Why is double?buffered input sometimes required?
What will be an ideal response?
The simplest input mechanism is to read the current value of the input. This has many problems, for example, the input may be changing. By latching the input in a flip?flop (e.g., a D?type), the data can be instantaneously captured. Once the data has been captured, it is read. A problem occurs if a new input arrives before the last value has been read. In that case either the new value is lost or the previous captured but not read data is lost. Double?buffering provides a solution. As soon as a value is captured, it is transferred from the input latch to the reading latch (a second flip?flop). This means that the input latch is ready for a new value.
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You can save files to the hard drive located inside your computer or to a(n) ________ drive such as a USB flash drive
A) internal B) external C) optional D) compact
Show the output of the following code:
``` public class Test { public static void main(String[] args) { int[] x = {1, 2, 3, 4, 5}; increase(x); int[] y = {1, 2, 3, 4, 5}; increase(y[0]); System.out.println(x[0] + " " + y[0]); } public static void increase(int[] x) { for (int i = 0; i < x.length; i++) x[i]++; } public static void increase(int y) { y++; } }``` a. 0 0 b. 1 1 c. 2 2 d. 2 1 e. 1 2