Analyze the following code:

try {
ocnsole.log("start");
} catch (error) {
console.log("error");
}
console.log("end");

What will happen as a result of its execution?

  • The operation of the program will be interrupted and the console will display the default message "Uncaught ReferenceError: ocnsole is not defined".
  • The console will display the following words on successive lines errorend.
  • The console will display the following words on successive lines: starterrorend.
  • The console will display the following words on successive lines: startend.
Explanation & Hint:

The error message “Uncaught ReferenceError: ocnsole is not defined” in JavaScript indicates that there is a reference to an undefined variable or identifier. In the specific case, the error is caused by the typo in the code where ocnsole.log(“start”); is written instead of console.log(“start”);. JavaScript is treating ocnsole as a variable or identifier, but it has not been defined anywhere in the code. As a result, a ReferenceError is thrown.

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Analyze the following code:

try {
ocnsole.log("start");
} catch (error) {
console.log("error");
} finally {
console.log("end");
}

What will happen as a result of its execution?

  • The words starterrorend will appear in the console on successive lines.
  • The word error will appear in the console.
  • The following words will appear in the console: errorend.
  • The following words will appear in the console: startend.
Explanation & Hint:

In the code, there is a typo in the console.log() function, which should be console.log(“start”);, not ocnsole.log(“start”);. The try block will be executed, which includes the statement console.log(“start”);. However, since there is a typo in the code, a ReferenceError will occur when trying to access the undefined variable ocnsole. This error will cause the execution of the try block to be interrupted. Since an error occurred within the try block, the execution will be transferred to the catch block. The statement console.log(“error”); will be executed, printing the message “error” to the console. This is a way to handle and report the error. After the catch block is executed, the finally block will be executed regardless of whether an error occurred or not. The statement console.log(“end”); will be executed, printing the message “end” to the console. In this case, even though an error occurred, the finally block is still executed, allowing you to perform necessary actions regardless of the error.

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JavaScript Essentials 1 | JSE1 – Module 6 Test Exam Answers Full Score 100%

 

Analyze the following code:

try {
console.log("start");
} catch (error) {
console.log("error");
} finally {
console.log("end");
}

What will happen as a result of its execution?

  • The word error will appear in the console.
  • The following words will appear in the console: startend.
  • The following words will appear in the console: starterrorend.
  • The following words will appear in the console: errorend.
Explanation & Hint:

The try block will be executed, which includes the statement console.log(“start”);. This statement will print “start” to the console without any errors. Since no error occurred within the try block, the catch block will be skipped entirely. After the try block is executed, the finally block will be executed regardless of whether an error occurred or not. The statement console.log(“end”); will be executed, printing “end” to the console. The finally block ensures that the specified code is always executed, regardless of whether an error occurs or not. It is commonly used to perform cleanup or finalize operations. In this case, since no error occurred, the finally block is still executed, allowing you to perform necessary actions at the end of the code execution.

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JavaScript Essentials 1 | JSE1 – Module 6 Test Exam Answers Full Score 100%