C ++ reference

C ++ reference
Referenced concepts

Reference: it is an alias of a variable (target). The operations for reference are exactly the same as those for direct operations on variables.

Declared method: type identifier & reference name = target variable name;

The reference ra is defined as the alias of variable.

Int;

　　　　Int & ra =;

(1) & here it is not an address operator, but an identifier.

(2) type identifier refers to the type of the target variable.

(3) When declaring a reference, it must be initialized at the same time.

(4) After the reference declaration is complete, it is equivalent that the target variable has two names: the original name and reference name of the target, and the reference name cannot be used as the alias of other variable names.

(5) declaring a reference is not a new variable. It only indicates that the reference name is an alias of the target variable name. It is not a data type, therefore, the reference itself does not occupy storage units, and the system does not allocate storage units to the reference. Therefore, finding the address for the reference is to find the address for the target variable. & Ra and &.

(6) arrays cannot be referenced. An array is a collection composed of several elements, so an array alias cannot be created.

 1 #include<iostream.h> 2 void main(){ 3     int a=5; 4     int &b=a; 5     b=6; 6     cout<<"a="<<a<<",b="<<b<<endl;//a=6,b=6 7     int c=7; 8     b=c; 9     cout<<"a="<<a<<",b="<<b<<endl;//a=7,b=710 }

1 # include <iostream. h> 2 void main () {3 int a [] = {1, 2, 3, 4}; 4 int & B = a; 5 // compilation error: cannot convert from 'int [4] 'to 'int &' 6}

 

Referenced Application

1. Reference as a parameter

An important role of reference is as a function parameter. In the previous C language, function parameter transfer was a value transfer. If a large piece of data is used as a parameter transfer, a pointer is often used, because this avoids the whole piece of data from being pushed to the stack, it can improve program efficiency. But now (in C ++), another option that is equally efficient (which is required in some special cases) is reference.

 

1 # include <iostream. h> 2 // The parameter p1 and p2 of the function here reference 3 void swap (int & p1, int & p2) {4 int p = p1; 5 p1 = p2; 6 p2 = p; 7}

To call this function in a program, the call point of the corresponding main function can be directly called using the variable as the real parameter without the real parameter.
Variables have any special requirements. For example, the swap function defined above can be written as follows:
8 void main () {9 int a, B; 10 cin> a> B; // enter the values of variables a and B 11 swap (a, B ); // call the swap function 12 cout directly using a and B as real parameters <"a =" <a <", B =" <B <endl; 13}

 

When the above program is running, if the input data is 10 20 and then press enter, the output result is a = 20, B = 10.

As shown in the preceding example:

(1) passing a reference to a function has the same effect as passing a pointer. In this case, the parameter of the called function becomes an actual parameter variable or an alias of the object in the original main function, therefore, the operations on the parameters in the called function are the operations on the corresponding target object (in the main function.

　　(2) using the reference parameter to pass the function does not generate a copy of the real parameter in the memory, it is directly to operate on the real parameter; while using the general variable to pass the function parameter, when a function call occurs, you need to allocate storage units to the parameters. The parameters are copies of the real variables. if the object is passed, the copy constructor will also be called. Therefore, when the data transmitted by a parameter is large, it is better to use reference than to transmit a parameter using a common variable.

(3) Although using pointers as function parameters can achieve the same effect as using references, in the called function, storage units must be allocated to the parameters, the \ "* pointer variable name \" format must be used repeatedly for calculation, which can easily produce errors and the program reading is poor. On the other hand, at the call point of the main function, the variable address must be used as the real parameter. References are easier to use and clearer.

 

　　If you want to use references to improve program efficiency and protect the data transmitted to the function from being changed in the function, you should use regular references.

2. Frequent reference

Common Reference declaration method: const type identifier & reference name = target variable name;

The reference declared in this way cannot be modified by reference to the value of the target variable, so that the referenced target becomes const, achieving the security of reference.

1 # include <iostream. h> 2 void main () {3 int a = 1; 4 int & B = a; 5 B = 2; 6 cout <"a =" <a <endl; // 2 7 int c = 1; 8 const int & d = c; 9 // d = 2; // compilation error C2166: l-value specifies const object10 c = 2; // correct 11}

This is not just to make the code more robust, but also has other needs.

[Example 4]: assume that the following function declaration is available:

String foo ();

Void bar (string & s );

The following expression is invalid:

Bar (foo ());

Bar ("hello world ");

The reason is that both foo () and "hello world" strings generate a temporary object,In C ++, temporary objects are of the const type.. Therefore, the above expression tries to convert a const type object to a non-const type, which is invalid.

　　The referenced parameter should be defined as const if it can be defined as const.

 

3. Reference as return value

To return a function value by reference, the function must be defined in the following format:

　　　　Type identifier & function name (parameter list and type description)

{Function body}

Note:

(1) to reference the returned function value, you must add &

　　(2) The biggest benefit of returning a function value with a reference is that a copy of the returned value is not generated in the memory.

[Example 5] The following program defines a common function fn1 (which returns the function value using the return value method), and another function fn2, which returns the function value by reference.

1 # include <iostream. h> 2 float temp; // defines the global variable temp 3 float fn1 (float r); // declares the function fn1 4 float & fn2 (float r ); // declare the function fn2 r 5 float fn1 (float r) {// define the function fn1, which returns the function value 6 temp = (float) in the return value method) (r * 3.14); 7 return temp; 8} 9 float & fn2 (float r) {// define the function fn2, it returns the function value 10 temp = (float) (r * 3.14); 11 return temp; 12} 13 void main () as a reference () {14 float a = fn1 (10.0); // in 1st cases, the system generates a copy (I .e. temporary variable) 15 // float & B = fn1 (10.0 ); // In 2nd cases, errors may occur (different C ++ systems have different rules) 16/* 17 compilation errors: cannot convert from 'float' to 'float & '18 A reference that is not to 'const' cannot be bound to a non-lvalue19 */20 // cannot return reference of temporary or local variables 21 float c = fn2 (10.0 ); // In 3rd cases, the system does not generate a copy of the returned value 22 // you can return a reference to a global variable from the called function 23 float & d = fn2 (10.0 ); // In 4th cases, the system does not generate a copy of the returned value 24 cout <"a =" <a <", c =" <c <", d = "<d <endl; 25 // a = 314, c = 314, d = 31426}

The reference must follow the following rules as the return value:

(1)Local variables cannot be referenced.For details, refer to Item 31 of Objective C ++ [1.The main reason is that local variables will be destroyed after the function returns, so the returned reference becomes a reference of "no finger", and the program enters the unknown state.For example, a compilation error occurs in 2nd cases in [Example 5.

(2)You cannot return a reference to the memory allocated by the new function.For details, refer to Item 31 of Objective C ++ [1. Although there is no passive destruction of local variables, this situation (returning a reference to the memory allocated by the new function) faces other embarrassing situations. For example, if a reference returned by a function only appears as a temporary variable and is not assigned to an actual variable, the space pointed to by the reference (allocated by new) cannot be released, cause memory leak.

(3)You can return a reference to a class member, but it is best to use const.This principle can be referred to Item 30 of Objective C ++ [1. The main reason is that when an object attribute is associated with a business rule, its value assignment is often related to some other attributes or the state of the object, therefore, it is necessary to encapsulate the value assignment operation in a business rule. If other objects can obtain the non-constant reference (or pointer) of this attribute, a simple value assignment to this attribute will damage the integrity of business rules.

(4) References and reloads of some operators: stream operators <and>. These operators are often used consecutively. For example: cout <\ "hello \" <endl; therefore, the return value of these two operators should be a stream reference that still supports these two operators. Other optional solutions include returning a stream object and returning a stream object pointer. However, for a returned Stream object, the program must re-(copy) to construct a new stream object. That is to say, two consecutive <operators are actually for different objects! This is unacceptable. If a stream pointer is returned, the <operator cannot be used consecutively. Therefore, returning a stream object reference is the only choice. This unique choice is critical. It illustrates the importance of reference and is irreplaceable. Maybe this is why the concept is introduced in C ++. Value assignment operator =. This operator can be used continuously like a stream operator, for example, x = j = 10; or (x = 10) = 100; the return value of the value assignment operator must be a left value, so that the value can be assigned. Therefore, it is referenced as the only return value choice of this operator.

[Example 6] Return the referenced function value as the left value of the value assignment expression.

1 # include <iostream. h> 2 int & put (int n); 3 int vals [10]; 4 int error =-1; 5 void main () {6 put (0) = 10; // use the put (0) function value as the left value, which is equivalent to vals [0] = 10; 7 put (9) = 20; // use put (9) function value as the left value, equivalent to vals [9] = 20; 8 cout <vals [0] <endl; // 10 9 cout <vals [9] <endl; // 2010} 11 int & put (int n) {12 if (n> = 0 & n <= 9) 13 return vals [n]; 14 else {15 cout <"subscript error"; 16 return error; 17} 18}

(5) among other operators, the reference: +-*/four Arithmetic Operators cannot be returned. They cannot return references. The Objective C ++ [1] Item23 discusses this issue in detail. The main reason is that these four operators do not have side effect. Therefore, they must construct an object as the return value. Optional solutions include: returning an object and returning a reference to a local variable, returns the reference of a newly assigned object and a static object reference. According to the preceding three rules that reference the returned value, both the 2nd and 3 schemes are rejected. Static object reference is caused by errors because (a + B) = (c + d) is always true. Therefore, only one object is returned.

4. References and Polymorphism

　　References are another method that can produce polymorphism effects except pointers. This means that a base class reference can point to its derived class instance.

[Example 7 ]:

Class;

Class B: public {......}

B B;

A & Ref = B; // use A derived class object to initialize A reference to A base class Object

Ref can only be used to access the members inherited from the base class in a derived class object. It is a base class reference pointing to a derived class. If A virtual function is defined in Class A, and this virtual function is rewritten in Class B, A multi-state effect can be generated through Ref.

 

Reference Summary

(1) In reference use, it is meaningless to simply give an alias to a variable,The purpose of reference is to solve the problem of transmission efficiency and space unsatisfactory of large data or objects during function parameter transfer.

　　(2) passing function parameters by reference can ensure that no copies are generated during parameter transfer and improve the transfer efficiency. The use of const ensures the security of reference transmission.

　　(3) The difference between a reference and a pointer is that after a pointer variable points to an object, it indirectly operates on the variable it points. When pointers are used in a program, the program has poor readability. The reference itself is the alias of the target variable, and the reference operation is the operation of the target variable.

　　(4) When to use the reference. Stream operators <and>, return values of the value assignment operator =, parameters of the copy constructor, parameters of the value assignment operator =, and references are recommended in other cases.

 

Reprinted from: http://www.cnblogs.com/gw811/archive/2012/10/20/2732687.html