin:/usr/local/bin, and you were "in" /tmp, then...
X11/xterm
...would happily run /tmp/X11/xterm if found, else /usr/bin/X11/xterm, 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.
Noteworthy differences:
1) shells tend to use the first match using the ordering in PATH, whereas C++ gives a compiler error when you've been ambiguous.
2) In C++, names without any leading scope can be matched in the current namespace, while most UNIX shells only do that if you put . in the PATH.
3) C++ always searches the global namespace (like having / implicitly your PATH).
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....

:: 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.

:: 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();

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
// ...
}

(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.

:: 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.

"::" represents scope resolution operator.
Functions/methods which have same name can be defined in two different classes. To access the methods of a particular class scope resolution operator is used.

How to use unary form of scope resolution operator with methods? [duplicate]

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;

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.

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;
}

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 your current directory or any element in your PATH environment variable, so if PATH=/usr/bin:/usr/local/bin, and you were "in" /tmp, then...
X11/xterm
...would happily run /tmp/X11/xterm if found, else /usr/bin/X11/xterm, 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.
Noteworthy differences:
1) shells tend to use the first match using the ordering in PATH, whereas C++ gives a compiler error when you've been ambiguous.
2) In C++, names without any leading scope can be matched in the current namespace, while most UNIX shells only do that if you put . in the PATH.
3) C++ always searches the global namespace (like having / implicitly your PATH).
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....

:: 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.

:: 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();

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
// ...
}

(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.

:: 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.

"::" represents scope resolution operator.
Functions/methods which have same name can be defined in two different classes. To access the methods of a particular class scope resolution operator is used.

What is the use of " :: " in Qt? [duplicate]

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;

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.

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;
}

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 your current directory or any element in your PATH environment variable, so if PATH=/usr/bin:/usr/local/bin, and you were "in" /tmp, then...
X11/xterm
...would happily run /tmp/X11/xterm if found, else /usr/bin/X11/xterm, 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.
Noteworthy differences:
1) shells tend to use the first match using the ordering in PATH, whereas C++ gives a compiler error when you've been ambiguous.
2) In C++, names without any leading scope can be matched in the current namespace, while most UNIX shells only do that if you put . in the PATH.
3) C++ always searches the global namespace (like having / implicitly your PATH).
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....

:: 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.

:: 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();

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
// ...
}

(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.

:: 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.

"::" represents scope resolution operator.
Functions/methods which have same name can be defined in two different classes. To access the methods of a particular class scope resolution operator is used.

class names after namespace in C++

New to C++
there is a namespace i.e. and right after it a couple of class names
namespace abc {
class Cursor;
class BufferAllocator;
....
....
}
What does the above class declaration of Cursor and BufferAllocator do here?

It simply means "these classes exists" in the namespace abc, without providing any informations on their implementations.
It's called forward declarations.
It can be useful for :
Avoiding cycles in header inclusions (When class A has a member of class B, and class B has a member of class A)
Reducing dependencies between classes (because you can have a member pointer to a forward-declared class, but can't have directly a member, as the compiler doesn't know what's the size of the class without its implementation details, but know the size of a pointer). This is used notably in the Pimpl idiom.
(There might be other uses for this, but these are the most obvious that come to mind).

It's a forward declaration. It tells the following code that "there is a class called Cursor. You don't need to know what's in it [because we're only using it as a pointer or reference in the code, until it has been defined]".

Cursor and BufferAllocator are simply being forward-declared in their namespace (so they can be used in pointer/reference contexts).

It's a forward declaration. It can be used to inform the compiler of the existence of types when you're only going to use a pointer or reference to that type. The size of a pointer or reference is invariant of the type that it refers to, so the compiler doesn't need to see the entire definition of the type in that case; it just needs to know that the type exists first.
This can be useful in cases where the header that normally declares the type is large (think headers that include a lot of declarations or template instantiations), in which case it can decrease your compile times (sometimes significantly). You can just forward-declare the type and skip including the header, so your compiler doesn't need to process it.

namespace are helpful in a way they avoid typing particular classname in front of every function.
As you are new you will mostly see using namespace std;
so now you can use cout directly if you do not use this statement then you have to write std::cout for every use of cout
hope this helps

What is the meaning of prepended double colon "::"?

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;

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.

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;
}

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 your current directory or any element in your PATH environment variable, so if PATH=/usr/bin:/usr/local/bin, and you were "in" /tmp, then...
X11/xterm
...would happily run /tmp/X11/xterm if found, else /usr/bin/X11/xterm, 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.
Noteworthy differences:
1) shells tend to use the first match using the ordering in PATH, whereas C++ gives a compiler error when you've been ambiguous.
2) In C++, names without any leading scope can be matched in the current namespace, while most UNIX shells only do that if you put . in the PATH.
3) C++ always searches the global namespace (like having / implicitly your PATH).
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....

:: 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.

:: 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();

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
// ...
}

(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.

:: 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.

"::" represents scope resolution operator.
Functions/methods which have same name can be defined in two different classes. To access the methods of a particular class scope resolution operator is used.