Function with optional paramaters

Question

I'm trying to create a function with optional parameters.

function RetvinkletTrekant(par1, par2, par3, par4, par5, par6){

    if (par1) {var a = par1}
    if (par2) {var b = par2}
    if (par3) {var c = par3}
    if (par4) {var A = par4}
    if (par5) {var B = par5}
    if (par6) {var C = par6}

    document.write(A);
}

RetvinkletTrekant(null, null, null, 4, null, null);

Thanks in advance.

I think you're looking for *optional* parameters, not *optimal* ones. — Bergi, Oct 15 '13 at 23:13
What's your question? Doesn't your code work as intended, what else did you expect? — Bergi, Oct 15 '13 at 23:14

score 1 · Answer 1 · answered Oct 15 '13 at 23:17

1

If you're actually asking about "optional parameters" rather than "optimal parameters" then the best way to do it would be to use a object like so.

function RetvinkletTrekant(a){

    if (a.par1) { var a = a.par1; }
    if (a.par2) { var b = a.par2; }
    if (a.par3) { var c = a.par3; }
    if (a.par4) { var A = a.par4; }
    if (a.par5) { var B = a.par5; }
    if (a.par6) { var C = a.par6; }

    document.write(A);
}

RetvinkletTrekant({
    par1: 4
});

answered Oct 15 '13 at 23:17

iConnor

19,997
14
62
97

You can (should) also omit those `if`-statements. If the properties don't exist, the variables would be `undefined` anyway. – Bergi Oct 15 '13 at 23:19

score 0 · Answer 2 · answered Oct 15 '13 at 23:17

Try to refactor function to use one parameter. Change

function RetvinkletTrekant(par1, par2, par3, par4, par5, par6) {
// ...
}

to something like this

function RetvinkletTrekant(config) {
// ...
}

where config is an object containing your optional parameters:

RetvinkletTrekant({par1: "foo", par5: "bar"});

Function with optional paramaters

2 Answers2