C++ from the Ground Up in Visual Studio .NET

Generator DataMatrix in Visual Studio .NET C++ from the Ground Up

C++ from the Ground Up
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values of i and j: "; << j << '\n'; call swap() with addresses of i and j values of i and j: "; << j << '\n';
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The output from the program is shown here:
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Initial values of i and j: 10 20 Swapped values of i and j: 20 10
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In this example, the variable i is assigned the value 10, and j the value 20 Then swap( ) is called with the addresses of i and j The unary operator & is used to produce the addresses of the variables Therefore, the addresses of i and j, not their values, are passed into the function swap( ) When swap( ) returns, i and j will have their values exchanged
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Reference Parameters
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A reference parameter automatically receives the address of its corresponding argument
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While it is possible to achieve a call-by-reference manually by using the pointer operators, this approach is rather clumsy First, it compels you to perform all operations through pointers Second, it requires that you remember to pass the addresses (rather than the values) of the arguments when calling the function Fortunately, in C++, it is possible to tell the compiler to automatically use call-by-reference rather than call-by-value for one or more parameters of a particular function You can accomplish this with a reference parameter When you use a reference parameter, the address (not the value) of an argument is automatically passed to the function Within the function, operations on the reference parameter are automatically de-referenced, so there is no need to use the pointer operators A reference parameter is declared by preceding the parameter name in the function s declaration with an & Operations performed on a reference parameter affect the argument used to call the function, not the reference parameter itself
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Functions, Part Two: References, Overloading, and Default Arguments
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To understand reference parameters, let s begin with a simple example In the following, the function f( ) takes one reference parameter of type int:
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// Using a reference parameter #include <iostream> using namespace std; void f(int &i); int main() { int val = 1; cout << "Old value for val: " << val << '\n'; f(val); // pass address of val to f() cout << "New value for val: " << val << '\n'; return 0; } void f(int &i) { i = 10; // this modifies calling argument }
This program displays the following output:
Old value for val: 1 New value for val: 10
Pay special attention to the definition of f( ), shown here:
void f(int &i) { i = 10; // this modifies calling argument }
Notice the declaration of i It is preceded by an &, which causes it to become a reference parameter (This declaration is also used in the function s prototype) Inside the function, the following statement
i = 10;
does not cause i to be given the value 10 Instead, it causes the variable referenced by i (in this case, val) to be assigned the value 10 Notice that this statement does not use the * pointer operator When you use a reference parameter, the C++ compiler automatically knows that it is an address (ie, a pointer) and de-references it for you In fact, using the * would be an error
C++ from the Ground Up
Since i has been declared as a reference parameter, the compiler will automatically pass f( ) the address of any argument it is called with Thus, in main( ), the statement
f(val); // pass address of val to f()
passes the address of val (not its value) to f( ) There is no need to precede val with the & operator (Doing so would be an error) Since f( ) receives the address of val in the form of a reference, it may modify the value of val To illustrate reference parameters in actual use and to fully demonstrate their benefits the swap( ) function is rewritten using references in the following program Look carefully at how swap( ) is declared and called
#include <iostream> using namespace std; // Declare swap() using reference parameters void swap(int &x, int &y); int main() { int i, j; i = 10; j = 20; cout << cout << swap(j, cout << cout << "Initial i << ' ' i); "Swapped i << ' ' values of i and j: "; << j << '\n'; values of i and j: "; << j << '\n';
return 0; } /* Here, swap() is defined as using call-by-reference, not call-by-value Thus, it can exchange the two arguments it is called with */ void swap(int &x, int &y) { int temp; temp = x; // save the value at address x x = y; // put y into x y = temp; // put x into y }
Notice again that by making x and y reference parameters, there is no need to use the * operator when exchanging values As explained, it would be an error to do so
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