Computer Programming: 7th Week Functions, Recursive Functions, Introduction to Pointers

Transcription

1 Computer Programming: 7th Week Functions, Recursive Functions, Introduction to Pointers Hazırlayan Asst. Prof. Dr. Tansu Filik

2 Introduction to Programming Languages Previously on Bil-200 Functions: Function calls, function definition

3 Introduction to Programming Languages Recursion Some mathematical problems and graph theoretical questions are inherently recursive : n! = n * (n-1)!... (n-1)! =(n-1) * (n-2)! Fibonacci series: 0, 1, 1, 2, 3, 5, 8... fib( n ) = fib( n - 1 ) + fib( n 2 ) In this case, the function must call itself! (n-2)! = (n-2)*(n-3)!

4 long fibonacci( long n ) if (n == 0 n == 1) /* en alt */ return n; else return fibonacci( n - 1) + fibonacci( n 2 ); f( 3 ) return f( 2 ) + f( 1 ) return f( 1 ) + f( 0 ) return 1 return 1 return 0

5 long fibonacci( long n ) printf("%d\n",n); if (n == 0 n == 1) /* en alt */ return n; else return fibonacci( n - 1) + fibonacci( n 2 ); f( 3 ) return f( 2 ) + f( 1 ) return f( 1 ) + f( 0 ) return return 1 return 0

6 int factorial(int a) if(a<0) fprintf(stderr, Gecersiz sayi\n ); return; else if ((a==0) (a==1)) return 1; else return (a*factorial(a-1));

7 Exercise: tanh 1 x x x x x Evaluate the above expression until the added numbers become smaller than a determined threshold.

8 Examples C code to find the addition of n numbers by recursion: #include<stdio.h> int getsum(int n); int main() int n,sum; printf("enter the value of n: "); scanf("%d",&n); int getsum(int n) static int sum = 0; if(n>0) sum = sum + n; getsum(n-1); sum = getsum(n); printf("sum of n numbers: %d",sum); return sum; return 0;

9 Examples C code to check a number is prime number or not by recursion: #include<stdio.h> int isprime(int num,int i); int main() int num,prime; int isprime(int num,int i) printf("enter a positive number: "); scanf("%d",&num); prime = isprime(num,num/2); if(prime==1) printf("%d is a prime number",num); else printf("%d is not a prime number",num); if(i==1) return 1; else if(num%i==0) return 0; else isprime(num,i-1); return 0;

10 void a(void) int x=10; printf( %d\n,x); x++; printf( %d\n,x); void b(void) static int x=10; printf( %d\n,x); x++; printf( %d\n,x); The value of this variable is assigned to 10 for the first call of function. Then, each time the function is called, the previous value is remembered. void main(void) a(); a(); a(); a(); b(); b(); b(); b(); a: b:

11 Exercises: What's wrong with the following listing? #include <stdio.h> void print_msg( void ); main() print_msg( "This is a message to print" ); return 0; void print_msg( void ) puts( "This is a message to print" ); return 0; What's wrong with the following function definition? int twice(int y); return (2 * y);

12 The values of local variables are unknown to outside the function. Different functions may have local variables with same names. Their values have nothing to do with other functions variables. The variables (parameters) that are sent to the function can change their values inside the function. BUT!!! The variable which is inside the function call does not change. The function makes a copy of the variable, and uses it. It does not use the original variable.

13 !! Example!! Write a function which swaps the values of its two parameters: int a=6, b=4; printf("%d %d\n", a, b); degistir(a, b); printf("%d %d\n", a, b); The first printf should write "6 4", the next should "4 6".

14 void degistir ( int a, int b ) int gecici; Solution (?) gecici = a; a = b; b = gecici; a and b are local variables, specific to the function only. The function is unable to change the values.

15 Why doesn t it work? Because you may send parameters to functions in two ways: by value by reference By value: While calling function, a copy of the variable value is sent. The original variable is not sent. By reference: While calling function, the real address of the variable is sent to the function. Inside the function, the values at that address can be changed. So the variable value also changes! (Remember scanf())

16 Function calls Remember function: toplam(). When it is called, the loal variables work, and when the function finishes, the local variables are lost. So how can we send parameters by reference? By address-pointer syntax. main() memory area i 12 j 25? copied copied toplam() memory area a 12 b 25

17 Two-way communication with functions #include <stdio.h> void swap(int *a,int *b); void main() int i = 1, j = 2; &i and &j, are memory locations of i and j. We send addresses to the swap() function. swap(&i, &j); void swap(int *a, int *b) int t; a and b are address (pointer) variables. t = *a; *a = *b; *b = t; By writing *a and *b, we can rach to the values in the given address.

18 Sending addresses to functions main() memory area &i toplam() memory area a 12 i 100 &j 100 b j This is the memory location of variable j

19 We already use such functions Remember scanf(); fscanf(); fopen(); etc... Think of function scanf(). What is critical here? Why do we send addresses of variables: scanf( %d, &i); Because we want to change the value of variable i. Why do we use scanf( %s, str);? We have not seen yet, but str is already an address!

20 scanf(&num) illustration &num &num 2 byte Read from keyboard: (Assume 10 is entered) 0 A

21 If you send the variable with & (address) sign - just like doing in scanf() we can change its value inside the function by writing thing to its memory location. Since the memory location corresponds to the container of the original variable, its value automatically changes.. So, what is the syntax (format) of a function that accepts addresses? We use *variable notation.

22 void my_func(int *num) This part is just like other functions. The argument listing part is different. int *num; is a variable whose address is num. We call the above function as: my_func(&i);

23 Example: If we want to print out the value of the parameter which was sent: void my_func(int *num) printf("\n%d", num); printf("\n%d\n", *num); void main(void) int *num; int i = 1; my_func(&i);

24 void my_func(int *num) printf( %d\n,*num); *num = 2; printf( %d\n,*num); void main(void) int a=3; printf( %d\n,a); my_func(&a); printf( %d\n,a); This is the int variable; num is not int! If you call the function like this (with address)...then the function argument must be declared with * (a pointer) indicator.

25 void degis(int *n, int *m) int k; printf( Degismeden once, n=%d, m=%d,,*n,*m); k=*n; *n=*m; *m=*k; printf( Degistikten sonra, n=%d, m=%d,,*n,*m); void main(void) int n1=10,n2=20; printf( Fonksiyon oncesi n1=%d, n2=%d,n1,n2); degis(&n1,&n2); printf( Fonksiyon sonrasi n1=%d, n2=%d,n1,n2);

26 = &n1 = n *n=10 &n2 m *m=

27 If you want to use address/pointer convention, but keep the value unchanged, then use const int *const num; /* for instance... */

28 Class Exercise Write a function which reads three float variables from the user, and returns them. The name would be: read_return(). The main program must be like: #include <stdio.h> main() float a, b, c; read_return(&a, &b, &c); printf( The numbers: %f %f %f\n",a,b,c);

29 Alternative uses void getnum(int *num) int k; printf( bir sayi girin: ); scanf( %d,&k); num=k; void getnum(int *num) printf( bir sayi girin: ); scanf( %d,num); void main(void) int k; getnum(&k); scanf( %d,k);