Home >Java >javaTutorial >Why Does Modifying a Variable in a Finally Block Not Affect the Returned Value?
Why Doesn't Modifying a Variable in a Finally Block Affect the Returned Value?
Java provides the try-finally statement, where code in the finally block is always executed, regardless of whether the try block completes normally or abruptly. However, this behavior can lead to unexpected results when a variable is modified in the finally block after a return statement is executed in the try block.
To illustrate this concept, consider the following Java class:
public class Test { private String s; public String foo() { try { s = "dev"; return s; // Return the value of s } finally { s = "override variable s"; System.out.println("Entry in finally Block"); } } public static void main(String[] xyz) { Test obj = new Test(); System.out.println(obj.foo()); } }
In this example, the foo method returns the string "dev" in the try block. However, the finally block modifies the s variable to "override variable s" and prints a message to indicate that it was executed.
Surprisingly, the output of this code is:
Entry in finally Block dev
This result may seem counterintuitive because we would expect the value of s to be "override variable s" after the finally block is executed. However, this is not the case because:
It's important to note that this behavior only applies to changes to the value of the s variable itself. If the s variable is a reference to a mutable object, changes to the contents of that object made in the finally block will be reflected in the returned value.
This detailed explanation clarifies why modifying a variable in a finally block does not change the return value of a method, even though the finally block is always executed.
The above is the detailed content of Why Does Modifying a Variable in a Finally Block Not Affect the Returned Value?. For more information, please follow other related articles on the PHP Chinese website!