Within the Flags detail is a flag titled recursion desired. This flag shows whether or not the local DNS should continue to query other DNSs if it is not able to resolve the current query. As DNS is local, it may or may not have the enough information to allow the address to be resolved. If the recursion flag is set, the local <span>DNS will continue to query higher level DNSs until it is able to resolve the address. In short, t</span>he condition is when a flag is raised and it doesn’t have enough <span>information to allow the request.</span>
A string is generally considered as a data type and is often implemented as an array data structure of bytes (or words) that stores a sequence of elements, typically characters, using some character encoding. String may also denote more general arrays or other sequence (or list) data types and structures.
Answer:
And clicking the security tab option.
Explanation:
Lets explain what an object's ACL is. I will use an example to best explain this. Let's suppose that user Bob would want to access a folder in a Windows environment. What supposedly will happen is that Windows will need to determine whether Bob has rights to access the folder or not. In order to do this, an ACE with the security identity of John will be created. These ACEs are the ones that grant John access to the folder and the ACLs of this particular folder that John is trying to access is a list of permissions of everyone who is allowed to access this folder. What this folder will do is the to compare the security identity of John with the folders ACL and determine whether John has Full control of the folder or not.
By right clicking the folder and selecting the security tab, John will be in a position to see a list of the permissions (ACLs) granted to him by the folder.
Answer:
OC
Explanation:
They were adapted from the Coleco company.
Answer:
- public static String bothStart(String text1, String text2){
- String s = "";
-
- if(text1.length() > text2.length()) {
- for (int i = 0; i < text2.length(); i++) {
- if (text1.charAt(i) == text2.charAt(i)) {
- s += text1.charAt(i);
- }else{
- break;
- }
- }
- return s;
- }else{
- for (int i = 0; i < text1.length(); i++) {
- if (text1.charAt(i) == text2.charAt(i)) {
- s += text1.charAt(i);
- }else{
- break;
- }
- }
- return s;
- }
- }
Explanation:
Let's start with creating a static method <em>bothStart()</em> with two String type parameters, <em>text1 </em>&<em> text2</em> (Line 1).
<em />
Create a String type variable, <em>s,</em> which will hold the value of the longest substring that both inputs start with the same character (Line 2).
There are two possible situation here: either <em>text1 </em>longer than<em> text2 </em>or vice versa. Hence, we need to create if-else statements to handle these two position conditions (Line 4 & Line 13).
If the length of<em> text1</em> is longer than <em>text2</em>, the for-loop should only traverse both of strings up to the length of the <em>text2 </em>(Line 5). Within the for-loop, we can use<em> charAt()</em> method to extract individual character from the<em> text1</em> & <em>text2 </em>and compare with each other (Line 15). If they are matched, the character should be joined with the string s (Line 16). If not, break the loop.
The program logic from (Line 14 - 20) is similar to the code segment above (Line 4 -12) except for-loop traverse up to the length of <em>text1 .</em>
<em />
At the end, return the s as output (Line 21).
