JavaScript question

Question

Hi

var a = {
atr: 12
}

var b = 37;

function change(obj, x) {
x++;
a.atr++;
}

change(a, b);

console.log(a.atr + "" + b); // 1337

Why is the output 1337? Should itnot br 1338, since parameter x is being incremented, and in the function call that argument is variable b?

Answer 1

// Because x is a local variable of the function the code should looks like:

var a = {
atr: 12
}
var b = 37;

function change(x) {
x++;
a.atr++;
return x
}

b=change(b);
console.log(a.atr + "" + b);

Answer 2

when you called the function you are not actually passing the variables but you are just passing the values of those variables, and hence the variables 'b' will remain unchanged because you incremented the 1 into value of b and not the actual variable b.

Answer 3

Just a little edit
var a = {
atr: 12
}

var b = 37;

function change(obj, x) {
x++;
obj++;
return obj + "" + x
}

console.log(change(a.atr, b)); // 1338