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Understanding Pointers in C: Concepts and Applications

This lecture, part of the ECE Application Programming course by Dr. Michael Geiger, focuses on pointers in C programming. It covers essential concepts such as pointer declaration, initialization, dereferencing, and passing arguments by address. Learn through examples illustrating the use of functions to manipulate data via pointers, including returning multiple values. Additionally, reminders for upcoming programming assignments and exams are discussed. This lecture serves as a vital resource for understanding how pointers enhance functionality in programming.

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Understanding Pointers in C: Concepts and Applications

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  1. 16.216ECE Application Programming Instructor: Dr. Michael Geiger Fall 2011 Lecture 22: Pointers

  2. Lecture outline • Announcements/reminders • Coming up: • PE4 on Monday • Program 6 to be posted Monday, due Monday 11/7 • Program 5 grading to be done this weekend; regrades due by Tuesday 11/1 • Exam 2: Wednesday 11/9 • Midterm grades posted only for those at risk • FN—Failing (never attended) • IDF—In danger of failing • Today • Pointers, pointer arguments ECE Application Programming: Lecture 22

  3. Review: functions • Used to break programs into smaller pieces • Useful when code sequences repeated • Functions have: • An optional return value • Return type is always required • If function returns nothing, return type is void • A name • Optional arguments • Must be prototyped or written completely prior to use ECE Application Programming: Lecture 22

  4. Example 1 • What does the following print? int f(int a, int b); int main() { int x = 1; int y = 2; int result1, result2, result3; result1 = f(x, y); result2 = f(y, result1); result3 = f(result1, result2); printf("x = %d, y = %d\n", x, y); printf("Result 1: %d\n", result1); printf("Result 2: %d\n", result2); printf("Result 3: %d\n", result3); return 0; } int f(int a, int b) { int i; // Loop index int r = 0; // Result for (i = 0; i < a; i++) r += b; return r; } ECE Application Programming: Lecture 22

  5. Example 1 solution x = 1, y = 2 Result 1: 2 Result 2: 4 Result 3: 8 ECE Application Programming: Lecture 22

  6. Example 2: Functions • Write a function that: • Prints a series of LINE_LENGTH dashes on a single line • LINE_LENGTH is a predefined constant (using #define) • Reads an integer value from the console input and returns 1 if the value is even, 0 if it’s odd • Takes four double-precision numbers as arguments and returns their average ECE Application Programming: Lecture 22

  7. Example 2 solutions • Write a function that: prints a series of LINE_LENGTH dashes on a single line • LINE_LENGTH is a predefined constant (using #define) void printLine() { inti; for (i = 0; i < LINE_LENGTH; i++) printf(“-”); } ECE Application Programming: Lecture 22

  8. Example 2 solutions (cont.) • Write a function that: reads an integer value from the console input and returns 1 if the value is even, 0 if it’s odd intcheckEven() { int value; printf(“Enter integer value: ”); scanf(“%d”, &value); if ((value % 2) == 0) return 1; else return 0; } ECE Application Programming: Lecture 22

  9. Example 2 solutions (cont) • Write a function that: takes four double-precision numbers as arguments and returns their average double avgFour(double a, double b, double c, double d) { return (a + b + c + d) / 4.0; } ECE Application Programming: Lecture 22

  10. Justifying pass by address • May want the ability to “return” multiple values from function • Functions can only return at most one value • Functions can take multiple arguments ... • ... but, as we’ve discussed so far, passing by value just copies arguments • No way to change arguments and have change reflected outside of function • Solution uses pointers ECE Application Programming: Lecture 22

  11. Pointers • Pointer: address of a variable • Can get address of existing object using & • Can get value of existing pointer using * • Pointer declaration: <base type>* <pointer name> • Base type determines how reference is interpreted • Be careful when declaring multiple pointers • What types do p1, p2, and p3 have below? • int *p1, p2, p3; • Be sure to initialize pointer before use • Always good to set pointer = NULL until it’s used • *p doesn’t make sense if you don’t know where p points! ECE Application Programming: Lecture 22

  12. Basic pointer example int *iPtr, i=6; double *dPtr, d=1.25; iPtr dPtr i d 6 1.25 ECE Application Programming: Lecture 22

  13. x xp ip 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; ECE Application Programming: Lecture 22

  14. x xp ip Dereferencing pointers • Dereference: access variable using pointer • Use the * operator • Expanding previous example: int x, *xp, *ip; xp = &x; ip = xp; *xp = 7; // x == 7 *ip = 18; // x == 18 ECE Application Programming: Lecture 22

  15. Example: using pointers • What does the following print? void main() { int x = 7, y = 10; int *xp = &x; int *yp = &y; int *p; int i; for (i = 0; i < 5; i++) { (*xp)++; (*yp)--; if ((i % 3) == 0) p = xp; else p = yp; *p = *xp + *yp; } printf("x = %d, y = %d\n", x, y); } ECE Application Programming: Lecture 22

  16. Example solution Walk through all 5 loop iterations: First iteration (i == 0): (*xp)++  x++  x = 7+1 = 8 (*yp)--  y--  y = 10-1 = 9 (i % 3) == 0  p = xp  *p = *xp + *yp  x = x + y = 8 + 9 = 17 Second iteration (i == 1): (*xp)++  x++  x = 17+1 = 18 (*yp)--  y--  y = 9-1 = 8 (i % 3) != 0  p = yp  *p = *xp + *yp  y = x + y = 18 + 8 = 26 ECE Application Programming: Lecture 22

  17. Example solution (cont) Walk through all 5 loop iterations: Third iteration (i == 2): (*xp)++  x++  x = 18+1 = 19 (*yp)--  y--  y = 26-1 = 25 (i % 3) != 0  p = yp  *p = *xp + *yp  y = x + y = 25 + 19 = 44 Fourth iteration (i == 3): (*xp)++  x++  x = 19+1 = 20 (*yp)--  y--  y = 44-1 = 43 (i % 3) == 0  p = xp  *p = *xp + *yp  x = x + y = 20 + 43 = 63 ECE Application Programming: Lecture 22

  18. Example solution (cont) Walk through all 5 loop iterations: Fifth iteration (i == 4): (*xp)++  x++  x = 63+1 = 64 (*yp)--  y--  y = 43-1= 42 (i % 3) != 0  p = yp  *p = *xp + *yp  y = x + y = 64 + 42 = 106 Final output x = 64, y = 106 ECE Application Programming: Lecture 22

  19. Pointer arguments • Passing pointer gives ability to modify data at that address • In prototype/definition—argument has pointer type • For example: int f(int *addr_x); • When calling function, can pass explicit pointer or use address operator (&<var>) • Examples: int x = 3; int y = 2; int *xPtr = &x; int result1, result2; result1 = f(xPtr); result2 = f(&y); ECE Application Programming: Lecture 22

  20. Functions - pass by address #include <stdio.h>#include <math.h>void get_r_theta(double a, double b, double *adr_r, double *adr_th);void main(){ double x,y,h,r,th; printf("Enter x, y components of vector: "); scanf("%lf %lf",&x,&y); get_r_theta(x,y,&r,&th); printf("Vector with x=%lf and y=%lf has r=%lf, theta=%lf\n",x,y,r,th); } void get_r_theta(double a, double b, double *adr_r, double *adr_th) { double sum; sum = pow(a,2)+pow(b,2); //or a*a+b*b; *adr_r = sqrt(sum); *adr_th = atan2(y,x);} x 4600 y 4608 r 4610 th 4618 ECE Application Programming: Lecture 22

  21. Functions - pass by address #include <stdio.h>#include <math.h>void get_r_theta(double a, double b, double *adr_r, double *adr_th);void main(){ double x,y,h; printf("Enter x, y components of vector: "); scanf("%lf %lf",&x,&y); // user enters 3,4 get_r_theta(x,y,&r,&th); printf("Vector with x=%lf and y=%lf has r=%lf, theta=%lf\n",x,y,r,th); } void get_r_theta(double a, double b, double *adr_r, double *adr_th) { double sum; sum = pow(a,2)+pow(b,2); //or a*a+b*b; *adr_r = sqrt(sum); *adr_th = atan2(y,x);} x 3.0 4600 y 4.0 4608 r ? 4610 th ? 4618 ECE Application Programming: Lecture 22

  22. Functions - pass by address #include <stdio.h>#include <math.h>void get_r_theta(double a, double b, double *adr_r, double *adr_th);void main(){ double x,y,h; printf("Enter x, y components of vector: "); scanf("%lf %lf",&x,&y); // user enters 3,4 get_r_theta(x,y,&r,&th); printf("Vector with x=%lf and y=%lf has r=%lf, theta=%lf\n",x,y,r,th); } void get_r_theta(double a, double b, double *adr_r, double *adr_th) { double sum; sum = pow(a,2)+pow(b,2); //or a*a+b*b; *adr_r = sqrt(sum); *adr_th = atan2(b,a);} x 3.0 4600 y 4.0 4608 r 4610 th 4618 a 3.0 7380 b 4.0 7385 adr_r 4610 7388 adr_th 4618 738c sum 7380 ECE Application Programming: Lecture 22

  23. Functions - pass by address #include <stdio.h>#include <math.h>void get_r_theta(double a, double b, double *adr_r, double *adr_th);void main(){ double x,y,h; printf("Enter x, y components of vector: "); scanf("%lf %lf",&x,&y); // user enters 3,4 get_r_theta(x,y,&r,&th); printf("Vector with x=%lf and y=%lf has r=%lf, theta=%lf\n",x,y,r,th); } void get_r_theta(double a, double b, double *adr_r, double *adr_th) { double sum; sum = pow(a,2)+pow(b,2); //or a*a+b*b; *adr_r = sqrt(sum); *adr_th = atan2(b,a);} x 3.0 4600 y 4.0 4608 r ? 4610 th ? 4618 a 3.0 7380 b 4.0 7385 adr_r 4610 7388 adr_th 4618 738c sum 25.0 7380 ECE Application Programming: Lecture 22

  24. Functions - pass by address #include <stdio.h>#include <math.h>void get_r_theta(double a, double b, double *adr_r, double *adr_th);void main(){ double x,y,h; printf("Enter x, y components of vector: "); scanf("%lf %lf",&x,&y); // user enters 3,4 get_r_theta(x,y,&r,&th); printf("Vector with x=%lf and y=%lf has r=%lf, theta=%lf\n",x,y,r,th); } void get_r_theta(double a, double b, double *adr_r, double *adr_th) { double sum; sum = pow(a,2)+pow(b,2); //or a*a+b*b; *adr_r = sqrt(sum); *adr_th = atan2(b,a);} x 3.0 4600 y 4.0 4608 r 5.0 4610 th ? 4618 a 3.0 7380 b 4.0 7385 adr_r 4610 7388 adr_th 4618 738c sum 25.0 7380 ECE Application Programming: Lecture 22

  25. Functions - pass by address #include <stdio.h>#include <math.h>void get_r_theta(double a, double b, double *adr_r, double *adr_th);void main(){ double x,y,h; printf("Enter x, y components of vector: "); scanf("%lf %lf",&x,&y); // user enters 3,4 get_r_theta(x,y,&r,&th); printf("Vector with x=%lf and y=%lf has r=%lf, theta=%lf\n",x,y,r,th); } void get_r_theta(double a, double b, double *adr_r, double *adr_th) { double sum; sum = pow(a,2)+pow(b,2); //or a*a+b*b; *adr_r = sqrt(sum); *adr_th = atan2(b,a);} x 3.0 4600 y 4.0 4608 r 5.0 4610 th 36.87 4618 a 3.0 7380 b 4.0 7388 adr_r 4610 7390 adr_th 4618 7394 sum 25.0 7398 ECE Application Programming: Lecture 22

  26. Functions - pass by address #include <stdio.h>#include <math.h>void get_r_theta(double a, double b, double *adr_r, double *adr_th);void main(){ double x,y,h; printf("Enter x, y components of vector: "); scanf("%lf %lf",&x,&y); // user enters 3,4 get_r_theta(x,y,&r,&th); printf("Vector with x=%lf and y=%lf has r=%lf, theta=%lf\n",x,y,r,th); } void get_r_theta(double a, double b, double *adr_r, double *adr_th) { double sum; sum = pow(a,2)+pow(b,2); //or a*a+b*b; *adr_r = sqrt(sum); *adr_th = atan2(b,a);} x 3.0 4600 y 4.0 4608 r 5.0 4610 th 36.87 4618 ECE Application Programming: Lecture 22

  27. Next time • More pointer examples • PE4—functions ECE Application Programming: Lecture 22

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