A .) Sort
more ... To arrange or group in a special way (such as by size, type or alphabetically).
It's an example of a poorly written email, it looks like somebody is angry and yelling at you, these types of emails can be described as a poor etiquette.
Answer:
<em>The registers that are compared are instructions 3 and 4</em>
<em>Explanation:</em>
<em>From the question given,</em>
<em>Recall that we need to explain what the hazard detection unit is doing during the 5th cycle of execution and which registers are being compared.</em>
<em>Now,</em>
<em>The instructions on the 5th cycle, at the stage ID/EX and IF/ID:</em>
<em>The instruction values are in ID/EX : sub $t2, $t3, $t6 (instruction 3)</em>
<em>The instruction values are in IF/ID: sub $t3, $t1 $t5 (instruction 4)</em>
<em>The register $t3 is compared in the instructions 3 and 4</em>
<em>The hazard detection unit between instruction 4 and 5t o be compared, it need to find out the values of $t1</em>
<em />
Since both arrays are already sorted, that means that the first int of one of the arrays will be smaller than all the ints that come after it in the same array. We also know that if the first int of arr1 is smaller than the first int of arr2, then by the same logic, the first int of arr1 is smaller than all the ints in arr2 since arr2 is also sorted.
public static int[] merge(int[] arr1, int[] arr2) {
int i = 0; //current index of arr1
int j = 0; //current index of arr2
int[] result = new int[arr1.length+arr2.length]
while(i < arr1.length && j < arr2.length) {
result[i+j] = Math.min(arr1[i], arr2[j]);
if(arr1[i] < arr2[j]) {
i++;
} else {
j++;
}
}
boolean isArr1 = i+1 < arr1.length;
for(int index = isArr1 ? i : j; index < isArr1 ? arr1.length : arr2.length; index++) {
result[i+j+index] = isArr1 ? arr1[index] : arr2[index]
}
return result;
}
So this implementation is kind of confusing, but it's the first way I thought to do it so I ran with it. There is probably an easier way, but that's the beauty of programming.
A quick explanation:
We first loop through the arrays comparing the first elements of each array, adding whichever is the smallest to the result array. Each time we do so, we increment the index value (i or j) for the array that had the smaller number. Now the next time we are comparing the NEXT element in that array to the PREVIOUS element of the other array. We do this until we reach the end of either arr1 or arr2 so that we don't get an out of bounds exception.
The second step in our method is to tack on the remaining integers to the resulting array. We need to do this because when we reach the end of one array, there will still be at least one more integer in the other array. The boolean isArr1 is telling us whether arr1 is the array with leftovers. If so, we loop through the remaining indices of arr1 and add them to the result. Otherwise, we do the same for arr2. All of this is done using ternary operations to determine which array to use, but if we wanted to we could split the code into two for loops using an if statement.
<span>The statement that P2P networks are most commonly used in home networks is true.
</span>P2P stands for peer-to-peer communication network. It is an example of local administration. <span> Two or more PCs share files and access to devices such as printers without requiring a separate server computer or server software with P2P type of network.</span>
