public class App { String myStr = "7007"; public void doStuff(String str) { int myNum = 0; try { String myStr = str; myNum = Integer.parseInt(myStr); } catch (NumberFormatException ne) {…

Question

Options

• AmyStr: 9009, myNum: 9009
• BmyStr: 7007, myNum: 7007
• CmyStr: 7007, myNum: 9009
• DCompilation falls

• A
  12% (4)
• B
  3% (1)
• C
  76% (25)
• D
  9% (3)

Explanation

Option C is correct because of variable scoping: the local String myStr = str is declared inside the try block, so it only exists within that block's scope. When System.out.println executes outside the try block, Java resolves myStr to the instance variable "7007", not the local one. myNum, however, is declared outside the try block and gets successfully assigned Integer.parseInt("9009") = 9009, so it prints as 9009.

• A is wrong because myStr in the print statement resolves to the instance variable "7007", not the local "9009".
• B is wrong because myNum is correctly parsed from "9009" and assigned 9009, not 7007.
• D is wrong because the code compiles fine - shadowing an instance variable with a local one is legal in Java.

Memory tip: When you see a variable declared inside a block (try, if, for), it dies when that block closes - the outer scope's variable with the same name "comes back to life" for any code that follows.

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