1. EECE.2160 ECE Application Programming
Instructor:
Dr. Michael Geiger
Spring 2019
Lecture 21
Arrays

2. ECE Application Programming: Lecture 21
Lecture outline
Announcements/reminders
Program 5 due 4/5
Exam 2: Monday, April 1
Will again be allowed one 8.5” x 11” note sheet
Today’s lecture
One-dimensional arrays
Two-dimensional arrays
6/26/2019
ECE Application Programming: Lecture 21

3. Review of scalar variables
Variables (up to now) have: name type (int, float, double, char) address value
N 28C4 (int)
q 28C8 (float)
r 28CC (float) 35
3.14
8.9
e.g. Name type address value
N integer 28C4 35
q float 28C8 3.14
r float 28CC 8.9
6/26/2019
ECE Application Programming: Lecture 21

4. ECE Application Programming: Lecture 21
Intro to Arrays
Any single element of x may be used like any other scalar variable
x[0] 45
3044
x[1] 55
3048
x[2] 25
304C
x[3] 85
3050
x[4] 75
3054
printf("%d",x[3]); // prints 85
printf("%d",x[7]+x[1]); // prints 115
x[5] 65
3058
x[6]
100
305C
x[7] 60
3060
6/26/2019
ECE Application Programming: Lecture 21

5. ECE Application Programming: Lecture 21
Declaring Arrays
Define an 8 element array:
int x[8];
Elements numbered 0 to 7
Arrays in C always start with location 0 (zero based)
The initial value of each array element is unknown (just like scalar variables)
x[0] ?
3044
x[1] ?
3048
x[2] ?
304C
x[3] ?
3050
x[4] ?
3054
x[5] ?
3058
x[6] ?
305C
x[7] ?
3060
6/26/2019
ECE Application Programming: Lecture 21

6. Declaring/defining Arrays
double A[]={ 1.23, , 2.718, };
A[0]
1.23
4430
You can also define the values to be held in the array and instruct the compiler to figure out how many elements are needed.
Not putting a value within the [] tells the compiler to determine how many locations are needed.
A[1]
4438
A[2]
2.718
4440
A[3]
0.7071
4448
6/26/2019
ECE Application Programming: Lecture 21

7. Working with Arrays (input)
#include <stdio.h>
void main(void) { int x[8]; int i;
// get 8 values into x[]
for (i=0; i<8; i++) { printf("Enter test %d:",i+1); scanf("%d",&x[i]); } }
6/26/2019
ECE Application Programming: Lecture 21

8. Working with Arrays (input)
Sample run (user input underlined):
Enter test 1:80 Enter test 2:75 Enter test 3:90 Enter test 4:100 Enter test 5:65 Enter test 6:88 Enter test 7:40 Enter test 8:90
x[0] 80
x[1] 75
x[2] 90
x[3]
100
x[4] 65
x[5] 88
x[6] 40
x[7] 90
6/26/2019
ECE Application Programming: Lecture 21

9. ECE Application Programming: Lecture 21
Pitfalls
What happens if we change previous code to:
#include <stdio.h>
void main(void) { int x[8]; int i; float sum, avg; // used later
// get 8 values into x[]
for (i=0; i<=8; i++) { printf("Enter test %d:",i+1); scanf("%d",&x[i]); } }
6/26/2019
ECE Application Programming: Lecture 21

10. ECE Application Programming: Lecture 21
Pitfalls (cont.)
Although x has 8 elements, x[8] is not one of those elements!
Compiler will not stop you from accessing elements outside the array
Must make sure you know the size of the array
6/26/2019
ECE Application Programming: Lecture 21

11. ECE Application Programming: Lecture 21
Example
What does the following program print?
int main() {
int arr[10];
int i;
printf("First loop:\n");
for (i = 0; i < 10; i++) {
arr[i] = i * 2;
printf("arr[%d] = %d\n", i, arr[i]); }
printf("\nSecond loop:\n");
for (i = 0; i < 9; i++) {
arr[i] = arr[i] + arr[i + 1];
return 0;
6/26/2019
ECE Application Programming: Lecture 21

12. ECE Application Programming: Lecture 21
Example solution
First loop: arr[0] = 0 arr[1] = 2 arr[2] = 4 arr[3] = 6 arr[4] = 8 arr[5] = 10 arr[6] = 12 arr[7] = 14 arr[8] = 16 arr[9] = 18
Output continued: Second loop: arr[0] = 2 arr[1] = 6 arr[2] = 10 arr[3] = 14 arr[4] = 18 arr[5] = 22 arr[6] = 26 arr[7] = 30 arr[8] = 34
6/26/2019
ECE Application Programming: Lecture 21

13. Two-dimensional arrays
Two-dimensional arrays: can be used to represent tabular data
Declaration: <type> <name>[<rows>][<cols>]
Example (see below): int x[3][4];
Index elements similarly to 1-D arrays
Col. 0
Col. 1
Col. 2
Col. 3
Row 0
x[0][0]
x[0][1]
x[0][2]
x[0][3]
Row 1
x[1][0]
x[1][1]
x[1][2]
x[1][3]
Row 2
x[2][0]
x[2][1]
x[2][2]
x[2][3]
6/26/2019
ECE Application Programming: Lecture 21

14. ECE Application Programming: Lecture 21
Initializing 2D arrays
Can initialize similarly to 1D arrays, but must specify dimensions
Each row treated like a 1D array; rows separated by commas:
int y[3][4] = { {1, 2, 3, 4},
{5, 6, 7, 8},
{9, 10, 11, 12} }; 1 2 3 4 5 6 7 8 9 10 11 12
6/26/2019
ECE Application Programming: Lecture 21

15. ECE Application Programming: Lecture 21
2D arrays and loops
Typically use nested loops to work with 2-D arrays
One loop inside another:
for (i = 0; i < 3; i++) {
for (j = 0; j < 4; j++) {
x[i][j] = y[i][j] * 2; }
Be careful in loop body—switching your loop indices will cause trouble
Using x[j][i] would take you outside of the array!
6/26/2019
ECE Application Programming: Lecture 21

16. Example: Working with 2-D arrays
Complete this program, which counts the # of negative values in each row of a 2-D array (assume the necessary #includes are done): #define NRows 3 // # of rows #define NCols 4 // # of columns int main() { double x[NRows][NCols] = // 2-D array { { 10, 2.5, 0, 1.5}, {-2.3, -1.1, -0.2, 0}, {10.5, -6.1, 23.4, -9.2} }; int negCnt[NRows] = {0}; // Initialize entire row count array to 0 int i, j; // Row and column indices /* INSERT CODE HERE--Visit every element in array x and count the number of negative values in each row */ // Now print the row counts for (i = 0; i < NRows; i++) printf(“Row %d has %d negative values.\n”, i, negCnt[i]); return 0; }
6/26/2019
ECE Application Programming: Lecture 21

17. ECE Application Programming: Lecture 21
Example solution
/* Code to be added to visit every element in array x and count the number of negative values in each row */ for (i = 0; i < NRows; i++) for (j = 0; j < NCols; j++) if (x[i][j] < 0) negCnt[i]++;
6/26/2019
ECE Application Programming: Lecture 21

18. ECE Application Programming: Lecture 21
Final notes
Next time
Arrays and functions
Reminders:
Program 5 due 4/5
Exam 2: Monday, April 1
Will again be allowed one 8.5” x 11” note sheet
6/26/2019
ECE Application Programming: Lecture 21