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I tried to use Typed arrays instead of arrays, to reduce memory:

function createarrayInt8(numrows,numcols,number){
       
	var arr = new Int8Array(numrows);
         
	for (var i = 0; i < numrows; ++i){
		var columns = new Int8Array(numcols);
		for (var j = 0; j < numcols; ++j){
			columns[j] = number;
		}
		arr[i] = columns;
	}
  
	return arr; 
}

I tried to use Typed arrays instead of arrays, to reduce memory:

function createarrayInt8(numrows,numcols,number){
       
	var arr = new Int8Array(numrows);
         
	for (var i = 0; i < numrows; ++i){
		var columns = new Int8Array(numcols);
		for (var j = 0; j < numcols; ++j){
			columns[j] = number;
		}
		arr[i] = columns;
	}
  
	return arr; 
}

But i can't create multidimensional Typed array. Why? Do i have to cast only the "number" var to Int8?

Share Improve this question asked Jul 14, 2016 at 18:10 Matthias MaMatthias Ma 1031 silver badge6 bronze badges 2
  • 1 Well, a typed array can only store values of its type. A uint8 array can therefore only store unsigned 8 bit integers, but not arrays (of uints). – le_m Commented Jul 14, 2016 at 18:18
  • i almost thought so :). But how can it get a multidimensional array that stores only unsigned 8 bit integers to reduce the used memory? – Matthias Ma Commented Jul 14, 2016 at 18:25
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1 Answer 1

Reset to default 11

A typed Int8Array can only hold 8-bit integers. So arr[i] = columns won't work since columns is of type Int8Array which cannot be converted to and stored (in any meaningful way) as a an 8-bit integer.

Solution: Either make arr a generic Array whose elements can be arrays or - probably the more advanced but usually more performant solution - store your multidimensional array as a single flat array of size numrows * numcols and access an element via arr[column + row * numcols]:

var numrows = 5, numcols = 4;
var arr = new Int8Array(numrows * numcols).fill(0);

arr[3 + 1 * numrows] = 1; // col = 3, row = 1

console.log (arr);

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