Complete Question:
The method shown below,
public void makeJunk()
{
new File("junk").createNewFile();
}
generates compile-time error:
"unreported exception java.io.IOException; must be caught or declared to be thrown".
Without changing the behavior (any part of the method body) and without changing the number of arguments the function takes or its visibility, add the necessary code that will permit this code to compile without error.
Answer:
public void makeJunk() throws IOException {
new File("junk").createNewFile();
}
Explanation:
In Order to gain a better understanding of this answer let first explain some terms.
Exception:
An exception is an event, which occurs during the execution of a program, that disrupts the normal flow of the program's instructions. When an error occurs within a method, the method creates an object and hands it off to the run-time system.The object, called an exception object, contains information about the error, including its type and the state of the program when the error occurred. Creating an exception object and handing it to the run-time system is called throwing an exception.After a method throws an exception, the run-time system attempts to find something to handle it. The set of possible "somethings" to handle the exception is the ordered list of methods that had been called to get to the method where the error occurred
java.io.IOException:
This exception is related to Input and Output operations in the Java code. This exception happens when there is a failure during reading, writing and searching file or directory operations.
The Compile- time error "unreported exception java.io.IOException; must be caught or declared to be thrown", means that the exception that would handle any errors that occurred during creating of the new file was not thrown in the first code so in order to solve the problem Without changing the behavior(any part of the method body) The exception that would handle that error was added to the solution code.
