varargs

Learn flexible parameters with int....

varargs: flexible parameter count

Sometimes you do not know in advance how many arguments will come. By writing int... you declare a parameter that accepts any number of int values. This is called varargs.

static int sum(int... nums) {
  int total = 0;
  for (int n : nums) total += n;
  return total;
}
sum(1, 2, 3);    // 6
sum();           // 0

Inside the method, nums behaves like an int[]. You can walk it with for-each or read nums.length. A varargs parameter must be the last parameter of the method.

varargs and for-each

To iterate over varargs, for-each is very handy. Because nums behaves like an array; it gives elements one by one.

static int max(int... nums) {
  int best = Integer.MIN_VALUE;
  for (int n : nums) {
    if (n > best) best = n;
  }
  return best;
}

Here Integer.MIN_VALUE is the smallest value an int can hold. So even on an empty call, best starts at a sensible value. After the loop, the largest element is returned.

Exercise Flexible sum method

Define a static int method called sum; parameter int... nums. The body should sum all elements and return. Call sum(1, 2, 3, 4) from main and print the result.

Show example solution

static int sum(int... nums) {
  int total = 0;
  for (int n : nums) total += n;
  return total;
}
// inside main:
System.out.println(sum(1, 2, 3, 4));

Test yourself

Where can a varargs parameter sit in a method?
- Only in the last position
- In the first position
- Only in single-parameter methods
- In any position
Show answer

In any position — varargs must always be the last parameter; otherwise it is ambiguous.
Match call to nums content

sum(1, 2, 3)

{1, 2, 3}
sum()

{}
sum(7)

{7}
sum(1, 1, 1, 1)

{1, 1, 1, 1}

Fill in the blank: varargs declaration

static int sum(int____ nums) {
  int total = 0;
  for (int n : nums) total += n;
  return total;
}

Show answer

static int sum(int... nums) {
  int total = 0;
  for (int n : nums) total += n;
  return total;
}

Next lesson: Multiple values via array

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