box callback functions returning the same string in Rascal

Question

I'm trying to draw some boxes in Rascal and trying to give each box its own callback function. On entering the box with the mouse the corresponding string should get displayed in the text element (so hovering box1 should display box1 etc.).

However, at the moment the text does pop up but just displays "box3" for each of the 3 boxes.

Any ideas?

strings = ["box1", "box2", "box3"];

boxes = [ box(
    size(100, 100),
    onMouseEnter(void() {
        output = s;
    })
) | s <- strings];

render(hcat([
    vcat(boxes),
    text(str () {return output;})
]));

score 2 · Accepted Answer · answered Jan 20 '19 at 17:15

Good question, classical problem. The essence of the problem is that Rascal uses "non-capturing closures": this means that functions that are returned from another function share the same context. In your case this is the variable s introduced by s <- strings. This nearly always happens when you create function values in a loop (as you do here). The solution is to wrap another function layer around the returned function.

Here is a simple example:

list[int()] makeClosures()
    =  [ int() {return i;} | i <- [0,1,2]];

void wrong(){
    lst = makeClosures();
    println(lst[0]());
    println(lst[1]());
    println(lst[2]());
}

which will print surprisingly the values 2,2and2`. The solution is, as said, to introduce another function level:

int() makeClosure(int i) 
    = int() { return i;};

list[int()] makeClosuresOK()
    = [ makeClosure(i) | i <- [0,1,2]];

void right(){
    lst = makeClosuresOK();
    println(lst[0]());
    println(lst[1]());
    println(lst[2]());
}

now calling right() will print 1, 2, and 3 as expected.

I leave it as an exercise how this is done in your example, but I am prepared to give a solution when you ask for it. Good luck!

This shorter hacky version also works: `[ int() { return j; } | i <- [0,1,2], j := i ]` — Jurgen Vinju, Jan 20 '19 at 19:59
that works because due to the semantics of the `,` comma in comprehensions which allocate new environments to the right. So at `j := ...` for every `i`, a new context/environment will be created in which every `j` is then different variable in a different environment. Contrast this with `i <- [..]` where the i is _updated_ in its own environment while the generator still has values. — Jurgen Vinju, Jan 21 '19 at 12:31
Paul, thanks for the explanation and good to know about the closures in Rascal. For this particular problem I went with the solution suggested by Jurgen as this proved to be an easy fix. Thanks both! — metatron, Jan 21 '19 at 15:03

box callback functions returning the same string in Rascal

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