1. Arrays as Function Arguments Array can be used as a function argument. E.g., #include int sum(int b[], int n) { int i, res; res = 0; for(i = 0; i < n; ++i) res += b[i]; return res; } main() { int a[] = {0, 1, 2, 3, 4, 5, 6}; printf("%d\n", sum(a, 2)); printf("%d\n", sum(a, 4)); } The printouts are 1 and 6.

2. Passing an Array v.s. Passing a Single Variable void func(int a[], int n) { a[0] += 1; ++n; } main() { int b[2] = {0, 0}; int n = 0; printf("%d %d %d\n", b[0], b[1], n); func(b, n); printf("%d %d %d\n", b[0], b[1], n); } 0 0 0 1 0 0 Passing an array is to pass an address. Passing a single variable is to pass a copy of the value.

3. String-Handling Functions String Concatenation With the declaration and initialization: s1[10] ="The "; s2[4] = "CZ"; The function (with string.h included) strcat(s1, s2); append the string s2 to the end of string s1, i.e., s1 now becomes "The CZ".

4. String-Handling Functions String Comparison With the declaration and initialization: s1[10] ="The "; s2[4] = "CZ"; s3[5] = "The "; The function (with string.h included) strcmp(s1, s2) returns a positive value (for 'T' is after 'C' according to ASCII table). strcmp(s1, s3) returns 0 (for the two strings are the same).

5. String-Handling Functions String Copy/String Length With the declaration and initialization: s1[10] ="The "; s2[10] = "CZ"; The function (with string.h included) strcpy(s1, s2) copy the second string s2 into the first string s1. strlen(s1) returns 4 (for there are four characters before the null character '\0').

6. String-Handling Functions String in String With the declaration and initialization: s1[10] ="The "; s2[10] = "e"; The function (with string.h included) strstr(s1, s2) return a pointer to character where s2 is part of s1 (s2 is a substring of s1). For this example, it returns the address of character 'e' is string s1.

7. Example on Strings #include main() { char s[] = "We study"; char t[] = "uoga uoga"; char v[15] = "did"; printf("%d\n", strlen(s)); printf("%d\n", strcmp(s,t)); printf("%s\n", strcpy(s,v)); printf("%s\t%s\n", s, v); printf("%s\n", strcat(v, t)); printf("%s\n", strstr(t, "og")); } The program prints: 8 -30 did diduoga uoga oga uoga

8. Computing a String's Length int length(char s[]) { int cnt; for(cnt = 0; s[cnt] != '\0'; ++cnt) ; return cnt; } #include main() { printf("\"otter\"\t%d\n", length("otter")); printf("\"\"\t%d\n", length("")); printf("\"a\"\t%d\n", length("a")); } The outputs are: "otter" 5 "" 0 "a" 1

9. Multidimensional Arrays The dimension of an array is the number of indices of the array. float temps[10]; –one dimensional array with 10 elements. float t[12][50]; –two dimensional array with 12*50=600 elements. The elements are t[0][0], t[0][1], t[0][2],..., t[0][49], t[1][0], t[1][1],...,..., t[11][48], t[11][49].

10. Point of Views of Multidimensional Array You can think of one dimensional array as a vector, and two dimensional array as a matrix (only the index starts from 0). You can view a multidimensional array as (one-dimensional) array of array. For example, t[12][50] is an array t[12], each of them is an array of 50 elements.

11. Initializing Multidimensional Arrays Multidimensional array can be initialized element by element: int nums[2][3]; nums[0][0] = 2; nums[0][1] = 4; nums[0][2] = 6; nums[1][0] = -9; nums[1][1] = -7; nums[1][2] = -5; Or initialized during declaration: int nums[2][3] = { {2, 4, 6}, {-9, -7, -5}}

12. Multidimensional Array as Arguments The dimensions of the array must be specified in function definition. E.g., print_table(int t[10][4]) {... } main() { int t[10][4];... Print_table(t); } The first size can be omitted.

13. Matrix Multiplication Problem: Provide a general function for matrix multiplication. The function should multiply two n by n matrices and store the product in a third n by n matrix. The product, Z, of two matrices, X and Y, is defined by Z(i,j) =  k X(i,k) Y(k,j) where 0  i, j  n - 1, and summation index k runs from 0 to n-1.

14. Sample Input/Output Input matrix size: 3 Input first matrix by row 1 2 4 3 -1 6 4 -6 2 Matrix m1: 1 2 4 3 -1 6 4 -6 2 Input second matrix by row 0 3 -5 1 1 -2 5 -2 7 Matrix m2: 0 3 -5 1 1 -2 5 -2 7 Product m3: 22 -3 19 29 -4 29 4 2 6 For example, 22 of the (0,0) element in m3 is obtained by 1*0 + 2*1 + 4*5 = 22

15. The Prototype and Main Program #include #define MAXSIZE 20 void store(int m[][MAXSIZE], int n), mult(int m1[][MAXSIZE], int m2[][MAXSIZE], int m3[][MAXSIZE], int n), print(int m[][MAXSIZE], int n); main() { int n; int m1[MAXSIZE][MAXSIZE]; int m2[MAXSIZE][MAXSIZE]; int m3[MAXSIZE][MAXSIZE]; printf("Input matrix size:"); scanf("%d", &n); printf("Input first matrix by row\n"); store(m1, n); printf("\nMatrix m1:\n"); print(m1, n); printf("Input second matrix by row\n"); store(m2, n); printf("\nMatrix m2:\n"); print(m2, n); mult(m1, m2, m3, n); printf("\nProduct m3:\n"); print(m3, n); }

16. Function for Matrix Multiplication /* Multiply matrix m1 by m2, store the product in matrix m3 */ void mult(int m1[][MAXSIZE], int m2[][MAXSIZE], int m3[][MAXSIZE], int n) { int i, j, k; for(i = 0; i < n; ++i) for(j = 0; j < n; ++j) { m3[i][j] = 0; for(k = 0; k < n; ++k) m3[i][j] += m1[i][k] * m2[k][j]; }

17. Read/Print Matrix /* Store/read data in matrix by row */ void store(int m[][MAXSIZE], int n) { int i, j; for(i = 0; i < n; ++i) for(j = 0; j < n; ++j) scanf("%d", &m[i][j]); } /* Print the matrix */ void print(int m[][MAXSIZE], int n) { int i, j; for(i = 0; i < n; ++i) { for(j = 0; j < n; ++j) printf("%d ", m[i][j]); putchar('\n'); }

18. Computational Complexity Time Complexity –The number of operations needed as a function of the problem size. –In our matrix computation example, we need about n 3 computation for matrices of size n. We say our program (algorithm) has a time computational complexity of order n 3. n Time Units1 28 327 5125 101,000 1001,000,000 10001,000,000,000

19. Computational Complexity Space Complexity –The number of memory cells needed as a function of the problem size. –In our matrix computation example, we need about n 2 memory space for matrices of size n. We say our program (algorithm) has a space computational complexity of order n 2. A good algorithm should have smallest CPU time and memory space usage.

20. Reading/Home Working Read Chapter 6, page 251 to 282. Work on Problems –Section 6.5, page 253, exercise 1, 3. –Section 6.6, page 261, exercise 1, 3. –Section 6.8, page 271, exercise 1, 3. Check your answers in the back of the textbook. Do not hand in.