EBK ENHANCED DISCOVERING COMPUTERS & MI
1st Edition
ISBN: 9780100606920
Author: Vermaat
Publisher: YUZU
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Chapter 7, Problem 30SG
Explanation of Solution
The differences between LCD, CCF, LED, TFT, OLED, and AMOLED technologies are as follows:
| LCD | CCF | LED | TFT | OLED | AMOLED |
| LCD stands for Liquid Crystal Display. | CCF stands for Cold Cathode Fluorescent Lamp. | LED stands for Light-Emitting Diode. | TFT stands for Thin-Film Transistor. | OLED stands for Organic LED. | AMOLED stands for Active-Matrix OLED. |
| LCD contains a liquid compound between the two sheets of a material. | CCF is used in the light source called backlight. | LED display is used in the backlight. | TFT is used as separate transistor to apply charges to each liquid crystal cell... |
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Rate Category Rate per MinuteDaytime (6:00 a.m. through 5:59 p.m.) $0.07Evening (6:00 p.m. through 11:59 p.m.) $0.12Off-Peak (midnight through 5:59 a.m.) $0.05Write a GUI application that allows the user to select a rate category (from a set of radio buttons), and enter the number of minutes of the call into an Entry widget. An info dialog box should display the charge for the call.
Name and Address
The Name and Address Problem
Write a GUI program that displays your name and address when a button is clicked. The program’s window should appear as the sketch on the left side of Figure 13-61 when it runs. When the user clicks the Show Info button, the program should display your name and address, as shown in the sketch on the right of the figure.
Exercise 1 Function and Structure [30 pts]
Please debug the following program and answer the following questions. There is a cycle in a linked
list if some node in the list can be reached again by continuously following the next pointer.
#include
typedef struct node {
int value;
struct node *next;
} node;
int 11_has_cycle (node *first)
if (first
==
node *head =
{
NULL) return 0;
first;
while (head->next != NULL) {
}
if (head
first) {
return 1; }
head = head->next;
return 0;
void test ll_has_cycle () {
int i;
node nodes [6];
for (i = 0; i < 6; i++) {
nodes [i] .next = NULL;
nodes [i].value = i;
}
nodes [0] .next
=
&nodes [1];
nodes [1] .next
=
&nodes [2];
nodes [2] .next
=
&nodes [3];
nodes [3] .next
nodes [4] .next
&nodes [4];
NULL;
nodes [5] .next = &nodes [0];
printf("1. Checking first list for cycles. \n Function 11_has_cycle says it
has s cycle\n\n", 11_has_cycle (&nodes [0])?"a":"no");
printf("2. Checking length-zero list for cycles. \n Function 11_has_cycle
says it has %s…
Chapter 7 Solutions
EBK ENHANCED DISCOVERING COMPUTERS & MI
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