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A developer wrote the following code: 01 let x = object.value; 02 03 try { 04 handleObjectValue(x); 05 } catch(error) { 06 handleError(error); 07 } The developer has a getNextValue function to execute after handleObjectValue(), but does not want to execute getNextValue() if an error occurs. How can the developer change the code to ensure this behavior?
Correct Answer: B
Comprehensive and Detailed Explanation From Exact Extract JavaScript Knowledge: Requirement: getNextValue() should run only if handleObjectValue(x) does not throw an error. If an error occurs, handleError(error) should run, and getNextValue() should be skipped. Option B: try { handleObjectValue(x); getNextValue(); } catch (error) { handleError(error); } Behavior: If handleObjectValue(x) succeeds (no error): Execution continues to getNextValue();. If handleObjectValue(x) throws: Control jumps directly to catch, getNextValue() is skipped. handleError(error); is called. This matches the requirement perfectly. Why others are incorrect: A: } then { is invalid JavaScript syntax. C: try { handleObjectValue(x); } catch (error) { handleError(error); } getNextValue(); getNextValue() is called after the try...catch regardless of whether an error occurred. Not acceptable. D: finally always runs: try { handleObjectValue(x); } catch (error) { handleError(error); } finally { getNextValue(); } getNextValue() will execute in both success and error cases. Therefore, the correct approach is B. Concepts: try/catch control flow, finally semantics, placing code inside vs outside try/catch blocks. ________________________________________