240-222 cpt: functions/31 240-222 computer programming techniques semester 1, 1998 objectives of...

240-222 CPT: Functions/3 1

240-222 Computer Programming Techniques240-222 Computer Programming TechniquesSemester 1, 1998Semester 1, 1998

Objectives of these slides:– to describe C functions

– illustrate recursion

3. Functions


OverviewOverview

1. Examples

2. Type Coercion

3. Call by Value

4. Scope Rules

5. Recursion

6. A Skeleton with Functions


1. Examples1. Examples/* Using a square function (Fig. 5.3) */#include <stdio.h>int square(int y);

int main(){ int x;

for (x = 1; x <= 10; x++) printf("%d ", square(x));

printf("\n"); return 0; }

continued

function prototype


int square(int y)/* returns the square of y */{ return y * y;}

function definition


/* Finding the maximum of three integers (Fig. 5.4) */#include <stdio.h>

int maximum(int x, int y, int z);

int main(){ int a, b, c;

printf("Enter three integers: "); scanf("%d%d%d", &a, &b, &c); printf("Maximum is: %d\n",

maximum(a, b, c)); return 0; }

continued


int maximum(int x, int y, int z)/* returns the biggest of x, y and z */{ if (x > y && x > z) return x; else if (y > z) return y; else return z;}


ExecutionExecution

Enter three integers: 22 85 17Maximum is 85


2. Type Coercion2. Type Coercion

#include <stdio.h>

double sqroot(double n);int square(int y);

int main(){ printf("%f %f\n",

sqroot(4.0), sqroot(4)); printf("%d %d\n",

square(4), square(4.5)); return 0;}

continued


double sqroot(double n)/* Use Newton Raphson to return the square root of n */{ float x0, eps = 0.000001;

x0 = n; while ( abs(x0*x0 - n)/n > eps) x0 = (x0 + n/x0)/2; return x0;}

int square(int y){ return y*y; }


Type Ordering (simplified)Type Ordering (simplified)

long double

double

float

int

char

can be coerced to


Why Simplfied?Why Simplfied?

Simplified in the sense that int can also be prefixed with the keywords:– short– long– unsigned

These refer to the size and range of the integer.


3. Call By Value3. Call By Value

#include <stdio.h>

int compute_sum(int x);

int main(){ int n = 3, sum;

printf("%d\n",n); sum = compute_sum(n); printf("%d\n",n); printf("%d\n",sum); return 0;}

continued


int compute_sum(int x)/* sum the integers from 1 to x */{ int tot = 0; for ( ; x > 0 ; --x) tot += x;

printf("%d\n", x);

return tot;}


4. Scope Rules4. Scope Rules

A block is a compound statement with declarations:

{ int x; s1; s2;}


An identifier is accessible only within the block where it is declared.

The scope of an identifier is that part of the program where the identifier is accessible.


ExamplesExamples

int main(){ int a = 2; printf("%d\n", a);

{ int a = 7; printf("%d\n", a); }

printf("%d\n", ++a); return 0;}


A similar piece of code:A similar piece of code:

int main(){ int a_outer = 2; printf("%d\n", a_outer);

{ int a_inner = 7; printf("%d\n", a_inner); }

printf("%d\n", ++a_outer); return 0;}


#include <stdio.h>

int compute_sum(int n);

int main(){ int n = 3, sum;

printf("%d\n",n); sum = compute_sum(n); printf("%d\n",n); printf("%d\n",sum); return 0;}

continued


int compute_sum(int n)/* sum the integers from 1 to n */{ int sum = 0; for ( ; n > 0 ; --n) sum += n;

printf("%d\n", n);

return sum;}


5. Recursion 5. Recursion Sec. 5.13Sec. 5.13

5.1. Factorial

5.2. The Fibonacci Series

5.3. Depth Tester


5.1. Factorial5.1. Factorial

Mathematical Definition:

fac n= 1, if n <= 1= n * fac (n - 1),

otherwise


The C version:The C version:

long int factorial(long int n){ if (n <= 1) return 1; else return (n * factorial(n - 1));}


/* full program (fig. 5.14) */

#include <stdio.h>

long int factorial(long int n);

int main(){ int i;

for (i = 1; i <= 10; i++) printf("%ld\n", factorial(i)); return 0;}

continued


long int factorial(long int n)/* return the factorial of n */{ if (n <= 1) return 1; else return (n * factorial(n - 1));}


Recursion Uses Lots of MemoryRecursion Uses Lots of Memory

fact(10)

fact(9)

fact(8)

fact(1)

3628800

362880

40320

1


A Common ErrorA Common Error

long factorial(long n){ if (n <= 1) return 1; else return (n * factorial(--n));}


5.2. The Fibonacci Series5.2. The Fibonacci Series

0, 1, 1, 2, 3, 5, 8, 13, 21, . . .

Mathematical definition:

fib n = 0 if n = 0= 1 if n = 1= fib (n-1) + fib (n-2) if n > 1


The C versionThe C version

long int fib(long int n){ if (n == 0 || n == 1) return n; else return fib(n - 1) + fib(n - 2);}


/* full program (fig. 5.15)#include <stdio.h>

long int fib(long int n);

int main(){ long int result, number;

printf("Enter an integer: "); scanf("%ld", &number); result = fib(number); printf("Fibonacci(%ld) = %ld\n",

number, result); return 0;}

continued


long int fib(long int n)/* return the nth fibonacci number */{ if (n == 0 || n == 1) return n; else return fib(n - 1) + fib(n - 2);}


ExecutionsExecutions

Enter an integer: 5Fibonacci(5) = 5

Enter an integer: 20Fibonacci(20) = 6765


Recursion can be InefficientRecursion can be Inefficient

fib(5)

fib(4) fib(3)

fib(2)

fib(1)

fib(3)

fib(2)

fib(1) fib(0)

1 0

1

fib(1) fib(0)

1 0

fib(1)fib(2)

fib(1) fib(0)

1 0

1

+

+

+

+ +

+

+


5.3. Depth Tester5.3. Depth Tester

/* Test the depth of recursion for sum() in steps of 100 */

#include <stdio.h>

long int sum(long int n);

int main(){ long int n = 0; for ( ; ; n +=100) printf("Recursion Test: n = %ld

sum = %ld\n", n, sum(n)); return 0;}


long int sum(long int n)/* sum the integers between 1 and n */{ if ( n <= 1) return n; else return ( n + sum(n - 1));}


Results on a PCResults on a PC

::

Recursion Test: n = 7900 sum = 31208950Recursion Test: n = 8000 sum = 32004000Recursion Test: n = %ld sum = %ld> /* DOS prompt */


Execution StackExecution Stack

32004000

31996001

31988003

sum(8001)

sum(8000)

sum(7999)

sum(7998)

sum(1) 1

?


But on a workstationBut on a workstation

:Recursion Test: n = 65100 sum = 2119037550Recursion Test: n = 65200 sum = 2125552600Recursion Test: n = 65300 sum = 2132077650Recursion Test: n = 65400 sum = 2138612700Recursion Test: n = 65500 sum = 2145157750Recursion Test: n = 65600 sum = -2143254496Recursion Test: n = 65700 sum = -2136689446Recursion Test: n = 65800 sum = -2130114396Recursion Test: n = 65900 sum = -2123529346Recursion Test: n = 66000 sum = -2116934296Recursion Test: n = 66100 sum = -2110329246

::


6. A Skeleton with Functions6. A Skeleton with Functions

/* Author & program details */#include <stdio.h>

int func1(int x);double func2(double y);/* ... more prototypes */

int main(){ /* declare variables */ /* do something with functions */ return 0;}

continued


int func1(int x)/* some comments */{ /* do something */ return /* an integer */ ;}

continued


double func2(double y)/* some comments */{ /* do something */ return /* a double */ ;}

/* ... more function definitions */

240-222 cpt: functions/31 240-222 computer programming techniques semester 1, 1998 objectives of...

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