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In the above code, two strings are evaluated using the equivalency operator. Why can this be a problem, and what other methods are offered by Java for working with characters and strings?
What will be an ideal response?
The problem stems from the fact that in Java, String is a class, and each created String is an object. As an object, a String variable name is not a simple data type. It is a reference; that is, a variable that holds a memory address. Therefore, when you compare two String objects using the == operator, you are comparing not their values but the computer memory locations.?Programmers want to compare the contents of memory locations (the values stored there)more frequently than they want to compare the addresses of the locations. Fortunately, the creators of Java have provided three classes that you can use when working with text data; these classes provide you with many methods that make working with characters and strings easier:?Character - A class whose instances can hold a single character value and whose methods manipulate and inspect single-character data?String - A class for working with fixed-string data; unchanging data composed of multiple characters?StringBuilder and StringBuffer - Classes for storing and manipulating changeable data composed of multiple characters
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There are two printers on the network: a shared printer connected to a computer called PrintStation and a printer with its own IP address called IndyPrint. If PrintStation stops working, what is the status of IndyPrint?
a. Offline b. Waiting to connect c. Available to all computers d. Available only to computers on the IndyPrint network.
Which of the following generates a random number between 0 and 10?
A. var randomNum = Math.random() * 10; randomNum = Math.floor(randomNum)+ 1; document.write(randomNum); B. var randomNum = Calculate.random() * 10; document.write(randomNum); C. var randomNum = Number.random() * 10; document.write(randomNum); D. var randomNum = Math.randomNumber() * 10; document.write(randomNum);