qr code size in c# Functions, Part Two: References, Overloading, and Default Arguments in .NET framework

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Functions, Part Two: References, Overloading, and Default Arguments
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Remember, the compiler automatically generates the addresses of the arguments used to call swap( ), and automatically de-references x and y Let s review When you create a reference parameter, that parameter automatically refers to (ie, implicitly points to) the argument used to call the function Further, there is no need to apply the & operator to an argument Also, inside the function, the reference parameter is used directly; the * operator is not necessary or, in fact, correct All operations involving the reference parameter automatically refer to the argument used in the call to the function
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EMEMBER: When you assign a value to a reference, you are actually assigning that value to the variable that the reference is pointing to In the case of function parameters, this will be the variable used in the call to the function
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Declaring Reference Parameters
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When Bjarne Stroustrup wrote The C++ Programming Language (in which he first described C++) in 1986, he introduced a style of declaring reference parameters, which some other programmers have adopted In this approach, the & is associated with the type name rather than the variable name For example, here is another way to write the prototype to swap( ):
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void swap(int& x, int& y);
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As you can see, the & is immediately adjacent to int and not to x Further, some programmers also specify pointers by associating the * with the type rather than the variable, as shown here:
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float* p;
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These types of declarations reflect the desire by some programmers for C++ to contain a separate reference or pointer type However, the trouble with associating the & or * with the type rather than the variable is that, according to the formal C++ syntax, neither the & nor the * is distributive over a list of variables, and this can lead to confusing declarations For example, the following declaration creates one, not two, integer pointers
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int* a, b;
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C++ from the Ground Up
Here, b is declared as an integer (not an integer pointer) because, as specified by the C++ syntax, when used in a declaration, an * or an & is linked to the individual variable that it precedes, not to the type that it follows It is important to understand that, as far as the C++ compiler is concerned, it doesn t matter whether you write int *p or int* p Thus, if you prefer to associate the * or & with the type rather than the variable, feel free to do so However, to avoid confusion, this book will continue to associate the * and the & with the variable name that each modifies, rather than the type name
IP: The C language does not support references Thus, the only way to create a call-by-reference in C is to use pointers, as shown earlier in the first version of swap( ) When converting C code to C++, you will want to convert these types of parameters to references, where feasible
Returning References
A function can return a reference In C++ programming, there are several uses for reference return values You will see some of these later in this book when you learn about operator overloading However, reference return values have other important applications that you can use now When a function returns a reference, it returns an implicit pointer to its return value This gives rise to a rather startling possibility: The function can be used on the left side of an assignment statement! For example, consider this simple program:
// Returning a reference #include <iostream> using namespace std; double &f(); double val = 1000; int main() { double newval; cout << f() << '\n'; // display val's value newval = f(); // assign value of val to newval cout << newval << '\n'; // display newval's value
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