CODEWITHSUNDEEP
javascriptarraysfunctionloops
Jake
Jake

Reputation: 254

confusing execution relating to function, loop and array

Refer to the given below code:
I am not able to understand that what's happening under the hood that let the confusion between the two points mentioned below:

  1. when revArrayInPlace(a) executes, it change the original value of variable a but
  2. when revArray(a) executes, it doesn't change the original value of variable a

In revArrayInPlace(a), if(x[someIndex] = someValue) replace the original value of x[someIndex]with somevalue , then logically x = newArray should also replace the original value of array xwith new value newArray when revArray(a) executes 

a = [1,2,3,4,5]

function revArray(x) {
    var result = [];
    for (var i = x.length -1; i >= 0 ; i = i-1) {
        result.push(x[i]);
    }
    x = result; 
}



function revArrayInPlace(x) {
  for (var i = 0; i<Math.floor(x.length/2); i=i+1) {
    var old = x[i];
    x[i] = x[x.length - 1 - i];
    x[x.length -1 -i] = old;
  }
  return x;
}

Upvotes: 0

Views: 50

Answers (1)

T.J. Crowder
T.J. Crowder

Reputation: 1075209

At the end of revArray, you're doing this:

x = result;

That does nothing to modify the original array or the original variable (a, I think) referring to it (but it's a common misunderstanding, you're not the only one!). All it does is modify the value in x, changing it from being a reference to the original array to being a reference to the new array. Since the next thing you do is exit the function, that doesn't accomplish anything.

In JavaScript, only values are passed into functions, not variables. So:

var a = [1, 2, 3, 4, 5];
revArray(a);

...reads the value from a and passes that value into revArray. That value tells the JavaScript engine where the array is, but it has no ongoing connection to the a variable at all. If you think about it, it has to be that way, because you can pass the result of an expression into a function; then what would assigning to the argument do? :-)

To make revArray work, return the result:

return result;

...and call it like this:

a = revArray(a);

Example:

var a = [1,2,3,4,5];

function revArray(x) {
    var result = [];
    for (var i = x.length -1; i >= 0 ; i = i-1) {
        result.push(x[i]);
    }
    return result;
}

a = revArray(a);
console.log(a);

Upvotes: 4

Related Questions