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Computer Science II " Lecture 7O =C++ Structures & PointersWagner Truppel Lecturer, Dept. of Computer Science & Engineering UC Riverside wagner@cs.ucr.edu http://www.cs.ucr.edu/~wagner http://www.cs.ucr.edu/cs12O  O`a~9Today s TopicsNC++ Structures Pointers Static and Dynamic Memory Allocation Dynamic arraysOO 4C++ structuresvWe ve seen Built-in data types int, float, double, bool, char, etc They are very simple Arrays Also  built-in (kind of& ) Collection of same-type elements Is it possible to have a collection of different-type elements that we can treat as a single entity ? Yes! But why would you need or want a creature like that ? ZZ9ZZ<ZZ <  ' q,5C++ structuresIs it possible to have a collection of different-type elements that we can treat as a single entity? Yes! But why would you need or want a creature like that ? Easier explained with an example Suppose you have to write an employee management program You need to have some data structure to store employees and their data Each employee has First and last names Social Security Numbers Address and phone numbers Salary, etc Z!ZZSZZ' q!S6C++ structureskEach employee has First and last names Social Security Numbers Address and phone numbers Salary, etc These are data of different types Names are strings SSNs are integer numbers Addresses are strings Phone numbers are integer numbers Salary are real numbers It would be nice to be able to refer to an employee and his/her information directly, as a single entity ZSZ"Z{ZjZS"{i7C++ structuresIt would be nice to refer to an employee and his/her information directly, as a single entity Example: instead of having An array of first names An array of last names An array of SSNs An array of addresses An array of phone numbers An array of salaries It s far more convenient to have simply An array of Employees C++ lets you do that with something called a structure yZZ(ZZ8Zy(78C++ structuresBstruct Employee { std::string firstName; std::string lastName; int ssNumber; Address address; int phoneNumber; double salary; }; // ` do not forget this  ; struct Address { std::string street; int number; std::string city; std::string state; int zip; }; // ` do not forget this  ; ZlZZll  2$:C++ structuresNote: you can have a struct inside of another struct How do you access a struct s member variable? Employee wagner; wagner. firstName =  Wagner ; wagner.lastName =  Truppel ; wagner.ssNumber = 123456789; // not really& wagner.phoneNumber = 9095551234; // not really& wagner.address.zip = 92507; wagner.address.state =  CA ; Why is the semi-colon  ; required after the { } ? Because you can declare structure variables at the same time you define your structures eZZ3ZYZR 3W       ;C++ structuresWhy is the semi-colon  ; required after the { } ? Because you can declare structure variables at the same time you define your structures struct Employee { std::string firstName; std::string lastName; int ssNumber; Address address; int phoneNumber; double salary; } wagner, brian, peter; The above declares 3 Employee variables but their member variables haven t been assigned yet.3ZZZZlZZ^Z3Wl@   h<C++ structuresFunctions can take structures as arguments ! using call by value: int getSSNumber(Employee emp) { return emp.ssNumber; } using call by reference: void setSalary(Employee & emp, double salary) { emp.salary = salary; } What s the difference once again ? Functions can return structures too ! Address getAddress(Employee & emp) { return emp.address; } @-ZZ7ZZGZ$Z&Z<Z-   &$  &    3    A  "  f    =C++ structures Initializing structures Similar to how you can initialize static arrays struct Date { int month; // 1 = Jan, & , 12 = Dec int day; // 1..31 int year; // 2000, 2001, etc }; Date myBirthday = { 3, 25, 1703 }; // yes, I am 300 years old !! :)Z1ZZRZJZ0 R+I" >C++ structures&Useful example: a stack of int s struct Stack { int top; // -1 indicates empty stack int array[MAX_STACK_SIZE]; }; void push(Stack & s, const int value) { s.array[++s.top] = value; } int pop(Stack & s) { return s.array[s.top--]; } These functions apply to any stack now !"ZZ@ZyZ*Z@ b!5&V@Pointers5A little review of how variables are stored in memoryAPointersCan we refer to a variable s address rather than to its value ? Yes ! The variable s address is stored in another variable, a pointer variable.HABPointersPointer variables are type-dependent int * px; px = &x; \\ px = 346 (sort of& ) char * pc; pc = &c; \\ pc = 350 (sort of& )\%W*+Z% =CPointersA pointer variable holds the address of the variable it points to (plus type information) Details are platform dependent What matters is that a pointer variable points to a (regular) variable int x = 123; // declares an int variable x and sets its value to 123 int * px; // declares a pointer variable px pointing to an int px = &x; // sets the pointer s target to the int variable xDZZ\%!* DPointersThe notation is confusing at first& int x; // declares a regular variable int *px; // declares a pointer variable px = &x; // sets the target of the pointer *px = 456; // sets the value of the variable pointed to by px (in this case, x) So& type * pointer declares a pointer variable pointing to a variable of type type &var returns the address of variable var (addressing operator) * pointer accesses the value of the variable pointed to by pointer (dereferencing operator)%ZZZZ$  3    %"!)7O \EPointersA surprising  but typical  example: what s the output of this block of code ? char c =  A ; std::cout << c << std::endl; char * p; p = &c; std::cout << *p << std::endl; *p =  R ; std::cout << *p << std::endl; std::cout << c << std::endl;FQZZP e  FPointersYou re used to declaring and setting a regular variable in one step: int x = 123; You ve just learned to declare and then define a pointer variable, in two steps: int * p; p = &x; Can you declare and define a pointer variable in one single step too ? Yes, but it s a bit counter-intuitive& int * p = x; // is it like this ? EZ ZZQZZZGZJZE Q  G  J  HE[|GPointersYou re used to declaring and setting a regular variable in one step: int x = 123; You ve just learned to declare and then define a pointer variable, in two steps: int * p; p = &x; Can you declare and define a pointer variable in one single step too ? Yes, but it s a bit counter-intuitive& int * p = x; // is it like this ? No ! int * p = &x; // it s like this !EZ ZZQZZZGZpZE Q  G  p  PE[|#HPointersrMore tricky stuff& What happens when you write p1 = p2; ?0/Z Z/ IPointersrMore tricky stuff& What about when you write *p1 = *p2; ?0-Z Z- JPointers KThe new operator$ zWhat are pointers useful for anyway ? Recall that all variables you re used to defining so far (other than globals) are allocated in the stack area they re always local variables in some block of code or function How do you allocate variables in the heap area ? You use the operator new new always returns a pointer, so that s the answer: pointer variables are useful b/c they re the means by which you refer to variables allocated in the heapZAZZ& c A%  A@kKLThe new operator$ *int * p; // declares a pointer to an int p = new int; The above line of code: Allocates space in the heap for an int Makes p point to that int But does not set that int s value (*p is undefined) p = new int(123); // same but also sets the value of the new int *p equals 123OZcZZZAZZOJ A!>2MThe new operator$ 2You can use new to allocate any kind of variable struct Employee { std::string name; int ssn; }; Employee * empPtr = new Employee( John Doe , 123456789); How to access member variables ? int hisSSN = (*empPtr).ssn; // these two ways int hisSSN = empPtr->ssn; // are equivalent 2ZZZZZ;Z"ZZZ    ;  !    Z  2 H  N$Static and Dynamic Memory Allocation&%$$0d Regular variables are typically allocated in the stack area  always local to some block of code The variables created with new are always allocated in the heap area Variables allocated in the stack have a well defined life span  they get popped off the stack when the block of code ends Variables allocated in the heap exist until someone deletes them  it s your responsibility to delete them ! What if you don t ? Memory leaks !Z385 OThe delete operator$ You use delete to de-allocate a variable created with new Employee * empPtr = new Employee( John Doe , 123456789); delete empPtr; // now *empPtr is undefined The space in the heap previously reserved for the employee record above is now available for other variables That means: any other pointers to that same area are also undefined Undefined pointers are also called dangling pointers:ZhZZ(] y&.>F0 PDangling pointersIf you try this: int * p1 = new int(5); int * p2 = p1; // p2 points to the same int std::cout << *p1 << std::endl; // prints 5 std::cout << *p2 << std::endl; // also prints 5 delete p1; // that int is now gone ! std::cout << *p1 << std::endl; The result is unpredictable This will also be unpredictable: std::cout << *p2 << std::endl; C++ has no built-in method to determine if a pointer is danglingZZ>Z ZAZ C [ %   ! 7( $CDQMore on new & delete6R Regular variables are allocated in the stack area Dynamic variables are allocated in the heap using new, and deleted from the heap using delete What happens when you call new and there s no more space in the heap ? New fails& Older compilers: A failing new returns NULL Newer compilers following the standard definition of C++ : a failing new raises an exception (which, if not dealt with, terminates the program) we ll talk about exceptions later in the course NULL is a special value& any pointer can be set to NULL, regardless of the pointer type ZZZ;Z"Z4Z0ZXZZ   R;     5  0"mRNULL & dangling pointers(A good practice to decrease the chances of running into dangling pointers is to set all pointers to dynamic variables you re done with to NULL then you can test for NULL pointers If (p == NULL) { /* p is dangling */ } else { /* p is ok */ } Java is nicer than C++ when it comes to memory management It automatically deletes dynamic variables when no one points to them There s a price though: speed But you still need to test for dangling pointersPZdZ>ZZ:ZZP3  > :SMore syntax: typedef( It s often useful to create aliases ( nicknames ) for types that you create This is particularly useful for pointer variables typedef int* IntPtr; int* and IntPtr are identical type names and can be used in each other s place IntPtr p = new int(123); // completely int * p = new int(123); // equivalent This has two major advantages: You declare your pointer variables just as you declare regular variables (no * to worry about) The other has to do with function arguments (more on this later)~ZZOZNZZIZZAZ~OM   ~ ? ~WTPointers & functions(Functions can have pointer arguments Functions can return pointers too ! char* SomeFunction(char * p); Note that even though p is a pointer, it s passed by value This passes the pointer p by reference char* SomeFunction(char * & p); Confusing& used typedef insteadxIZ ZcZ"Z ZI c" >P v UPointers & functions(This passes the pointer p by reference char* SomeFunction(char * & p); Confusing& use typedef instead typedef char* CharPtr; char* SomeFunction(CharPtr & p); CharPtr SomeFunction(CharPtr & p);`'Z"ZZ^Z'"^-   VPointers & functions(Alertness check: what s the output of the code section below ? typedef char* CharPtr; void f(CharPtr pc) { *pc =  Z ; } CharPtr p = new char( A ); f(p); Std::cout << *p << std::endl; Note that pc is a call-by-value argument D?Z{Z*Z?{*t@ -WPointers & functions(typedef char* CharPtr; void f(CharPtr pc) { *pc =  Z ; } CharPtr p = new char( A ); f(p); Std::cout << *p << std::endl; Note that pc is a call-by-value argument, so p itself gets copied into the local variable pc inside the function Being a copy of p, that local variable still points to the same variable as p So the result is that the value of the variable pointed to by p changes from  A to  Z despite the fact that p is not passed by reference p never changed, but the value of its target did! RzZ~Zyl  XPointers & functions(typedef char* CharPtr; void f(CharPtr pc) { *pc =  Z ; } CharPtr p = new char( A ); f(p); Std::cout << *p << std::endl; "zZyl  YPointers & Static Arrays(Recall that array variables are interpreted as addresses to the first array element So array variables are pointer variables ! Well, not quite, but almost. For all practical purposes in CS 12, you can think of array variables as pointer variables int intA[100]; // static array ! int * p; // declares pointer to an int p = intA; // this is ok !!! p[0] = 7; That s the same as intA[0] = 7;ZJZZ(ZZZ J ( !t ZDynamic Arrays(You define a dynamic array the same way you define a pointer for a dynamic variable char * vowels = new char[5]; Compare with the static version char vowels[5]; Recall that I said once that functions cannot return arrays but that there s a way around it& Here it is: char* f(); This function returns a pointer to a char and that s what an array of chars is !TZZ ZZjZ ZRZT j &,{][Dynamic Arrays(4Since dynamic arrays are dynamic variables, how do you delete them from the heap ? Using the delete operator& But the syntax is a little different than what we saw before: delete p; // deletes the variable pointed to by the pointer p delete [] p; // deletes the array variable phZlZ^K 6 !\Pointer Arithmetic(Since a pointer is type dependent, it knows how large its underlying type is An array variable is a pointer variable so we can perform some arithmetic on the pointer variable itself float *speed = new float[10]; speed is the address of speed[0] speed + 1 = address of speed[1] & speed + 6 = address of speed[6] speed is an address addresses are numbers But (speed + n) does not follow the usual arithmetic: what gets added to the address in speed is n times the size of the base typeZ/PsxHH@Rg(HH(dh   ` www3ff` 3fff` ___>?" dF@,?n<d@uA @ " d`  n?" dd@   @@``PR   @ ` `P0p>> - 2003 WL Truppel ZCS 12: Intro. Computer Science II " Lecture 7 X( = 2  Ng?"! :    NPfgֳgֳ ?"`  T Click to edit Master title style! !<  Hfgֳgֳ ?"  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