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Address-of Operator: &

& cast-expression

The unary address-of operator (&) takes the address of its operand. The operand of the address-of operator can be either a function designator or an l-value that designates an object that is not a bit field and is not declared with the register storage-class specifier.

The address-of operator can only be applied to variables with fundamental, structure, class, or union types that are declared at the file-scope level, or to subscripted array references. In these expressions, a constant expression that does not include the address-of operator can be added to or subtracted from the address-of expression.

When applied to functions or l-values, the result of the expression is a pointer type (an r-value) derived from the type of the operand. For example, if the operand is of type char, the result of the expression is of type pointer to char. The address-of operator, applied to const or volatile objects, evaluates to const type * or volatile type *, where type is the type of the original object.

When the address-of operator is applied to a qualified name, the result depends on whether the qualified-name specifies a static member. If so, the result is a pointer to the type specified in the declaration of the member. If the member is not static, the result is a pointer to the member name of the class indicated by qualified-class-name. (See Primary Expressions for more about qualified-class-name.) The following code fragment shows how the result differs, depending on whether the member is static:

// expre_Address_Of_Operator.cpp
// C2440 expected
class PTM {
public:
           int   iValue;
    static float fValue;
};

int main() {
   int   PTM::*piValue = &PTM::iValue;  // OK: non-static
   float PTM::*pfValue = &PTM::fValue;  // C2440 error: static
   float *spfValue     = &PTM::fValue;  // OK
}

In this example, the expression &PTM::fValue yields type float * instead of type float PTM::* because fValue is a static member.

The address of an overloaded function can be taken only when it is clear which version of the function is being referenced. See Address of Overloaded Functions for information about how to obtain the address of a particular overloaded function.

Applying the address-of operator to a reference type gives the same result as applying the operator to the object to which the reference is bound. For example:

// expre_Address_Of_Operator2.cpp
// compile with: /EHsc
#include <iostream>
using namespace std;
int main() {
   double d;        // Define an object of type double.
   double& rd = d;  // Define a reference to the object.

   // Obtain and compare their addresses
   if( &d == &rd )
      cout << "&d equals &rd" << endl;
}

&d equals &rd

The following example uses the address-of operator to pass a pointer argument to a function:

// expre_Address_Of_Operator3.cpp
// compile with: /EHsc
// Demonstrate address-of operator &

#include <iostream>
using namespace std;

// Function argument is pointer to type int
int square( int *n ) {
   return (*n) * (*n);
}

int main() {
   int mynum = 5;
   cout << square( &mynum ) << endl;   // pass address of int
}

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