EM108 Software Development for Engineers

Transcription

1 EE108 Section 2 Arrays page 1 of 19 EM108 Software Development for Engineers Section 2 - Arrays 1) Introduction 2) Definition 3) Using Arrays 4) Array Initialisation 5) Searching Within Arrays 6) Array Parsing/Processing 7) Multi-dimensional Arrays 8) Array as Function Parameters 9) Complex example 10) Laboratory examples

2 EE108 Section 2 Arrays page 2 of Introduction Array: Data structure Stores a collection of data items All data items have to be of the same type Idea: I t e m 1 I t e m 2 I t e m 3 I t e m N Examples: array of integer numbers array of real numbers Components: Items type Array name Number of items Items order

3 EE108 Section 2 Arrays page 3 of Definition <array_item_type> <array_name> [ <array_size> ] Components: Items type Name: Effect: <array_item_type> = <type> <array_name> = <name> Maximum number of items (array size): <array_size> = <const_expression> <const_expression> = [ <constant> + <operator> + <const_expression>] Order of items (implicit) from 0 (first item) to <array_size>-1 (last item) Computer allocates a contiguous memory block of size <array_size> * sizeof(<array_item_type>) Usage: int marks[6]; 4 * 6 = 24 B char ascii[255]; 1 * 255 = 255 B float avg_day_temps[366]; 8 * 366 = 2928 B

4 EE108 Section 2 Arrays page 4 of 19 Memory allocation: marks[0] marks[1] marks[2] marks[3] marks[4] marks[5] The contiguous memory block consists of memory locations allocated as each array item had been allocated separately The size of each such memory location is equal to the size required to store a variable of type <array_type> Notes: There is no length limit for array names - usually they should not be very long (prone to errors!) There is no limit for the array size - depends on the available memory There is no general accepted naming convention - usually names relate to what the arrays are for The size of arrays has to be indicated statically, when writing the program - if not fully used waist of resources - if required more space impossible to extend

5 EE108 Section 2 Arrays page 5 of Using Arrays Components: <array_name> [ <index> ] Array name: <array_name> - must be defined prior to usage Array index: <index> <index> = <int_expression> - must be evaluated to an integer Example: marks[0] ascii[201] avg_day_temps[10] Notes: marks marks[0] Although allocated, it is not required to use every location within an array - however the locations used are usually contiguous C compiler does not check or enforce array bounds - e.g. marks[-1] and ascii[255] will cause run errors An array item is treated as any other variable of scalar type (int, float, char) - e.g. marks[2] = 65; Regular operations with arrays include: - initialisation, linear search and parsing/processing

6 EE108 Section 2 Arrays page 6 of Array Initialisation Types of initialisation: At declaration - complete initialisation (all items of the array) e.g. int vals[5] = 2, 3, 4, 5, 6; - partial initialisation (some items only) e.g. int vals[5] = 2, 3, 4; During execution - with user input - complete initialisation (all items of the array) e.g. int vals[5]; for (index = 0; index < 5; index++) scanf( %d, &vals[index]); - partial initialisation (some items of the array) e.g. int vals[5]; scanf( %d%d, &vals[0], &vals[3]); - without user input - complete initialisation (all items of the array) e.g. int vals[5]; for (index = 0; index < 5; index++) vals[index] = 0; - partial initialisation (some items of the array) e.g. int vals[5]; vals[0] = 3; vals[3] = x + y/4;

7 EE108 Section 2 Arrays page 7 of 19 Most Common Initialisation: <array_item_type> <array_name> [ <array_size> ] ; int index; for (index = 0; index < <array_size>; index++) <array_name> [ index ] = <expression>; Example: #include <stdlib.h> int main() int marks[6]; int index; for (index = 0; index < 6; index++) marks[index] = 0; 3 6 return (EXIT_SUCCESS); 21 Effect: /* init */ index = 0; N o t e s : /* test */ (0 < 6) /* true */ /* body */ marks[0] = 0; /* update */ index L = 0 + 1; /* test */ (1 < 6) /* true */ /* body */ marks[1] = 0; /* update */ index = 1 + 1; /* test */ (5 < 6) /* true */ /* body */ marks[5] = 0; /* update */ index = 5 + 1; /* test */ (6 < 6) /* false */

8 EE108 Section 2 Arrays page 8 of Linear Search Through Arrays Task Description: search for an element or a number of elements within a given array that satisfy a certain condition linear search examines all the items systematically search may stop or not when the first item that satisfies the condition was found Examples: Ex1: search for all items with value less than 40 int marks[6]; int index; for (index = 0; index < 6; index++) if (marks[index] < 40) printf( Found: %d, marks[index]); Output: marks [0] [1] [2] [3] [4] [5] Found: 35 Found: 25

9 EE108 Section 2 Arrays page 9 of 19 Ex 2: search for the first item with value equal to 40 int marks[6]; int index; for (index = 0; index < 6; index++) if (marks[index] = = 40) printf( Found: %d, marks[index]); How it works? marks [0] [1] [2] [3] [4] [5] /* init */ index = 0; /* test: index < 6 */ (0 < 6) => TRUE /* body: if(...) */ marks[0] == 40? => FALSE /* update: index++ */ 4 index = 0 + 1; /* test: index < 6 */ (1 < 6) => TRUE /* body: if(...) */ marks[1] == 40? => TRUE /* body: printf (...)*/ 7 Found: 40 /* update: index++ */ index = 1 + 1; /* test: index < 6 */ (2 < 6) => TRUE /* body: if(...) */ marks[2] == 40? => FALSE /* update: index++*/ index = 2 + 1; /* test: index < 6 */ (3 < 6) => TRUE /* body: if(...) */ marks[3] == 40? => TRUE /* body: printf (...)*/ Found: 40 /* update: index++ */ index = 3 + 1; /* test: index < 6 */ (4 < 6) => TRUE /* body: if(...) */ marks[4] == 40? => FALSE /* update: index++ */ index = 5 + 1; /* test: index < 6 */ (6 < 6) => FALSE problem: second example prints all the items and does not print anything if the item was not found

10 EE108 Section 2 Arrays page 10 of 19 Output: SOLUTION 1: using break int marks[6]; int index; for (index = 0; index < 6; index++) if (marks[index] == 40) printf( Found: %d in position: %d, marks[index],index); 7 break; 8 if (index == 6) 9 printf( Not found in the array! ); Found: 40 in position: 1 3 How it works? /* init */ index = 0; /* test: index < 6 */ 2 (0 < 6) => TRUE /* body: if(...) */ marks[0] == 40? => FALSE /* update: index++ */ 4 index = 0 + 1; /* test: index < 6 */ (1 < 6) => TRUE /* body: if(...) */ marks[1] == 40? => TRUE /* body: printf (...)*/ Found: 40 in position: 1 /* body: break */ 8 break; /* END of FOR statement */ /* if (index==6) */ (1= = 6)?=> FALSE

11 EE108 Section 2 Arrays page 11 of 19 SOLUTION 2: using while int marks[6]; int index; index = 0; 1 while (index < 6 && marks[index]!= 40) index++; if (index == 6) 5 printf( Not found in the array! ); else printf( Found: %d in position: %d\n, marks[index], index); Output: Found: 40 in position: 1 How it works? /* init */ index = 0; /* while test */ 2 (0<6) => TRUE && marks[0]!= 40? => TRUE /* update: index++ */ index = 0 + 1; /* while test */ 4 (1<6) => TRUE && marks[1]!= 40? => FALSE /* END of WHILE statement */ 1 3 /* if (index==6) */ (1= = 6)?=> FALSE 5 /* printf( ) */ 6 Found: 40 in position: 1 6 Notes: When using a loop the test can be used for both testing the array s index and its items for certain value

12 EE108 Section 2 Arrays page 12 of Parsing/Processing Arrays Task Description: parse all the array item by item apply some processing to each item some items may be processed differently Examples: The simplest processing is printing all the items: int marks[6]; int index; for (index = 0; index < 6; index++) printf( marks[%d] = %d\n, index, marks[index]); Sometimes processing of some items only is required: int marks[6]; int index; for (index = 0; index < 6; index++) if (marks[index] >= 40) printf( marks[%d] = %d\n, index, marks[index]); Sometimes processing is more complex: int marks[6]; int index; for (index = 0; index < 6; index++) marks[index] = complex_proc(index);

13 EE108 Section 2 Arrays page 13 of 19 Notes: 2.7 Multiple-dimensional Arrays Arrays can be multiple dimensional and their parsing is performed line-by-line, on each dimension In order to parse a N-dimensional array N index variables are required, each for one of its dimensions A particular case of an N-dimensional array si a 2- dimensional array that was denoted matrix In order to parse a matrix (NxM), two indexes are required: i and j, where i = 0, N 1 and j = 0, M 1 r o w s i J [0] [1] [2] [3] [4] columns [0] [1] [2] [3] [5] table [3, 2] Example: int table[4][6]; int i,j; for (i = 0; i < 4; i++) for (j = 0; j < 6; j++) printf( table[%d, %d] = %d\n, i, j, table[i][j]);

14 EE108 Section 2 Arrays page 14 of 19 Notes: 2.8 Arrays as Function Parameters Arrays are passed as parameters to any function using by-address strategy In consequence functions receive only the reference to the beginning of the contiguous memory zone allocated for the array As there is no indication for the functions about the size of the arrays, there must be other parameters that indicate the size Example: Function that computes the average value of a onedimensional array: float compute_avg(int marks[], int size) int i; float avg = 0.0; for (i = 0; i < size; i++) avg = avg + marks[i]; avg = avg / size; return avg; int main() int marks[6]; float average; average = compute_avg(marks, 6); return (EXIT_SUCCESS);

15 EE108 Section 2 Arrays page 15 of 19 Example: 2.9 Complex Example Maintain the record of average marks for the students from a school given their studentids #include <stdio.h> #include <stdlib.h> /* declarations */ /* function that initialises student records and returns the number of registered students */ int init_records (int sids[], float avgm[], int maxsize); /* function that searches for the record position given a studentid and returns it */ int search_student (int sids[], int studid, int size); /* function that sets the average mark in the records given a position */ void set_mark (float avgm[], int pos, float avgmark); /* function that lists the studentids and average marks for all registered students */ void list_marks (int sids[], float avgm[], int size); /* main function */ int main() /* 1-dimensional array that stores studentids for all registered students */ int studentids[5000]; /* 1-dimensional array that stores average marks for all registered students */ float avgmarks[5000]; /* no current registered students (no records) */ int no_recs = 0;

16 EE108 Section 2 Arrays page 16 of 19 /* Other variables */ int studid; float mark; /* Welcome message */ printf( Welcome to the program!\n ); /* Init. records by calling init_records() */ no_recs = init_records (studentids, avgmarks, 5000); /* Ask user for input data */ printf( Enter student ID: ); scanf( %d, &studid); printf( \nenter the mark ); scanf( %f, &mark); /* Update the records with user data */ set_mark (avgmarks, search_student (studentids, no_recs, studid), mark); /* Print studentids and students marks */ list_marks (studentids, avgmarks, no_recs); printf( Goodbye!\n ); return (EXIT_SUCCESS);

17 EE108 Section 2 Arrays page 17 of 19 /* function definitions */ int init_records (int sids[], float avgm[], int maxsize) int size = 0; int ans = 1; while (ans!= 0 && size < maxsize) printf( New student registration:\n ); /* Read student ID */ printf( Enter student ID:\n ); scanf( %d, &sids[size]); /* Initialise average mark with 0.0 */ avgm[size] = 0.0; /* Update no of records */ size++; /* Check for new student registration */ printf( Continue (yes=1, no=0):\n ); scanf( %d, &ans); return size; int search_student (int sids[], int size, int studid) int i; for (i = 0; i < size; i++) if (sids[i] == studid) return i; return 1;

18 EE108 Section 2 Arrays page 18 of 19 void set_mark (float avgm[], int pos, float avgmark) if (pos >= 0) avgm[pos] = avgmark; void list_marks (int sids[], float avgm[], int size) int i; printf( StudentID Avg Mark \n ); for (i = 0; i < size; i++) printf( %8d %8.2f \n, sids[i], avgm[i]); [0] [1] [no_recs -1] studentids [0] [1] [no_recs -1] avgmarks Main tasks: Initialise the records (studentids and avgmarks arrays) o Read each student ID o Place 0 in avgmarks for each student Read one student ID and the average mark Update the corresponding record in avgmarks for that student Print the whole list of marks and student IDs

19 EE108 Section 2 Arrays page 19 of Laboratory exercises Read a text letter-by-letter, store it into an array and then print it letter-by-letter /* letters.c */ #include <stdlib.h> #include <stdio.h> int main(void) char letters[50], ch; int size, i; printf("enter no of letters (max 50):\n"); scanf("%d", &size); printf("enter the array letter by letter:\n"); for (i = 0; i < size; i++) printf("enter letters[%d]: ", i); scanf("%c%c", &letters[i], &ch); for (i = 0; i < size; i++) printf("letters[%d]=%c\n", i, letters[i]); printf("goodbye!\n"); return(exit_success); Re-write the example using functions