Practically speaking, a file for a C++ program ends with.cpp, but not.c.

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1 Chapter 8 C++ C++ is an object-oriented extension of C. Since you have all got, or will get, in touch with Java, an object oriented programming language, in CS2381 Data Structure and Intermediate Programming, we will not discuss the object-oriented part here. Practically speaking, a file for a C++ program ends with.cpp, but not.c. The way to compile a C++ program in turing is quite similar to what we have been doing with C. Instead of using cc, we use g++. For example, /users/faculty/zshen > g++ Hello.cpp Notice that, in C++, the main function has to return something, while it does not have to in C. 1

2 The syntactic difference C++ allows unformatted input, cin, and output, cout, through <iostream>, which is similar to <stdio.h> for C, in both its nature and location. /home/plymouth/zshen > more Hello.cpp #include <iostream> //for C++ //#include <stdio.h> //for c using namespace std; //? int main(){ //printf("hello, World\n"); cout << "Hello, World\n"; return 0; /home/plymouth/zshen > g++ Hello.cpp /home/plymouth/zshen >./a.out Hello, World Question: What about this using namespace std; stuff? 2

3 A bit more on name space You have to add a using directive in your program if you use the C++ header files to specify which name space you want to use for your program. Again, for this pre-internet language, various vendors use this notion to package their code. For example, if you use a function test() from two vendors, you can specify you want to use the one sent over by vendor1. using namespace vendor1::test(); If you always use the stuff as supplied by vendor1, you can say using namespace vendor1; In our case, nothing specific but the standard stuff. We will see an application example later. 3

4 A little data structure C++ allows us to declare a class, quite similar to Java, in a.h file. class customer { private: char name[max_name+1]; int age; public: void print(void); void read(void); We can then provide a definition of the read() function as follows, in the associated.cpp file: void customer::read(void){ cout << "Name? "; cin.get(name, MAX_NAME, \n ); cout << "Age? "; cin >> age; 4

5 A small example /home/plymouth/zshen > more customer.cpp #include <iostream> #define MAX_NAME 40 using namespace std; class customer { private: char name[max_name+1]; int age; public: void print(void); void read(void); ; //This :: is called a scope operator. void customer::print(void){ cout << "Name: :" << name << "\nage: " << age << \n ; void customer::read(void){ cout << "Name? "; cin.get(name, MAX_NAME, \n ); cout << "Age? "; cin >> age; int main(int argc, char *argv[]){ customer customer1; customer1.read(); cout << "\nthis is what you just entered\n"; customer1.print(); return 0; 5

6 Compile and run /home/plymouth/zshen > g++ customer.cpp /home/plymouth/zshen >./a.out Name? James Age? 54 This is what you just entered Name: :James Age: 54 Definitely not a surprise.... 6

7 Reference C++ provides yet another way to pass in parameters, reference, which is again the same as Java. Or, shall we say, that the stuff in Java follows suit in C++? For example, all the following output lines print out the same values of x. int x, &r=x, *p=&x; cout << x << \n ; cout << r << \n ; cout << *p << \n ; Here r is a reference, nickname(?) of x, while p points to x. 7

8 Watch out A reference is essentially a pointer in disguise. /home/zshen > more testref.cpp #include <iostream> using namespace std; int main(void){ int x = 1, y = 2, &r=x; cout << "x=" << x << " " << "y=" << y << " " << "r=" << r << "\n"; //change reference r=y; cout << "x=" << x << " " << "y=" << y << " " << "r=" << r << "\n"; /home/zshen > g++ testref.cpp /home/zshen >./a.out x=1 y=2 r=1 x=2 y=2 r=2 Thus, once a reference is initialized, it should not change. In Java, it could not change. Check out the differences in reference variables between Java and C++ on the course page. 8

9 A positive example We saw earlier how to swap in Chapter 5 with the help of pointers. We now see another example in C++ that swaps things with the help of references. void swap(int &a, int &b){ int temp; temp=a; a=b; b=temp; The following prints out 2 and 1, in this order. #include <iostream> using namespace std; int main(void){ int x = 1, y = 2; swap(x, y); cout << x << " " << y << "\n"; return 0; Notice that reference does not work in C. (Cf. Page9 of the pointer chapter) 9

10 Test it out /home/plymouth/zshen > more refswap.cpp #include <iostream> using namespace std; void swap(int&, int&); int main(void){ int x = 1, y = 2; cout << "before swap:\n"; cout << x << " " << y << "\n"; swap(x, y); cout << "after swap:\n"; cout << x << " " << y << "\n"; return 0; void swap(int& a, int& b){ int temp; temp=a; a=b; b=temp; /home/plymouth/zshen > g++ refswap.cpp /home/plymouth/zshen >./a.out before swap: 1 2 after swap:

11 Dynamic space allocation C++ provides a pair of operators to do the memory allocation and deallocation. double *p; p=new double;... delete p; If what you want is an array, we have to do something special. double *list list=new double[10];... delete [] list; The issue is that if you don t apply delete explicitly, C++ won t do an automatic garbage collection, which leads to a memory leak. 11

12 An example /home/plymouth/zshen > more animal.cpp #include <iostream> #include <string> using namespace std; int main( ) string *animals[3000]; bool run = true; int i=0, j=0, k=0, counter=0; for(i=0; i<3000 && run; i++){ cout << "Enter the name of an animal or \"exit\" to quit" << endl; animals[i]= new string; getline(cin, *animals[i]); counter++; if(*animals[i]=="exit"){ run = false; i--; cout << "This is what you just entered\n"; for(j=0; j<=counter-2; j++) cout << *animals[j] << "\t"; cout << "\n"; for(k=0; k<=counter-1; k++) delete animals[k]; // delete [] animals; Why not? return 0; 12

13 Cut it open... string *animals[3000] It declares an array of 3000 pointers to strings, as defined in string.h. animals[i]=new string; It gets a dynamically allocated space for a string type, and puts a pointer to such a space in animals[i]. if(*animals[i]=="exit") run = false; It will terminate the for loop if a string of exit was just entered. 13

14 for(j=0; j<=counter-2; j++) cout << *animals[j] << "\t"; cout << "\n"; The index for animals starts with 0, but counter starts with 1; and the last cell of animals keeps the string exit. The last string, other than exit is kept in animals[counter-2]. If exit is the only string entered, counter=1, the loop will not be run even once. for(k=0; k<=counter-1; k++) delete animals[k]; The array animals[3000] itself is statically allocated, thus can t be deallocated. But, all the string type elements actually gets allocated can be deallocated. All such space ends at the index counter-1, where the string exit is kept. 14

15 Let s try this out /home/plymouth/zshen > g++ animal.cpp /home/plymouth/zshen >./a.out Enter the name of an animal or "exit" to quit dog Enter the name of an animal or "exit" to quit cat Enter the name of an animal or "exit" to quit bunny Enter the name of an animal or "exit" to quit exit This is what you just entered dog cat bunny /home/plymouth/zshen > 15

16 How about a vector? It might be easier to work with, as compared with an array. /home/zshen > more stlvector.cpp #include <iostream> #include <vector> using namespace std; main(void){ int iindex; //Space for vector of int, called ivector vector <int> ivektor; //we will add the elements at the end of vector do { char crespond; //space for next element of the vector cout<<"next Element->"; int ielement; cin>>ielement; //Add the value of ielement at the back end //To remove the element from the back use pop_back ivektor.push_back(ielement); //How many have we got so far? cout<<"size of the vector is ="<<ivektor.size() <<endl 16

17 //How many could we get? <<"Capacity of the vector="<<ivektor.capacity( <<endl; //What to get more? cout<<"more elements yes(y)/no(n)->"; cin>>crespond; if((crespond == y ) (crespond== Y )) continue; //No more. Let s have a break break; while(true); //display all the elements from begin to end //Notice the empty loop body for(vector<int>::iterator i=ivektor.begin(); i!= ivektor.end(); cout<<*i++<<endl); cout<<endl; //display elements from end to begin for(vector<int>::reverse_iterator r=ivektor.rbegin(); r!= ivektor.rend(); cout<<*r++<<endl); cout<<endl; 17

18 //Which one do you want? cout<<"you wish to see the element->"; cin>>iindex; //first method to display n-th element of a vector cout<<"at("<<iindex<<")=" <<ivektor.at(iindex)<<endl <<"or like this="; //Another approach to display one of vector elements //Notice that p is a pointer vector<int>::iterator p=ivektor.begin()+iindex; cout<<*p<<endl; cout <<"Good bye!" << endl; return 0; 18

19 Let s try this out /home/zshen > g++ stlvector.cpp /home/zshen >./a.out Next Element->11 Size of the vector is =1 Capacity of the vector=1 More elements yes(y)/no(n)->y Next Element->12 Size of the vector is =2 Capacity of the vector=2 More elements yes(y)/no(n)->y Next Element->13 Size of the vector is =3 Capacity of the vector=4 More elements yes(y)/no(n)->n You wish to see the element->1 At(1)=12 or like this=12 Good bye! Question: How does it change its capacity? 19

20 C, C++ and Java 1. C is a minimal language, while both C++ and Java are significantly more extensive. 2. C is a procedural language, while both C++ and Java are object oriented. You still have a choice with C++, and you are stuck with Java. 3. C is really fast, Java is not: interpreter, and some overhead such as garbage collection. 4. C gets to the bottom, via, e.g., bitwise operators and malloc; while Java is aiming high. 5. Java is a language of the Internet stage, neither C nor C++ is. Particularly, Java is portable, but not C++ (C). 6. C++, actually a bit extension of C as we know it, is the basis of ros as we will do with the Robotics program. You should be ready. 20

21 A final lab (MoAL?) In parallel processing, it is required that data as kept in a complete binary tree T h (?) be retrieved in parallel without memory conflicts. To make a long story short, if we use color to indicate the number of data cells, we can implement such an allocation strategy by assigning colors to data in the following way: Assume that nodes in T h 1 are colored using colors from the set {0,1,..., h 1, then we color the root node in T h with h. While the color of a node in LT h 1, the left subtree of T h,, is the same as its color in T h 1, that of a node in RT h 1 will be the opposite of that in T h 1. As a result, the most frequently used color in LT h 1 will be the least frequently used one in RT h 1. 21

22 Mathematically speaking... Technically, c R (u), the color of the node u in RT h 1, is h 1 c L (u),where c L (u) is its color in LT h 1. Below shows an example of T 2, where h = 2. If we always assign h to the root of T h, let s i, i [0, h 1], be the number of nodes in T h, with color i, the frequency sequence, f h, is the following: f 0 = 1, and let f h 1 be (s 0, s 1,..., s h 1 ), f h = ( 1, sorted sequence of ( s i + s h 1 i ) h 1 i=0 For example, the sequence for T 2 is (1,3,3). 22 ).

23 Generate f h, h 0,... Write a program in C++ to generate a sequence f(h), h [1, 20], which is defined as follows: f(0) = (1), and, for a given f(h), h [1,20], generate f(h) as follows: 1. reverse f(h); 2. add up f(h) and its reversal, item by item; 3. sort the result obtained for the previous step; 4. finally, add a 1 at the front. Thus, for h = 8, your program should print out f(8) as follows: (1, 47, 47, 65, 65, 65, 65, 78, 78) 23