What will be the output when executing this code? let x = 10; const test = (y = (x += 10)) => y; console.log(test());

What will be the output when executing this code? let x = 10; const test = (y = (x += 10)) => y; console.log(test());

• A. 10
• B. 20
• C. 30
• D. NaN

Answer: (B)

The code snippet initializes a variable `x` with the value `10`. The function `test` is defined as an arrow function that takes one parameter `y`. If no argument is passed when `test` is called, `y` will be assigned the result of the expression `(x += 10)`. The expression `x += 10` increments the value of `x` by `10`, changing `x` from `10` to `20`. Since `test()` is called without an argument, the default parameter `y` gets the value of `x`, which is now `20`. The function then returns `y`, which is also `20`. Thus, when `console.log(test())` is executed, it outputs `20`. The logic of how default parameters work in conjunction with variable updates here is key, as it demonstrates the order of operations and the implications of mutating the value of a variable within the context of a function call.

The code snippet initializes a variable x with the value 10. The function test is defined as an arrow function that takes one parameter y. If no argument is passed when test is called, y will be assigned the result of the expression (x += 10).

The expression x += 10 increments the value of x by 10, changing x from 10 to 20. Since test() is called without an argument, the default parameter y gets the value of x, which is now 20. The function then returns y, which is also 20.

Thus, when console.log(test()) is executed, it outputs 20. The logic of how default parameters work in conjunction with variable updates here is key, as it demonstrates the order of operations and the implications of mutating the value of a variable within the context of a function call.