Function with unknown amount of variables

Question

I saw someone creating a function that has an unknown number of arguments:

public static double calculator(double ... value){
    double result=0d;

    for(int i=0;i<?;i++){
        result+=value;
    }

    return result
}

Now I'm trying to create a for loop that will run the number of times as the number of arguments entered so:

double calc = calculator(1,2,3,4,5)

this will make the for loop run 5 times.

`?` --> `value.length`. And your `+= value` won't work, you can't add a `double[]` to a `double` — fge, Jan 09 '15 at 14:00
possible duplicate of [Java variable number or arguments for a method](http://stackoverflow.com/questions/2330942/java-variable-number-or-arguments-for-a-method) — tddmonkey, Jan 09 '15 at 14:02
Your parameter is named wrong. Should be `values`, not `value`. — AJMansfield, Jan 09 '15 at 14:02

Konstantin Yovkov · Accepted Answer · 2015-01-09T14:31:22.213

8

Internally, the value is an array, so you can treat it as such:

for(int i = 0; i < value.length; i++) {
    result += value[i];
}

edited Jan 09 '15 at 14:31

answered Jan 09 '15 at 13:59

Konstantin Yovkov

Worked, thank you very much! I can mark this as the answer only in 11 minutes from now ;) – Ishay Frenkel Jan 09 '15 at 14:02

score 2 · Answer 2 · answered Jan 09 '15 at 14:02

2

The ellipsis operator (...) is just syntactic sugaring for an array. In other words, the method itself interprets the arguments as follows:

public static double calculator(double[] value)

Once you understand that, the for loop becomes obvious - you just need to iterate up to value.length:

for (int i = 0; i <value.length; i++) {
    result += value[i];
}

Or better yet, just use an enhanced for loop:

for (v : value) {
    result += v;
}

answered Jan 09 '15 at 14:02

Mureinik

Wait.. Now when I'm trying to use `public static double calculator(double[] value)` it doesn't work with int's.. – Ishay Frenkel Jan 09 '15 at 14:13
1

The ellipsis operator is much more forgiving - it promotes each value separately. If you want to actually change the signature to receive a `double[]` (redundant, IMHO), you'll have to pass a `double[]` instance, even if you are filling its elements with `int`s. – Mureinik Jan 09 '15 at 14:15

score 1 · Answer 3 · answered Jan 09 '15 at 14:01

1

You could use advance for loop (a.k.a for each loop) as below:

for(int element : value){
    result+=element;
}

Its same as:

for(int i = 0;i < value.length; i++){
     result+=value[i];
}

answered Jan 09 '15 at 14:01

SMA

I don't really like for each loops, is there an advantage of using them? and how exactly do they work? – Ishay Frenkel Jan 09 '15 at 14:04

score 0 · Answer 4 · answered Jan 09 '15 at 14:02

0

you can also use for each loop,

 for (double d : value) {

 }

answered Jan 09 '15 at 14:02

atish shimpi

4 Answers4