Create a new list a (or array, or slice) of size n, where all elements are integers initialized with the value 0.
import java.util.*;
List<Integer> a = new ArrayList<Integer>(n); for (int i =0;i<n;i++){ a.add(0); }
import static java.util.stream.IntStream.range;
int a[] = range(0, n).map(i -> 0).toArray();
import static java.util.stream.IntStream.iterate;
int a[] = iterate(0, i -> 0).limit(n).toArray();
import static java.util.stream.IntStream.range; import java.util.ArrayList; import java.util.List;
List<Integer> a = new ArrayList<>(n); range(0, n).forEach(i -> a.add(0));
import static java.util.Arrays.fill;
Integer a[] = new Integer[n]; fill(a, 0);
int a[] = new int[n];
int[] a = new int[n];
import java.util.stream.IntStream;
int[] a = IntStream.generate(() -> 0) .limit(n) .toArray();
#include <stdlib.h>
int *a = calloc(n, sizeof(*a));
(def a (repeatedly n #(identity 0)))
std::vector<int> a(n, 0);
var a = new int[n];
var a = List.filled(n, 0);
integer, dimension(:), allocatable :: a allocate(a(n),source = 0)
a := make([]int, n)
a = replicate n 0
const a = new Array(n).fill(0);
local function new_zeroed_list(n) local ret={} for i=1,n do ret[i]=0 end return ret end local zeroed_list=new_zeroed_list(n)
a := nil; setlength(a, n);
my @a = (0) x $n;
a = [0] * n
from itertools import repeat
a = list(repeat(0, n))
a = Array.new(n, 0)
let a = vec![0; n];
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