What will be the output of the following code?var x = 10;function oute...
The 'inner' function is called before the 'x' variable is defined within the 'outer' function. Therefore, when 'console.log(x)' is executed, 'x' is undefined.
What will be the output of the following code?var x = 10;function oute...
Explanation:
The output of the code will be undefined.
Reason:
When the function outer() is called, it executes the inner function inner() which tries to print the value of x. However, at that point, the variable x is declared but not yet assigned a value, so its value is undefined.
Detailed Explanation:
1) The variable x is declared and assigned the value 10 outside the outer() function. This makes it a global variable.
2) The outer() function is declared and defined.
3) Inside the outer() function, the inner() function is declared and defined.
4) The inner() function tries to print the value of x. However, at this point, a variable named x is also declared inside the outer() function, which creates a new local variable with the same name. This local variable x shadows the global variable x within the scope of the outer() function.
5) Before the inner() function is called, the local variable x is declared and assigned the value 20. At this point, the global variable x is not affected.
6) The inner() function is called, and it tries to print the value of x. Since the local variable x is declared but not yet assigned a value, its value is undefined.
7) The outer() function completes execution, and the program terminates.
Thus, the output of the code is undefined.
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