Answer:
B. blocking I/O
Explanation:
Most of the input and output request placed to the computer considers the blocking request. It means that the controls given cannot be returned to the given application until and unless the input/output is complete.
Thus, blocking the input/output does not return till the input and output is complete.
With a blocking I/O, the process is moved to a wait queue when the I/O request is made, and it moved backs to the ready queue as soon as the request is complete, thereby allowing the other processes to run in the meantime.
Thus (B) blocking I/O is the answer.
Answer:
Explanation:
The following code is written in Java. It is hard to fully create the code without the rest of the needed code including the T class and the Measurable interface. Regardless the following code can be implemented if you have that code available.
public static T minmax(ArrayList<T> mylist) {
T min = new T();
T max = new T();
for (int x = 0; x < mylist.size(); x++) {
if (mylist.get(x) > max) {
max = mylist.get(x);
} else if (mylist.get(x) < min) {
min = mylist.get(x);
}
}
return (min, max);
}
The catalog contains a list of all the resources owned by the library.
- Mabel <3
Answer:
When the transmission exceeds 667 packets
Explanation:
In computer networking, a packet is a chunk of data transmitted across the network. The packet size of an Ethernet network is 1.5kilobytes, while the packet size of an IP packet payload is 64 kilobytes.
A switch is a physical network device that connects nodes or workstations while communicating the packets (or frames). The I/O bus size bandwidth is 1Gbps which allows approximately 667 packets. Once this packet size is crossed, the bus becomes a limiting factor or bottle neck.
int firstNumber,secondNumber = -1, duplicates = 0;
do {
cin >> firstNumber;
if ( secondNumber == -1) {
secondNumber = firstNumber;
}else {
if ( secondNumber == firstNumber )
duplicates++;
else
secondNumber = firstNumber;
}
} while(firstNumber > 0 );
cout << duplicates;
