Computer Programming for Engineering Applica4ons. Intro to Programming 10/2/13 ECE 175. Limita4ons of Reference by Value. The Concept of Pointers

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1 Computer Programming for Engineering Applica4ons ECE 175 Intro to Programming Limita4ons of Reference by Value Write a C func4on that does the following Argument: the radius of a circle r Output: circumference c and area a #include <stdio.h> #define PI 3.14 void circle(float r, float a, float c); int main(void) { float r, a=0, c=0; printf("enter the radius of the circle:"); scanf("%f",&r); circle(r, a, c); // arguments passed by value printf("area:%.1f Circumference:%.1f\n", a, c); return(0); } void circle(float r, float a, float c) { "a = PI*r*r; // computation of area "c = 2*PI*r; // computation of circumference } Reference by value does not allow the manipula4on of variables external to the func4on ECE The Concept of Pointers Pointer: A variable containing the address of another variable A label used to refer (point) to memory space Very frequently found in real life 138 Page numbers in a book, or powerpoint slides Page numbers are pointers to the contents of the par4cular page ECE

2 The Concept of Pointers Mail boxes at the post office Numbers are used to label each mailbox. The contents (levers) of a par4cular mailbox can be retrieved by referring to the mail box number. ECE Pointers in C Labels for referencing memory addresses Address a Contents ECE Why do we need pointers? Faster memory access and usage via memory referencing Cri4cal when C was developed S4ll required for programming resource- constrained devices such as microcontrollers Saving memory space No need to hold local copies of variables inside func4ons Can pass values to a func4on by reference Downside Pointers usually make program tracking/reading more complicated ECE

3 Declara4on of pointers Basic syntax Syntax: data type *name; Examples: FILE *inp; // pointer to a file " int *p; // pointer to an integer variable float *p; // pointer to a float variable We use the special character * to denote that a variable is a pointer Though pointers point to different data types, they are nothing more than memory addresses Compiler needs to know the data type for manipula4ng its contents ECE Pointers cont. There are two parts associated with a pointer: the address to the memory and the contents of the memory P var We can change both parts, that is, to which memory cell the pointer points (address), and the contents of the memory cell (value) ECE Changing the Address of a Pointer Use the address operator & Syntax: pointer = &var; Examples: int *p; // pointer to an integer int i; // an integer variable p = &i; // p points to variable i The type of pointer and type of the variable must match Example: if i is an integer variable, pointer p has to be of type integer ECE

4 Changing the contents of the memory Use the contents operator * Syntax: *pointer = value; variable = *pointer; Examples: int *p; // pointer to an integer *p = 10; // memory cell pointed by p is " " " assigned the value 10 var1 = *p; // variable var1 is assigned " " " the contents of memory cell " " " pointed by p Again, the pointer type and the value or variable must match ECE So Far we have silently used pointers To access files To scan for keyboard and file input FILE *inp; // declaration of a pointer of type FILE inp = fopen( filename, r ); // returns the address to the memory where the file is stored scanf( %d, &i); // scanf receives as input the address of variable i, and stores input in i ECE A Simple Example Using Pointers #include<stdio.h> int main(void) { int var1, var2; int *p; // declaration of a pointer of type integer p= &var1; // initialization of pointer to address of var1 *p=10; // initialization of var1 via pointer p var2 = *p+10; // var2 = var printf("%d %d %d\n", var1, var2, *p); printf("%p %p %p\n", p, &var1, &var2); // printing addresses return (0); } ECE

5 At Declara4on int var1, var2; int *p; // declaration of a pointer of type integer p: 0xffbfe5cc var1 556 var ECE Changing the address of point p= &var1; // initialization of pointer to address of var1 p: 0xffbff6cc var1 556 var2 345 Point is now poin4ng to the address of var1 The value of var1 has not changed ECE Changing the value pointed by point *p=10; // initialization of var1 via pointer p p: 0xffbff6cc var1 10 var2 345 The value of var1 is changed to 10 Equivalent to var1 = 10; May seem redundant, but we will see it is actually very useful ECE

6 Ini4aliza4on of a variable via p var2 = *p+10; // var2 = var p: 0xffbff6cc var1 10 var2 20 var2 becomes equal to the value pointed by point Equivalent to the assignment var2 = var ECE Prin4ng the contents printf("%d %d %d\n", var1, var2, *p); *p prints the value pointed by point Outcome 10, 20, 10 ECE Prin4ng the addresses printf("%p %p %p\n", p, &var1, &var2); // printing addresses Prints the address of the variable pointed by point And the address where point is stored point: &point: Integer variables reserve 4 bytes (32- bit addressing). var1 starts at var2 starts at , point starts at ECE

7 Looking at the memory ECE Common Programming Errors with Pointers Segmenta4on Faults #include <stdio.h> int main(void) { int *p; // pointer to an integer variable printf( %d, *p); // printing the contents of p return(0); } Pointer p is unini4alized and may point to a forbidden segment of the memory ECE Common Programming Errors with Pointers Other case of unini4alized pointer #include <stdio.h> int main(void) { int *p; // pointer to an integer variable *p = 10 // pointer p is uninitialized return(0); } Pointer p is unini4alized and may point to a forbidden segment of the memory ECE

8 Common Programming Errors with Pointers Mismatch between pointer type and variable type #include <stdio.h> int main(void) { float *p; // pointer to a float variable int i = 18; p = &i *p = 10.5 // pointer mismatch return(0); } i will store a value different than 10.5 ECE Mul4ple Pointers on the Same Address Three pointers p1, p2, p3, one variable c #include <stdio.h> int main(void) { char *p1, *p2, *p3; // three pointers of type char char c; // a char variable p1=&c; // p1 now points to c *p1='a'; // value of c changed to 'a' p2=p1; // p2 is assigned the arithmetic value of p1 p3=p2; // p3 is assigned the arithmetic value of p2 printf("%c %c %c\n",*p1, *p2, *p3); printf("%p %p %p\n", p1, p2, p3); printf( %p %p %p\n", &p1, &p2, &p3); return(0); } ECE Mul4ple Pointers on the Same Address All three pointers point to the same variable p1 p2 p3 c a ECE

9 Pointer Assignment The assignment p1 = p2 is valid if both are pointers of the same type Example inp = fopen( filename, r ) It is valid because fopen returns a pointer of type FILE. ECE Re- visi4ng the circle func4on #include<stdio.h> #define PI 3.14 void circle(float r, float *p_a, float *p_c); int main(void) { float r, a=0, c=0; printf("enter the radius of the circle:"); scanf("%f",&r); circle(r, &a, &c); // arguments passed by value printf("area:%.1f Circumference:%.1f\n", a, c); return(0); } void circle(float r, float *p_a, float *p_c) { "*p_a = PI*r*r; // computation of area "*p_c = 2*PI*r; // computation of circumference } ECE Create a Common Frac4on Calculator Problem Inputs: Frac4on 1: n1, d1 Operator: +, -, *,/ Frac4on 2: n2, d2 Problem Outputs: n, d in reduced form Algorithms: Addi4on: n = n1 * d2 + n2* d1, d = d1*d2 Subtrac4on: n = n1 * d2 - n2* d1, d = d1*d2 Mul4plica4on: n = n1 * n2, d = d1*d2 Division: n = n1 * d2, d_ans = d1*n2 Reduc4on: n = n /gcd(n, d), d = d/gcd(n, d) ECE

10 Addi4on Func4on n1 d1 n2 d2 Addi4on Func4on n d Func4on Prototype void add_fractions(int n1, int d1, int n2, int d2, int *p_n, int *p_d); Subtrac4on func4on use the addi4on func4on ECE Mul4plica4on Func4on n1 d1 n2 d2 Mul4plica4on Func4on n d Func4on Prototype void multiply_fractions(int n1, int d1, int n2, int d2, int *p_n, int *p_d); Division func4on use the mul4plica4on func4on ECE Reduc4on Func4on Func4on n d Reduc4on Func4on n d Func4on Prototype void reduce_fraction(int *p_n, int *p_d); int gcd(int x, int y); ECE

11 Func4ons Declara4ons #include<stdio.h> void scan_frac(int *p_n, int *p_d); char scan_op(void); int gcd(int x,int y); void add_fractions(int n1, int d1, int n2, int d2, int *p_n, int *p_d); void multiply_fractions(int n1, int d1, int n2, int d2, int *p_n, int *p_d); void reduce_fraction(int *p_n, int *p_d); void print_fractions(int n1, int d1, int n2, int d2, int n, int d, int op); ECE Main func4on int main(void) { int n1, d1, n2,d2, n, d; char op; scan_frac(&n1, &d1); // read the first fraction op=scan_op(); // read the operator scan_frac(&n2, &d2); // read the second fraction switch(op) { case '+': add_fractions(n1, d1, n2, d2, &n, &d); break; case '-': add_fractions(n1, d1, -n2, d2, &n, &d); break; case '/': multiply_fractions(n1, d1, n2, d2, &n, &d); break; case '*': multiply_fractions(n1, d1, d2, n2, &n, &d); break; default: printf("this is not a valid operator"); } reduce_fraction(&n, &d); // reduce the fraction print_fractions(n1,d1,n2,d2, n, d, op); // print result return(0); } ECE Scan Frac4on and Operator Func4ons void scan_frac(int *p_num, int *p_den) { printf("enter the nominator and the denominator of the fraction:"); scanf("%d%d", p_num, p_den); } char scan_op(void) { char c; printf("enter the operator:"); fflush(stdin); scanf("%c", &c); return c; } ECE

12 Greatest Common Denominator Func4on int gcd(int x,int y) { int rem = 1, gcd; while (rem=0) // when rem = 0, the loop terminates { " "if (y == 0) // if y is zero " "{ " " "gcd = x; // gcd becomes x " " "rem = 0; " "} " "else // if y is non-zero " "{ " " "gcd = y; // gcd becomes equal to y " " "rem = x%y; // remainder computation " " "x = y; // x becomes y " " "y = rem; // y becomes the remainder " "} "} return gcd; } ECE Frac4on Addi4on, Mul4plica4on, and Reduc4on void add_fractions(int n1, int d1, int n2, int d2, int *p_n, int *p_d) { *p_n = (n1 * d2) + (n2 * d1); *p_d = d1 * d2; } void multiply_fractions(int n1, int d1, int n2, int d2, int *p_n, int *p_d) { *p_n = n1 * n2; *p_d = d1 * d2; } void reduce_fraction(int *p_n, int *p_d) { int gd = gcd(*p_n, *p_d); *p_n = (*p_n) / gd; *p_d = (*p_d) / gd; } ECE Frac4on Prin4ng void print_fractions(int n1, int d1, int n2, int d2, int n, int d, int op) { printf("%d/%d",n1, d1); printf(" %c ", op); printf("%d/%d = ",n2, d2); printf("%d/%d",n, d); } ECE