Array values transfer between 2 functions

Question

I try to read an array using a function and use it in another function, however the second function does not seem to read it. What I am doing wrong?

  function calcDet () {   
            var A = [];     //generates the array
                for (var i = 0; i < k; i++) {
                    A[i] = [];
                    for (var j = 0; j < k; j++) {   
                        var id = "A" + (i + 1) + (j + 1);                   
                        A[i][j] =  parseFloat(document.getElementById(id).value);               
                    }
                }   
                alert (A);
            return (A); 
        }
        function calcRec() {            
                var s;
                var det;
                alert (A)
        }

thx for all the answers. I can't declare the variable globaly as it's length and values depend on user input. So I have first to read k and then read the array values. I will just use the second approach suggested (A = calcDet()). It works now so thanks a bunch — viktor
– viktor, Commented May 18, 2012 at 7:27

Neji · Accepted Answer · 2012-05-18 08:02:29Z

2

firstly your array is not declared outside the function. its scope is limited to function body. but as your function returns the array back then try using it for initializing your local variable in other array. also this will work. or try declaring it outside the functions

edited May 18, 2012 at 8:02

answered May 18, 2012 at 7:15

Neji

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3 Comments

Madara's Ghost Over a year ago

I expect better than that from a Byakugan.

Neji Over a year ago

nw does it complete u r expectations frm a byakugan:)

Madara's Ghost Over a year ago

Now it does. An example would be great

Pranay Rana · Accepted Answer · 2012-05-18 07:14:57Z

1

Frist way is

function function1()
{
  var valriable1=12;
  function2(val);
}

function function2(val)
{
  var variableoffunction1 = val;
}

Second way is

var globalvarialbe;
function function1()
{
  globalvarialbe=12;
  function2();
}

function function2()
{
  var local = globalvarialbe;
}

answered May 18, 2012 at 7:14

Pranay Rana

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Comments

Joseph · Accepted Answer · 2012-05-18 07:13:47Z

1

It's because JavaScript has a "functional scope". It means that whatever is declared inside the function cannot be seen by the outside, but what's outside can be seen from the inside.

In your case, you must declare the array outside the function.

var A = []

function calcDet(){...}

function calcRec(){
    alert(A);
}

or call the function and return A

function calcDet(){
    var A = [];
    ...
    return A;
}

function calcRec(){
    var A = calcDet(); //get A from calcDet
    alert(A);
}

answered May 18, 2012 at 7:13

Joseph

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1 Comment

Joseph Over a year ago

@Truth updated it. and the OP didn't mention about the code living in the global scope. It could be living in another function or in a closure.

Andreas Wong · Accepted Answer · 2012-05-18 07:14:35Z

1

You can't because you are declaring A locally in calcDet

One way is to declare A outside the function, or simply feed it into calcRec:

var A
function calcDet() {
   // code
}

function calcRec() {
   // code
}

or (better)

function calcRec() {            
   var s;
   var det;
   alert (calcDet())
}

answered May 18, 2012 at 7:14

Andreas Wong

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Comments

Madara's Ghost · Accepted Answer · 2012-05-18 07:14:59Z

1

See this link about scope.

The A variable, being declared inside of the function, is only scoped to that function. It will die after the function dies. You need to pass it as a parameter:

function calcRec(data) {            
        var s;
        var det;
        alert (data)
}
A = calcDet();
calcRec(A);

answered May 18, 2012 at 7:14

Madara's Ghost

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