1 / 48

CS1010: Programming Methodology comp.nus.sg/~cs1010/

CS1010: Programming Methodology http://www.comp.nus.edu.sg/~cs1010/. Week 8: Multidimensional Arrays. Objectives: Understand the concept and application of multidimensional arrays. Reference: Chapter 6: Numeric Arrays. Outline (1/2). One Dimensional Arrays (revision) 1.1 Print Array

flavio
Download Presentation

CS1010: Programming Methodology comp.nus.sg/~cs1010/

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. CS1010: Programming Methodologyhttp://www.comp.nus.edu.sg/~cs1010/

  2. Week 8: Multidimensional Arrays • Objectives: • Understand the concept and application of multidimensional arrays • Reference: • Chapter 6: Numeric Arrays CS1010 (AY2013/4 Semester 1)

  3. Outline (1/2) • One Dimensional Arrays (revision) 1.1 Print Array 1.2 Find Maximum Value 1.3 Sum Elements 1.4 Sum Alternate Elements 1.5 Sum Odd Elements 1.6 Sum Last 3 Elements 1.7 Minimum Pair Difference 1.8 Accessing 1D Array Elements in Function 1.9 Week 7 Exercise #4: Set Containment CS1010 (AY2013/4 Semester 1)

  4. Outline (2/2) • Multidimensional Arrays 2.1 Initializers 2.2 Example 2.3 Accessing 2D Array Elements in Function 2.4 Demo #1: Class Enrolment 2.5 Using UNIX input redirection 2.6 Exercise #1: Sum to Random Position • Matrices 3.1 Demo #2: Matrix addition 3.2 Exercise #2: Matrix multiplication • Maze Problem • Exercise #3: Valid path CS1010 (AY2013/4 Semester 1)

  5. 1. One-Dimensional Arrays (1/2) Element type Array • A collection of data, called elements, of homogeneous type Array name int a [6] Array size a[0] a[1] a[2] a[3] a[4] a[5] 12 20 25 8 36 9 CS1010 (AY2013/4 Semester 1)

  6. 1. One-Dimensional Arrays (2/2) • Preparing an array prior to processing: Initialization (if values are known beforehand): int main(void) { int numbers[] = { 20, 12, 25, 8, 36, 9 }; ... } some_fn(numbers, 6); Or, read data into array: int main(void) { int numbers[6], i; for (i = 0; i < 6; i++) scanf("%d", &numbers[i]); ... } some_fn(numbers, 6); CS1010 (AY2013/4 Semester 1)

  7. 1.1 Print Array voidprintArray(intarr[], int size) { inti; for (i = 0; i < size; i++) printf("%d ", arr[i]); printf("\n"); } Calling: int main(void) { int numbers[6]; ... printArray(numbers, 6); printArray(numbers, 3); } Value must not exceed actual array size. Print first 6 elements (all) Print first 3 elements CS1010 (AY2013/4 Semester 1)

  8. 1.2 Find Maximum Value • findMax(intarr[], int size) to return the maximum value in arr with size elements • Precond: size > 0 intfindMax(intarr[], int size) { inti, max; max = arr[0]; for (i = 1; i < size; i++) if (arr[i] > max) max = arr[i]; return max; } i arr max 1 2 6 4 5 3 20 25 36 20 12 25 8 36 9 CS1010 (AY2013/4 Semester 1)

  9. 1.3 Sum Elements • sum(intarr[], int size) to return the sum of elements • Precond: size > 0 int sum(intarr[], int size) { inti, sum = 0; for (i = 0; i < size; i++) sum += arr[i]; return sum; } i arr sum 5 4 3 2 1 0 6 20 110 20 57 65 101 32 0 12 25 8 36 9 CS1010 (AY2013/4 Semester 1)

  10. 1.4 Sum Alternate Elements • sumAlt(intarr[], int size) to return the sum of alternate elements (1st, 3rd, 5th, etc.) • Precond: size > 0 intsumAlt(intarr[], int size) { inti, sum = 0; for (i = 0; i < size; i+=2) sum += arr[i]; return sum; } int sum(intarr[], int size) { inti, sum = 0; for (i = 0; i < size; i++) sum += arr[i]; return sum; } i arr sum 0 2 6 4 20 45 81 0 20 12 25 8 36 9 CS1010 (AY2013/4 Semester 1)

  11. 1.5 Sum Odd Elements • sumOdd(intarr[], int size) to return the sum of elements that are odd numbers • Precond: size > 0 intsumOdd(intarr[], int size) { inti, sum = 0; for (i = 0; i < size; i++) if (arr[i]%2 == 1) sum += arr[i]; return sum; } int sum(intarr[], int size) { inti, sum = 0; for (i = 0; i < size; i++) sum += arr[i]; return sum; } i arr sum 5 6 1 0 3 4 2 20 25 0 34 12 25 8 36 9 CS1010 (AY2013/4 Semester 1)

  12. 1.6 Sum Last 3 Elements (1/3) • sumLast3(intarr[], int size) to return the sum of the last 3 elements among size elements • Precond: size  0 • Examples CS1010 (AY2013/4 Semester 1)

  13. 1.6 Sum Last 3 Elements (2/3) Thinking… • Last 3 elements of an array arr • arr[size – 1] • arr[size – 2] • arr[size – 3] • A loop to iterate 3 times (hence, need a counter) with index starting at size – 1 and decrementing in each iteration inti, count = 0; for (i = size - 1; count<3; i--) { . . . count++; } • But what if there are fewer than 3 elements in arr? CS1010 (AY2013/4 Semester 1) 

  14. 1.6 Sum Last 3 Elements (3/3) • Complete function (to be shown in class): CS1010 (AY2013/4 Semester 1) 

  15. 1.7 Minimum Pair Difference (1/3) • All problems on 1D arrays can be solved by single loop? Of course not! • Write a function minPairDiff(intarr[], int size) that computes the minimum possible difference of any pair of elements in arr. • For simplicity, assume size > 1 (i.e. there are at least 2 elements in array). CS1010 (AY2013/4 Semester 1)

  16. 1.7 Minimum Pair Difference (2/3) Thinking… Eg: size = 5. Need to compute difference of Outer loop index i from 0 to size-2 Inner loop index j from i+1 to size-1 arr[1] arr[2] arr[0] Inner loop index j from 1 to size-1 arr[3] arr[4] arr[2] Inner loop index j from 2 to size-1 arr[1] arr[3] arr[4] arr[3] Inner loop index j from 3 to size-1 arr[2] arr[4] Inner loop index j from 4 to size-1 arr[3] arr[4] CS1010 (AY2013/4 Semester 1)

  17. 1.7 Minimum Pair Difference (3/3) The code… Outer loop index i from 0 to size-2 Inner loop index j from i+1 to size-1 intminPairDiff(intarr[], int size) { inti, j, diff, minDiff; minDiff = abs(arr[0] – arr[1]); // init min diff. for (i = 0; i < size-1; i++) for (j = i+1; j < size; j++) { diff = abs(arr[i] – arr[j]); if (diff < minDiff) minDiff = diff; } returnminDiff; } CS1010 (AY2013/4 Semester 1)

  18. Codes Provided • Week8_FindMax.c: • Section 1.2 Find Maximum Element • Week8_SumElements.c: • Section 1.3 Sum Elements • Section 1.4 Sum Alternate Elements • Section 1.5 Sum Odd Elements • Section 1.6 Sum Last 3 Elements • Week8_MinPairDiff.c: • Section 1.7 Minimum Pair Difference CS1010 (AY2013/4 Semester 1)

  19. 1.8 Accessing 1D Array Elements in Function (1/2) A function header with array parameter, int sum(int a[ ], int size) Why is it not necessary to have a value in here to indicate the “real” size? • A value is not necessary (and is ignored by compiler if provided) as accessing a particular array element requires only the following information • The address of the first element of the array • The size of each element • Both information are known • For example, when the above function is called with ans = sum(numbers, 6); in the main(), the address of the first element, &numbers[0], is copied into the parameter a • The size of each element is determined since the element type (int) is given (in sunfire, an integer takes up 4 bytes) CS1010 (AY2013/4 Semester 1)

  20. 1.8 Accessing 1D Array Elements in Function (2/2) A function header with array parameter, int sum(int a[ ], int size) Why is it not necessary to have a value in here to indicate the “real” size? • With this, the system is able to calculate the effective address of the required element, say a[2], by the following formula: • Address of a[2] = base address + (2  size of each element) • where base address is the address of the first element • Hence, suppose the base address is 2400, then address of a[2] is calculated to be 2408. a[0] a[1] a[2] a[3] 19 5 12 7 ... CS1010 (AY2013/4 Semester 1)

  21. 1.9 Week 7 Exercise #4: Set Containment • Consider two arrays, arrA and arrB, of int values, where their sizes are sizeA and sizeB respectively. • Write a functionintisSubset(intarrA[], intsizeA, intarrB[], intsizeB) to determine if the set of numbers in arrA is a subset of the set of numbers in arrB. • The function returns 1 if arrA is a subset of arrB, or 0 otherwise. You may assume there are no duplicate numbers in each set. • Example: If arrA[ ] = {14, 5, 1, 9} and arrB[ ] = {2, 9, 3, 14, 5, 6, 1} • isSubset(arrA, 4, arrB, 7) returns 1 • isSubset(arrA, 4, arrB, 6) returns 0 • An incomplete program Week7_SetContainment.c is given. Complete the program. This is your take-home exercise. Also mounted on CodeCrunch. CS1010 (AY2013/4 Semester 1)

  22. 2. Multidimensional Arrays (1/2) • In general, an array can have any number of dimensions • Example of a 2-dimensional (2D) array: // array with 3 rows, 5 columns inta[3][5]; a[0][0] = 2; a[2][4] = 9; a[1][0] = a[2][4] + 7; 0 1 2 3 4 0 2 1 16 2 9 • Arrays are stored in row-major order a[0][0] … a[0][4] a[1][0] … a[1][4] a[2][0] … a[2][4] row 0 row 1 row 2 CS1010 (AY2013/4 Semester 1)

  23. 2. Multidimensional Arrays (2/2)  1 2 3 30 31 • Examples of applications Jan 32.1 31.8 31.9 32.3 32.4 Feb 32.6 32.6 33.0 0 0  : matrix[3][3] 31.8 32.3 30.9 31.6 32.2 Dec Daily temperatures: temperatures[12][31] Suise Jerna Zass Emily Ex1 Ex1 Ex1 Ex1 Ex2 Ex2 Ex2 Ex2 Ex3 Ex3 Ex3 Ex3 Lab1 Lab1 Lab1 Lab1 76 79 59 52 75 68 50 80 60 45 66 62 Lab2 Lab2 Lab2 Lab2 0 60 57 90 83 60 72 0 63 48 77 0 Lab3 Lab3 Lab3 Lab3 52 76 67 81 73 59 71 80 66 75 79 62 Lab4 Lab4 Lab4 Lab4 38 58 60 33 64 72 52 42 52 48 35 37 Lab5 Lab5 Lab5 Lab5 58 93 78 68 86 80 68 79 72 85 73 82 Students’ lab marks: marks[4][5][3] CS1010 (AY2013/4 Semester 1)

  24. 2.1 Multidimensional Array Initializers // nesting one-dimensional initializers inta[3][5] = { {4, 2, 1, 0, 0}, {8, 3, 3, 1, 6}, {0, 0 ,0, 0, 0} }; // the first dimension can be unspecified int b[][5] = { {4, 2, 1, 0, 0}, {8, 3, 3, 1, 6}, {0, 0, 0, 0, 0} }; // initializer with implicit zero values int d[3][5] = { {4, 2, 1}, {8, 3, 3, 1, 6} }; What happens to the uninitialized elements? CS1010 (AY2013/4 Semester 1)

  25. 2.2 Multidimensional Array: Example Week8_2DArray.c #include <stdio.h> #define N5// number of columns in array intsumArray(int[][N], int); // function prototype int main(void) { int foo[][N] = { {3,7,1}, {2,1}, {4,6,2} }; printf("Sum is %d\n", sumArray(foo, 3)); printf("Sum is %d\n", sumArray(foo, 2)); return0; } // To sum all elements in arr intsumArray(intarr[][N], int rows) { inti, j, total = 0; for (i = 0; i < rows; i++) { for (j = 0; j < N; j++) { total += arr[i][j]; } } return total; } Second dimension must be specified; first dimension is not required. CS1010 (AY2013/4 Semester 1) 

  26. 2.3 Accessing 2D Array Elements in Function A function header with 2D array parameter, function(int a[][5], ...) Why second dimension must be specified, but not the first dimension? • To access an element in a 2D array, it must know the number of columns. It needs not know the number of rows. • For example, given the following two 2D-arrays: A 3-column 2D array: A 5-column 2D array: • As elements are stored linearly in memory in row-major order, element a[1][0] would be the 4th element in the 3-column array, whereas it would be the 6th element in the 5-column array. • Hence, to access a[1][0] correctly, we need to provide the number of columns in the array. • For multi-dimensional arrays, all but the first dimension must be specified in the array parameter. : : CS1010 (AY2013/4 Semester 1)

  27. 2.4 Demo #1: Class Enrolment (1/5) • A class enrolment system can be represented by a 2D array enrol, where the rows represent the classes, and columns the students. For simplicity, classes and students are identified by non-negative integers. • A ‘1’ in enrol[c][s] indicates student s is enrolled in class c; a ‘0’ means s is not enrolled in c. • Assume at most 10 classes and 30 students. • Example of an enrolment system with 3 classes and 8 students: 0 1 2 3 4 5 6 7 • Queries: • Name any class with the most number of students • Name all students who are enrolled in all the classes 1 0 1 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1 0 1 1 1 0 0 1 0 2 CS1010 (AY2013/4 Semester 1)

  28. 2.4 Demo #1: Class Enrolment (2/5) • Inputs: • Number of classes and students • Number of data entries • Each data entry consists of 2 integers s and c indicating that student s is enrolled in class c. • Sample input: Number of classes and students: 3 8 Number of data entries: 15 Enter 15 entries (student class): 3 1 0 0 0 1 1 2 2 0 2 1 2 2 3 2 7 1 6 0 5 0 4 1 4 0 6 2 6 1 0 1 2 3 4 5 6 7 1 0 1 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1 0 1 1 1 0 0 1 0 2 CS1010 (AY2013/4 Semester 1)

  29. 2.4 Demo #1: Class Enrolment (3/5) #define MAX_CLASSES 10 #define MAX_STUDENTS 30 int main(void) { intenrol[MAX_CLASSES][MAX_STUDENTS] = { {0} }, numClasses, numStudents; readInputs(enrol, &numClasses, &numStudents); return0; } 1 0 1 0 1 1 1 0 1 0 1 1 1 0 1 1 // Read data into array enrol voidreadInputs(intenrol[][MAX_STUDENTS], int *numClassesPtr, int *numStudentsPtr) { int entries; // number of data entries inti, class, student; printf("Number of classes and students: "); scanf("%d %d", numClassesPtr, numStudentsPtr); printf("Number of data entries: "); scanf("%d", &entries); printf("Enter %d data entries (student class): \n", entries); // Read data into array enrol for (i = 0; i < entries; i++) { scanf("%d %d", &student, &class); enrol[class][student] = 1; } } 0 1 1 1 0 0 1 0 3 8 15 3 1 0 0 0 1 1 2 2 0 2 1 2 2 3 2 7 1 6 0 5 0 4 1 4 0 6 2 6 1 0 1 2 3 4 5 6 7 0 1 2 CS1010 (AY2013/4 Semester 1)

  30. 2.4 Demo #1: Class Enrolment (4/5) • Query 1:Name any class with the most number of students Row sums intclassWithMostStudents (intenrol[][MAX_STUDENTS], intnumClasses, intnumStudents) { intclassSizes[MAX_CLASSES]; int r, c; // row and column indices intmaxClass, i; for (r = 0; r < numClasses; r++) classSizes[r] = 0; for (c = 0; c < numStudents; c++) { classSizes[r] += enrol[r][c]; } // find the one with most students maxClass = 0; // assume class 0 has most students for (i = 1; i < numClasses; i++) if (classSizes[i] > classSizes[maxClass]) maxClass = i; returnmaxClass; } 5 6 4 1 0 1 0 1 1 1 0 1 0 1 1 1 0 1 1 0 1 1 1 0 0 1 0 0 1 2 3 4 5 6 7 0 1 2 CS1010 (AY2013/4 Semester 1)

  31. 2.4 Demo #1: Class Enrolment (5/5) • Query 2:Name all students who are enrolled in all classes // Find students who are enrolled in all classes voidbusiestStudents(intenrol[][MAX_STUDENTS], intnumClasses, intnumStudents) { int sum; intr, c; printf("Students who take all classes: "); for (c = 0; c < numStudents; c++) { sum = 0; for (r = 0; r < numClasses; r++) { sum += enrol[r][c]; } if (sum == numClasses) printf("%d", c); } printf("\n"); } 1 0 1 0 1 1 1 0 1 0 1 1 1 0 1 1 0 1 1 1 0 0 1 0 2 1 3 2 2 1 3 1 Column sums 0 1 2 3 4 5 6 7 0 1 Refer to Week8_ClassEnrolment.c for complete program. 2 CS1010 (AY2013/4 Semester 1)

  32. 2.5 Using UNIX input redirection (1/2) • When running programs involving arrays, we usually need to read in substantial amount of data for the arrays. It is hence not convenient for the user to key this the data interactively. • Instead, we usually store the data in a file, and the program reads the data from this file. • We will do file processing later in this module. For the moment, there is a simple way – redirect the input from a file instead of keying in interactively. • We will take the previous demo program, Class Enrolment, as an example. CS1010 (AY2013/4 Semester 1)

  33. 2.5 Using UNIX input redirection (2/2) 3 8 15 3 1 0 0 0 1 1 2 2 0 2 1 2 2 3 2 7 1 6 0 5 0 4 1 4 0 6 2 6 1 • Using an editor, create a file to contain the required data. • Let’s call the file enrolment.in. Its content is shown here. • Use the UNIX input redirection <to redirect input from the file: $ gcc –Wall Week8_ClassEnrolment.c –o enrolment $ enrolment <enrolment.in CS1010 (AY2013/4 Semester 1)

  34. 2.6 Exercise #1: Sum to Random Position (1/4) • Write a program Week8_SumToRandomPos.c that reads in values (of type float) for a 2D array with at most 5 rows and 8 columns, generates a random position in the array and sums the elements from index [0][0] to that position, in row-major order. • Your program should contain the function sumPartial () to generate a random position and return the sum of the elements up to that position. It also passes back the position generated. • What are the parameters of sumPartial()? • The incomplete program Week8_SumToRandomPos.c is given. Study the function scanArray(). CS1010 (AY2013/4 Semester 1)

  35. 2.6 Exercise #1: Sum to Random Position (2/4) • The sum is printed in 2 decimal places. • To ease data input, create a file to store the input data, and use UNIX input redirection as shown in section 2.5. • Sample run: $ Enter rows and columns: 3 4 $ Enter 12 values: 5.1 4.2 -6.312.4 7.5 8.6-3.7 11.8 9.9 -20.0 17.1 10.2 Sum to position [1][2] = 27.80 CS1010 (AY2013/4 Semester 1)

  36. 2.6 Exercise #1: Sum to Random Position (3/4) Week8_SumToRandomPos.c #include <stdio.h> #include <stdlib.h> #include <time.h> #define MAX_ROWS5 #define MAX_COLS5 voidscanArray(float [][MAX_COLS], int *, int *); intmain(void) { float array[MAX_ROWS][MAX_COLS], answer; scanArray(array, &rows, &cols); // call sumPartial() function below printf("Sum to position [%d][%d] = %.2f\n"); // incomplete return0; } CS1010 (AY2013/4 Semester 1)

  37. 2.6 Exercise #1: Sum to Random Position (4/4) Week8_SumToRandomPos.c voidscanArray(floatarr[][MAX_COLS], int*rowsPtr, int *colsPtr) { int r, c; printf("Enter rows and columns: "); scanf("%d %d", rowsPtr, colsPtr); printf("Enter %d values:\n", *rowsPtr * *colsPtr); for (r=0; r < *rowsPtr; r++) for (c=0; c < *colsPtr; c++) scanf("%f", &arr[r][c]); } // Generate a random position and sum elements from // position [0][0] to that position. // Fill in sumPartial() function below. CS1010 (AY2013/4 Semester 1)

  38. 3. Matrices • A two-dimensional array where all rows have the same number of elements is sometimes known as a matrix because it resembles that mathematical concept. • A matrix A with m rows and ncolumns, also called an mn (m by n) matrix, is represented mathematically in the following manner. • Note that in implementing the matrix as an array in C, the row and column indices start at 0 instead of 1. CS1010 (AY2013/4 Semester 1)

  39. 3.1 Demo #2: Matrix Addition (1/2) • To add two matrices, both must have the same size (same number of rows and columns). • To compute C = A + B, where A, B, C are matrices • ci,j = ai,j + bi,j • Examples: CS1010 (AY2013/4 Semester 1)

  40. 3.1 Demo #2: Matrix Addition (2/2) Week8_MatrixOps.c // To sum mtxA and mtxB to obtain mtxC voidsumMatrix(floatmtxA[][MAX_COL], floatmtxB[][MAX_COL], floatmtxC[][MAX_COL], introw_size, intcol_size) { int row, col; for (row=0; row<row_size; row++) for (col=0; col<col_size; col++) mtxC[row][col] = mtxA[row][col] + mtxB[row][col]; } 10 21 7 9 3 7 18 20 ? 13 28 ? 25 ? 29 ? + = 4 6 14 5 6 5 8 15 ? 10 11 ? 22 ? ? 20 CS1010 (AY2013/4 Semester 1)

  41. 3.2 Exercise #2: Matrix Multiplication (1/3) • To multiply two matrices A and B, the number of columns in A must be the same as the number of rows in B. • The resulting matrix has same number of rows as A and number of columns as B. • For example, multiplying a 24 matrix with a 43 matrix gives a 23 matrix. m n matrix n p matrix  = m p matrix CS1010 (AY2013/4 Semester 1)

  42. 3.2 Exercise #2: Matrix Multiplication (2/3) • To compute C = A  B, where A, B, C are matrices • ci,j = (ai,1 b1,j ) + (ai,2 b2,j ) + . . . + (ai,nbn,j) • ci,j is sum of terms produced by multiplying the elements of A’s row i with B’s column j. • Examples: • Complete the prodMatrix() function in Week8_matrixOps.c CS1010 (AY2013/4 Semester 1)

  43. 3.2 Exercise #2: Matrix Multiplication (3/3) • Multiplying a 2  4 matrix with a 4  3 matrix. col 1 col 0 row 0, col 0 row 0, col 1 3 2 1 2 2 3 ? 17 ? ? 11 row 1 row 0 15  = 1 3 0 ? ? ? 13 13 6 2 1 3 row 1, col 0 2 1 3 2 6 + 2 + 3 + 4 = 15 3 0 2 1 4 + 2 + 2 + 9 = 17 9 + 0 + 2 + 2 = 13 CS1010 (AY2013/4 Semester 1)

  44. 4. Maze Problem (1/2) • Let’s consider a maze that is represented by a two-dimensional 6  6 integer array. • The value of each array element is either 0 (representing a wall) or 1 (representing a cell). • The starting and exit points in the maze are specified by the cells maze[0][0]and maze[5][5] respectively. • A path is represented by a single-dimensional character array with four possible element values representing the move directions: ‘N’ (for north), ‘S’ (for south), ‘E’ (for east), and ‘W’ (for west). Each path is defined with respect to the starting cell maze[0][0]. CS1010 (AY2013/4 Semester 1)

  45. 4. Maze Problem (2/2) Start 0 1 2 3 4 5 • Example of a 6  6 maze 0 1 2 3 4 5 Exit Cell Wall CS1010 (AY2013/4 Semester 1)

  46. 4. Exercise #3: Valid Path 0 1 2 3 4 5 • A path in a maze is defined to be valid if the path is within the maze and does not knock against any wall. • Examples: • Valid path: ‘E’, ‘E’, ‘S’, ‘N’, ‘E’, ‘E’, ‘S’ • Invalid path: ‘S’, ‘S’, ‘W’ • Invalid path: ‘S’, ‘S’, ‘S’, ‘E’ • Write a functionintisValid (int maze[][6], char path[])that takes in a 6  6 maze and a path with at most 10 characters. It returns 1 if path is valid in maze, or returns 0 otherwise. • This is a take-home exercise. • An incomplete program Week8_IsValid.c is given. It handles string input which is not covered yet. 0 1 2 3 4 5 CS1010 (AY2013/4 Semester 1)

  47. Announcements/Things-to-do • ReviseChapter 6 Numeric Arrays • Mid-Semester Test • 12 October 2013, Saturday • Please see http://www.comp.nus.edu.sg/~cs1010/3_ca/termtests.html for details • Next week’s lecture • Chapter 7: Strings and Pointers • Lessons 7.1 – 7.4 , Lessons 7.6 – 7.8 CS1010 (AY2013/4 Semester 1)

  48. End of File

More Related