Answer: The SFD is the eight-bit (one-byte) value that marks the end of the preamble, which is the first field of an Ethernet packet, and indicates the beginning of the Ethernet frame.
Answer:
The solution code is written in Java
- public static void checkCommonValues(int arr1[], int arr2[]){
- if(arr1.length < arr2.length){
- for(int i = 0; i < arr1.length; i++){
- for(int j = 0; j < arr2.length; j++){
- if(arr1[i] == arr2[j]){
- System.out.print(arr1[i] + " ");
- }
- }
- }
- }
- else{
- for(int i = 0; i < arr2.length; i++){
- for(int j = 0; j < arr1.length; j++){
- if(arr2[i] == arr1[j]){
- System.out.print(arr2[i] + " ");
- }
- }
- }
- }
- }
Explanation:
The key idea of this method is to repeated get a value from the shorter array to check against the all the values from a longer array. If any comparison result in True, the program shall display the integer.
Based on this idea, an if-else condition is defined (Line 2). Outer loop will traverse through the shorter array (Line 3, 12) and the inner loop will traverse the longer array (Line 4, 13). Within the inner loop, there is another if condition to check if the current value is equal to any value in the longer array, if so, print the common value (Line 5-7, 14-16).
Good luck and I hope u can find the right answer
Answer:
All
Explanation:
From the available options given in the question, once the while loop terminates any of the provided answers could be correct. This is because the arguments passed to the while loop indicate two different argument which both have to be true in order for the loop to continue. All of the provided options have either one or both of the arguments as false which would break the while loop. Even options a. and b. are included because both would indicate a false statement.
Answer:
CompTIA A+ there ya go!!!
