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Nov 28th, 2005

Scope Resolution Operator

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What exactly is the purpose of the scope resolution operator? I can't seem to find anything in my book telling exactly what it does. It just seems to dance around the subject somewhat.

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kahaj

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Nov 28th, 2005

Re: Scope Resolution Operator

years ago before anyone heard about namespaces programmers had a terrible time avoiding name conflicts in their own code and the libraries they used. That was the main reason (I think) for the introduction of namespaces -- you can have as many objects with the same name as you wish as long as each object is contained in a different namespace. you will commonly see code such as

C++ Syntax (Toggle Plain Text)
std::cout << "Hello Worod" << std::endl;
std::cout << "Hello Worod" << std::endl;

This is identifying the namespace (std) of cout and endl. For objects in global namespace, such as all the win32 api functions just preceed the object with two colons as in the following code. This is really more useful when there is another function called MessageBox() in another namespace that your program also might use. Its just telling the compiler which function to call.

C++ Syntax (Toggle Plain Text)
  ::MessageBox(0,"Hello World","Message",MB_OK);
  ::MessageBox(0,"Hello World","Message",MB_OK);

c++ classes have the same affect as formal namespace

C++ Syntax (Toggle Plain Text)
class MyClass
{
   MyClass();
   static void SayHello() {std::cout << "Hello World" << std::endl;}
};
 
// call the SayHello method
MyClass::SayHello();
class MyClass
{
   MyClass();
   static void SayHello() {std::cout << "Hello World" << std::endl;}
};

// call the SayHello method
MyClass::SayHello();

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Nov 28th, 2005

Re: Scope Resolution Operator

Lets say you have a global variable and a local variable with the same name. When you compile it, the compiler will use the local variable, and not the global variable. If you prefix the variable name with ::[variablename], then the compile uses the global variable.

C++ Syntax (Toggle Plain Text)
#include <iostream>
 
using namespace std;
 
int amount = 123;             // A global variable
 
int main()
{
   int amount = 456;          // A local variable
   cout  << ::amount << endl  // Print the global variable
         << amount << endl;   // Print the local variable
}
#include <iostream>

using namespace std;

int amount = 123;             // A global variable

int main()
{
   int amount = 456;          // A local variable
   cout  << ::amount << endl  // Print the global variable
         << amount << endl;   // Print the local variable
}

You can do this even for function names.
Note that you can use this to specify only global variables. Not the variables in the next outermost scope.

C++ Syntax (Toggle Plain Text)
#include <iostream>
 
using namespace std;
 
int amount = 123;             // A global variable
 
int main()
{
	int amount = 1234 ; // the local variable in the outermost scope
	{
		int amount = 456; // A local variable
		cout  << "Innermost Scope" << endl ;
		cout  << ::amount << endl  // Print the global variable not the value with 1234
			<< amount << endl;   // Print the local variable in  current scope
	}
	cout  << "Outermost Scope" << endl ;
	cout  << ::amount << endl  // Print the global variable 
		<< amount << endl;   // Print the local variable in  current scope
 
}
#include <iostream>

using namespace std;

int amount = 123;             // A global variable

int main()
{
	int amount = 1234 ; // the local variable in the outermost scope
	{
		int amount = 456; // A local variable
		cout  << "Innermost Scope" << endl ;
		cout  << ::amount << endl  // Print the global variable not the value with 1234
			<< amount << endl;   // Print the local variable in  current scope
	}
	cout  << "Outermost Scope" << endl ;
	cout  << ::amount << endl  // Print the global variable 
		<< amount << endl;   // Print the local variable in  current scope

}

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Nov 28th, 2005

Re: Scope Resolution Operator

operator :: known as scope resolution operator has been introduced to access an item that is outside the current scope. This operator is also used in distiguishing class members and defining class methods.

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