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  • Post category:Blog
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  • Post last modified:June 12, 2024

Review the following code (note the variable name):

let age = 32;
age = age + 1;
console.log(Age);

As a result of its execution, the following should appear in the console:

  • undefined
  • 32
  • error message: "uncaught ReferenceError: Age is not defined" .
  • 33
Answers Explanation & Hints:

In the code snippet provided, there is a mismatch in the variable names. The variable age is declared and assigned the value 32. Then, it is incremented by 1 using age = age + 1;. However, when attempting to console.log(Age);, it refers to a variable Age with a capital “A”, which has not been declared or assigned any value.

JavaScript is case-sensitive, so Age and age are considered as separate variables. Since Age is not defined, it will result in an “uncaught ReferenceError” when trying to access it.

For more Questions and Answers:

JavaScipt Essentials 1 (JSE) | JSE1 – Module 2 Test Exam Answers Full 100%

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