The code segment does not compile because the method 'reset' does not return a value that can be displayed.
A method refers to a block of that which only runs when it is called. It is permissible to pass data, referred to as parameters, into a method. Methods are used to accomplish certain actions, and they are also called functions. As per the given code segment that appears as a method in a class, it does not compile because the method named 'reset' is not returning a value that can be printed on the screen as an output.
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<em>Complete question:</em>
Consider the following code segment, which appears in a method in a class other than password.
Password p = new Password("password");
System.out.println("The new password is " + p.reset("password"));
The code segment does not compile because _____________ .
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Answer:
1 14 3
Explanation:
If that code fragment is put under a main() function and then it will yield the above output. According to the function something(), only changes in value second parameter will be reflected and first will be unchanged because second is passed by reference and the first one is passed as value so only address of 's' is passed, so, only its value is changed and the rest are same.
Answer:
The correct answer to the following question will be "Anonymous logon".
Explanation:
- Windows would never let anyone sign in interactively with an anonymous logon to the device, a person who has linked to the device without a login and password being given.
- So they won't have to verify to an account of the user just whether you are running any shares.r document, let everyone log into the machine collaboratively with such a logon.
Therefore, Anonymous logon is the right answer.
Answer:
office online
Explanation:
office online is a free online version of Microsoft office suite. it includes Microsoft word, Microsoft excel, Microsoft power point, one note. It allows users to create and edit files online.
Statement A is true. R4 is not part of the circuit since it has one terminal not connected. The other resistors are in series, so the same current flows through them. Hence they will dissipate equal power.