1. ECE 264 Object-Oriented Software Development Instructor: Dr. Honggang Wang Fall 2012 Lecture 22: Pointers

2. Midterm Survey Feedback Satisfied with Exam 1 format and difficulty level  Exam 2 will follow the similar format and same level of difficulties 11/14 Wednesday, 9:00-9:55 am  Exam 3 will also follow the similar format and same level of difficulties 12/10 Monday, 9:00-9:55 am; does not need 3 hour exam in the final week (No exam after 12/10) Ask you to focus on the labs and project Pace of class: a little bit slow  Take care of students’ diversity  The scheduled classes will be completed on time Post Lab on Monday afternoon, Due on next Tuesday  More time to prepare the lab  The difficulty levels of remaining Labs are gradually increased 12/16/2015 ECE 264: Lecture 22 2

3. Lecture outline Announcements / reminders  Lab 6 is due 10/31 (new date) Note that project design is due 11/19 (Monday) In Lab 7, you will be playing with Microsoft Visio for UML design (Due 11/07, Wednesday) Today  Pointers Basic usage Pointer arithmetic Common pointer problems Passing arguments by reference 12/16/2015 ECE 264: Lecture 22 3

4. Dynamic memory allocation Up until now, allocated memory statically  Assumed we knew data size at compile time What if data size is input-dependent and unknown until run time?  In C, dynamic memory allocation handled through malloc and free  In C++, we use new and delete 12/16/2015 ECE 264: Lecture 22 4

5. Refresher on pointers Allocators ( malloc, new ) return pointer to allocated space  Pointer: address of another object We implicitly use these when we pass function arguments by reference in C++ Can get address of existing object using & Can get value of existing pointer using * Pointer declaration: *  Base type determines how reference is interpreted  Be careful when declaring multiple pointers  Be sure to initialize pointer before use 12/16/2015 ECE 264: Lecture 22 5

6. Pointer example int *iPtr, i=6; char* s, str[] = "example"; double *dPtr, d=1.25; 12/16/2015 ECE 264: Lecture 22 6 iPtr s dPtr 6 "example" 1.25 i str d

7. Pointer assignment The assignment operator (=) is defined for pointers of the same base type. The right operand of the assignment operator can be any expression that evaluates to the same type as the left operand. Example: int x, *xp, *ip; xp = &x; ip = xp; 12/16/2015 ECE 264: Lecture 22 7 x xp ip

8. Arrays and pointers Array name is a pointer to first array element  Can use pointers and arrays interchangeably You can use [] to “index” a pointer Example: char myString[] = "This is a string"; char *str; str = myString; for(int i =0; str[i]; i++)//look for null cout << str[i];  What does this print? 12/16/2015 ECE 264: Lecture 22 8

9. Initialize the Pointer Both define the pointer student and initialize student to the address of the first element in section: int section[80]; int *student = section; Is equivalent to int section[80]; int *student = &section[0]; The pointer string is initialized to point to the character a in the string "abcd". char *string = "abcd"; 12/16/2015 ECE 264: Lecture 22 9

10. Pointer arithmetic When using pointers/arrays interchangeably, can make use of pointer arithmetic  Can’t change where array name points, but you can change pointer  If p is a pointer, p++ means “point to next element” “Next element” determined by base type  Can compare pointers p == NULL  pointer points nowhere p == q  p and q point to same location  Example int num[4] = {1,2,3,4}, *p; p = num;//same as p = &num[0]; cout << *p <<endl; ++p; cout << *p; 12/16/2015 ECE 264: Lecture 22 10

11. Practice! 12/16/2015 ECE 264: Lecture 22 11 int q=6; int *iPtr = &q; cout << "iPtr is " << iPtr << endl; cout << "*iPtr is " << *iPtr << endl; cout << "++*iPtr, is " << ++*iPtr << endl; cout << "q is " << q << endl; cout << "iPtr is " << iPtr << endl; cout << "*iPtr++ is " << *iPtr++ << endl; cout << "iPtr is " << iPtr << endl; cout << "q is " << q << endl; Complete the output: iPtr is 0x7fff2f14

12. 12/16/2015 ECE 264: Lecture 22 12 Result of Practice iPtr is 0x7fff2f14 *iPtr is 6 ++*iPtr is 7 q is 7 iPtr is 0x7fff2f14 *iPtr++ is 7 iPtr is 0x7fff2f18 q is 7

13. Practice #2 char myString[ ] = "This is a string"; char *strPtr; strPtr = myString; cout << *myString << endl; cout<<myString << endl; cout << *(myString + 1) << endl; strPtr++; cout << *++strPtr << endl; myString++;//not legal What does this print? 12/16/2015 ECE 264: Lecture 22 13 0xfff4c252 strPtr 0xfff4c252 myString Thisisastring\0

14. Practice #2 result 12/16/2015 ECE 264: Lecture 22 14 T This is a string h i

15. Common Pointer Problems Using uninitialized pointers int *iPtr; *iPtr = 100;  iPtr has not been initialized. The value 100 will be assigned to some memory location. Which one determines the error. Incorrect/unintended syntax. 12/16/2015 ECE 264: Lecture 22 15

16. Example #include int main() { char *aString = "What happens here?"; int len=0; while(*aString++ != '\0') len++; std::cout << len << ": " << aString; return 0; } Does this compile? If not, why? If it does, what is the output? Explain 12/16/2015 ECE 264: Lecture 22 16

17. Final notes Next time  Start dynamic allocation (time permitting) Acknowledgements: this lecture borrows heavily from lecture slides provided with the following texts: Deitel & Deitel, C++ How to Program, 8 th ed. Etter & Ingber, Engineering Problem Solving with C++, 2 nd ed. 12/16/2015 ECE 264: Lecture 22 17