问题
I want to write an average method in java such that it can consume N amount of items, returning the average of them:
My idea was:
public static int average(int[] args){
int total = 0;
for(int i=0;i<args.length;i++){
total = total + args[i];
}
return Math.round (total/args.length);
}
//test it
average(1,2,3) // s**hould return 2.
how can I change my method to consume any amount of parameters instead of int[] args so can work the way I want ? Cheers
回答1:
Java 5 supports varargs, which is what you want.
e.g.
public static int average(Integer... ints) {
for (Integer i : ints) {
// sum here...
}
}
回答2:
Since Java 5, there is a feature commonly called varargs which achieves what is desired.
Here's a little example:
public static int add(int... nums) {
int total = 0;
for (int n : nums)
total += n;
return total;
}
public static void main(String[] s) {
// The following prints "10"
System.out.println(add(1, 2, 3, 4));
}
回答3:
Here's the optional arguments version (almost no code changes...)
public class Spike2 {
public static final void main(String argv[]) {
System.out.println(average(1,2,3));
}
public static int average(int... args){
int total = 0;
for(int i=0;i<args.length;i++){
total = total + args[i];
}
return Math.round (total/args.length);
}
}
with iterator changes:
public class Spike2 {
public static final void main(String argv[]) {
System.out.println(average(1,2,3));
}
public static int average(int... args){
int total = 0;
for(int i: args){
total = total + i;
}
return Math.round (total/args.length);
}
}
回答4:
function average() {
var total = 0;
if(arguments.length > 0) {
for(var i = 0, n = arguments.length; i < n; i++) {
total += parseFloat(arguments[i]);
}
total /= arguments.length;
}
return total;
}
来源：https://stackoverflow.com/questions/1737350/java-optional-parameters