2D数组返回错误值

Question

var term = [,];
term[0,0]="0";
term[0,1]="1";
term[1,0]="2";
term[1,1]="3";
alert(term[0,1]);

Returns 3, and I don't know why. Logically, it should return 1, correct?

Answer 1

In JavaScript, when you have an expression like a, b , it will evaluate both a and b and the result of the expression will be b . You can confirm that like this

console.log((1, 2));
# 2
console.log((0, 1));
# 1
console.log((1, 0));
# 0

So,

term[0, 0] = "0";
term[0, 1] = "1";
term[1, 0] = "2";
term[1, 1] = "3";

was evaluated like this

term[0, 0] = term[0] = "0";
term[0, 1] = term[1] = "1";
term[1, 0] = term[0] = "2";
term[1, 1] = term[1] = "3";

So the actual array has got only 2 and 3 . Now, you are trying to access, 0, 1 , which is equivalent to

term[0, 1] = term[1]

That is why you are getting 3 .

To actually create a 2-D Array, you create a main array and keep adding subarrays to it, like this

var term = [];
term.push([0, 1]);
term.push([2, 3]);

Now, to access value at 0, 1 , you need to do like this

term[0][1]

Answer 2

By 2D it's meant [[]] , not [,] jsBin example

var term = [[],[]];
term[0][0]="0";
term[0][2]="1";
term[1][0]="2";
term[1][3]="3";

console.log( term );          // [["0", "1"], ["2", "3"]]
console.log( term[0][1] );    // "1"

Also you can insert/append keys into an array using Array.prototype.push()

Answer 3

JavaScript arrays don't work that way.

Here is an example that should work:

var term = []; // Create array
term[0] = ["0", "1"]; // What you referred to as [0,0] and [0,1]
term[1] = ["2", "3"]; // What you referred to as [1,0] and [1,1]

alert(term[0][1]); // Proper JavaScript way to access 2D array.

Here is the jsfiddle .