ARRAYS

1. ARRAYS Lecture 15 12.2.2002.

2. Arrays as parameters of functions • An array passed as a parameter is not copied. • An array name is a constant whose value serves as a reference to the first (index 0) item in the array.

3. Arrays as parameters of functions • Since constants cannot be changed, assignments to array variables are illegal. • Only the array name is passed as the value of a parameter, but the name can be used to change the array’s contents. • Empty brackets [] are used to indicate that the parameter is an array. The no of elements allocated for the storage associated with the array parameter does not need to be part of the array parameter.

4. #define MAXS 10 int main () { int numbers[MAXS]; int k, size = 10; for (k=0; k<size; k++) { numbers[k] = k+1; } printf (“Before : \n”) ; printarray (numbers, size) ; change (numbers, size) ; printf (“After : \n”) ; printarray (numbers, size) ; }

5. void change (int list[], int num) { int k; for (k=1; k<num; k++) { list[k] += list[k-1] ; } } void printarray (int list[], int num) { int k; for (k=0; k<num; k++) { printf (“%d “, list[k]) ; } printf (“\n”) ; }

6. Output Before: 1 2 3 4 5 6 7 8 9 10 After: 1 3 6 10 15 21 28 36 45 55

7. Array operations #define MAXS 100 int insert (int[], int, int, int) ; int delete (int[], int, int) ; int getelement (int[], int, int) ; int readarray (int[], int) ; main () { int a[MAXS]; int size; size = readarray (a, 10) ; size = insert (a, size, 4, 7) ; x = getelement (a, size, 3) ; size = delete (a, size, 3) ; }

8. Array operations int readarray (int x[], int size) { int i; for (i=0; i<size; i++) scanf(“%d”, &x[i]) ; return size; } int insert (int x[], int size, int pos. int val){ for (k=size; k>pos; k--) x[k] = x[k-1] ; x[pos] = val ; return size+1; } int getelement (int x[], int size, int pos){ if (pos <size) return x[pos] ; return -1; }

9. int delete (int x[], int size, int pos) { }

10. void reverse (int x[], int size) { } int findmax (int x[], int size) { }

11. void reverse (int x[], int size) { int i; for (i=0; i< (size/2); i++) temp = x[size-i-1] ; x[size-1-1] = x[i] ; x[i] = temp; } int findmax (int x[], int size) { int i, max; max = x[0]; for (i=1; i< size; i++) if (x[i] > max) max = x[i] ; return max; }

12. Strings • Strings are 1-dimensional arrays of type char. • By convention, a string in C is terminated by the end-of-string sentinel \0, or null character. • String constant : “abc” is a character array of size 4, with the last element being the null chaaracter \0. • char s[] = “abc” ; a b c \0

13. Searching an Array:Linear and Binary Search

14. Searching • Check if a given element (key) occurs in the array. • If the array is unsorted : • start at the beginning of the array • inspect every element to see if it matches the key

15. Linear Search /* If key appears in a[0..size-1], return its location, pos, s.t. a[pos] == key. If key is not found, return -1 */ int search (int a[], int size, int key) { int pos = 0; while (pos < size && a[pos] != key) pos++; if (pos<n) return pos; return -1; }

16. Linear Search int x= {12,-3, 78,67,6,50,19,10} ; Trace the following calls : search (x, 8,6) ; search (x,8,5) ;

17. Searching a sorted array • Binary search works if the array is sorted • Look for the target in the middle • If you don’t find it, you can ignore half of the array, and repeat the process with the other half.

18. Binary Search Strategy • What we want : Find split betwen values larger and smaller than x : 0 L R x: n <=key >key • Situation while searching : 0 L R n x: <=key ? >key • Step : Look at [(L+R)/2]. Move L or R to the middle depending on test.

19. Binary Search /* If key appears in x[0..size-1], return its location, pos s.t. x[pos]==key. If not found, return -1 */ int binsearch (int x[], int size, int key) { int L, R, mid; ______________________; while (_________________) { } ____________________ ; }

20. Binary Search /* If key appears in x[0..size-1], return its location, pos s.t. x[pos]==key. If not found, return -1 */ int binsearch (int x[], int size, int key) { int L, R, mid; ______________________; while (_________________) { mid = (L+R)/2; if (x[mid] <= key) L = mid; else R = mid; } ____________________ ; } mid = (L+R)/2; if (x[mid] <= key) L = mid; else R = mid;

21. Binary Search: loop termination /* If key appears in x[0..size-1], return its location, pos s.t. x[pos]==key. If not found, return -1 */ int binsearch (int x[], int size, int key) { int L, R, mid; ______________________; while (_________________) { mid = (L+R)/2; if (x[mid] <= key) L = mid; else R = mid; } ____________________ ; } L+1 != R

22. Binary Search: Return result /* If key appears in x[0..size-1], return its location, pos s.t. x[pos]==key. If not found, return -1 */ int binsearch (int x[], int size, int key) { int L, R, mid; ______________________; while ( L+1 != R) { mid = (L+R)/2; if (x[mid] <= key) L = mid; else R = mid; } } if (L>=0 && x[L]==key) return L; else return -1;

23. Binary Search: Initialization /* If key appears in x[0..size-1], return its location, pos s.t. x[pos]==key. If not found, return -1 */ int binsearch (int x[], int size, int key) { int L, R, mid; ______________________; while ( L+1 != R) { mid = (L+R)/2; if (x[mid] <= key) L = mid; else R = mid; } if (L>=0 && x[L]==key) return L; else return -1; } L=-1; R=size;

24. Binary Search Examples -17 -5 3 6 12 21 45 63 50 Trace : binsearch (x, 9, 3); binsearch (x, 9, 145); binsearch (x, 9, 45);

25. Is it worth the trouble ? • Suppose you had 1000 elements • Ordinary search (if key is a member of x) would require 500 comparisons on average. • Binary search • after 1st compare, left with 500 elements • after 2nd compare, left with 250 elements • After at most 10 steps, you are done. • What if you had 1 million elements ?

26. Sorting • Given an array x[0], x[1], ... , x[size-1] • reorder entries so that x[0]<=x[1]<= . . . <=x[size-1]

27. Sorting Problem • What we want : Data sorted in order sorted : x[0]<=x[1]<= . . . <=x[size-1] size 0 x: • Initial conditions : size 0 unsorted x:

28. Selection Sort • General situation : k 0 size x: smallest elements, sorted remainder, unsorted • Step : • Find smallest element, mval, in x[k..size-1] • Swap smallest element with x[k], then increase k. mval 0 k size x: smallest elements, sorted

29. Subproblem : : /* Yield location of smallest element int[x] in x[0 .. size-1];*/ int min_loc (int x[], int , int size) int j, pos; /* x[pos] is the smallest element found so far */ pos = k; for (j=k+1; j<size; j++) if (x[i] < x[pos]) pos = j; return pos; }

30. Selection Sort /* Sort x[0..size-1] in non-decreasing order */ int selsort (int x[], int size) { int k, m; for (k=0; k<size-1; k++) { m = min_loc(x, k, size); temp = a[k]; a[k] = a[m]; a[m] = temp; } }

31. Example x: x: 3 12 -5 6 142 21 -17 45 -17 -5 3 6 12 21 142 45 x: -17 12 -5 6 142 21 3 45 x: -17 -5 3 6 12 21 45 142 x: -17 -5 12 6 142 21 3 45 x: -17 -5 3 6 142 21 12 45 x: -17 -5 3 6 12 21 142 45

32. Analysis • How many steps are needed to sort n things ? • Total number of steps proportional to n2

33. Insertion Sort #define MAXN 100 void InsertSort (int list[MAXN], int size) ; main () { int index, size; int numbers[MAXN]; /* Get Input */ size = readarray (numbers) ; printarray (numbers, size) ; InsertSort (numbers, size) ; printarray (numbers, size) ; }

34. void InsertSort (int list[], int size) { for (i=1; i<size; i++) item = list[i] ; for (j=i-1; (j>=0)&& (list[j] > i); j--) list[j+1] = list[j]; list[j+1] = item ; } }

35. Common pitfalls with arrays in C • Exceeding the array bounds • int array[10];for (i=0; i<=10; i++) array[i] = 0; • C does not support array declaratiions with variable expressions. • void fun (int array, int size) { int temp[size] ; . . .}