Assign variable value through array in Haxe?

Question

How to assign variables a , b , c and d values 1 , 2 , 3 and 4 through an array?

 var a:Int = 0, b:Int = 0, c:Int = 0, d:Int = 0; var array1:Array<Int> = [a, b, c, d]; var array2:Array<Int> = [1, 2, 3, 4]; for (i in 0...4) array1[i] = array2[i]; //this does not work, it assigns array1 elements instead trace(a); // this prints 0 instead of 1 trace(b); // 0 instead of 2 trace(c); // 0 instead of 3 trace(d); // 0 instead of 4 

Answer 1

The reason this does not work is because basic types such as Int are immutable. In fact, the variables a , b , c and d may even exist, since they never change. If you look at the JS Source tab on try.haxe.org , you will notice that they are not present in the output:

 var array1 = [0,0,0,0]; var array2 = [1,2,3,4]; var _g = 0; while(_g < 4) { var i = _g++; array1[i] = array2[i]; } console.log(0); console.log(0); console.log(0); console.log(0); 

This is an analyzer optimization.

To do this, you will need some type of container, such as class or typedef . Here is an example of using the latter:

 class Test { static function main() { var a:IntContainer = { value: 0 }; var b:IntContainer = { value: 0 }; var c:IntContainer = { value: 0 }; var d:IntContainer = { value: 0 }; var array1:Array<IntContainer> = [a, b, c, d]; var array2:Array<Int> = [1, 2, 3, 4]; for (i in 0...4) array1[i].value = array2[i]; trace(a.value); trace(b.value); trace(c.value); trace(d.value); } } typedef IntContainer = { value:Int } 

Answer 2

The workaround provided by @ Gama11 is true that we need a container type, but the argument is incorrect. The problem is not the result of any optimization .

If we run the JS equivalent of the code in question, we get the same result:

 var a = 0, b = 0, c = 0, d = 0; var array1 = [a, b, c, d]; var array2 = [1, 2, 3, 4]; for (var i = 0; i < 4; i++) array1[i] = array2[i]; console.log(a); // 0 console.log(b); // 0 console.log(c); // 0 console.log(d); // 0 

In fact, this behavior is the same for most other languages.

The reason is that the Array declaration accepts input expression values instead of references . This means a , b , c and d are evaluated when setting array1 .

This behavior is the same for all expressions except for the function only:

 /* variable */ var a = "A"; var b = a; a = "B"; trace(b); // A /* anon. object */ var a = "A"; var c = { _c: a }; a = "C"; trace(c._c); // A /* Map */ var a = "A"; var d = ["_d" => a]; a = "D"; trace(d["_d"]); // A /* function The function body is evaluated every time when it is called. */ var a = "A"; var f = function () return a; a = "F"; trace(f()); // F <-- notice this