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  • Go

Idiom #54 Compute sum of integers

Calculate the sum s of the integer list or array x.

s := 0
for _, v := range x {
	s += v
}

Such explicit loops are idiomatic in Go.
for E of x loop
   S := S + E;
end loop;

x is an array.
S is initialized to be 0.
New implementation...
< > ⌨
deleplace 
s := 0
for _, v := range x {
	s += v
}
for E of x loop
   S := S + E;
end loop;
int sum = 0;
for (int i = 0; i < n; ++i) {
  sum += x[i];
}
int i,s;
for(i=s=0;i<n;i++)
{
	s+=x[i];
}

List.fold_left (+) 0 l
(defn sum [x] (reduce + x))

(sum [1 20 300])
;;=> 321
auto s = std::ranges::fold_left(x, 0, std::plus<int>{});

int s = std::accumulate(x.begin(), x.end(), 0, std::plus<int>());
var s = x.Sum();
auto s = x.sum();
var s = x.fold(0, (a, b) => a + b);
s = Enum.sum(x)
S = lists:sum(X).
   s = sum(x)
s = sum x
var s = x.reduce((a, b) => a + b, 0);
var s = x.reduce((a, b) => a + b)
int s = 0;
for (int i : x) {
  s += i;
}
BigInteger s = new BigInteger(valueOf(x[0]));
for (int i = 1, n = x.length; i < n; ++i)
    s = s.add(new BigInteger(valueOf(x[i])));
BigInteger s = stream(x)
    .mapToObj(String::valueOf)
    .map(BigInteger::new)
    .reduce(BigInteger::add)
    .get();
int i = 1, n = x.length, s = x[0];
while (i < n) s = s + x[i++];
int s = of(x).sum();
val numbers = listOf(1, 2, 3, 4)
val sum = numbers.sum()
(reduce #'+ x)
s = 0
for i,v in ipairs(x) do
   s = s + v
end
id s=[x valueForKeyPath:@"@sum.self"];
$s = array_sum($x);
var
  _x: Array of integer;
  _s, i: Integer;
begin
  _s := 0;
  //assume _x contains some values
  for i := Low(_x) to High(_x) do _s := _s + _x[i];
end;
$s = sum(@x);
s = sum(x)
s = reduce(add, x)
s = x.sum
s = x.reduce(:+)
let s = x.iter().sum::<i32>();
x.iter().sum()
val s = x.sum()
(define s (apply fx+ x))
(define s (foldr + 0 x))
(define s (apply + x))
(define s (apply + x))
(define s (fold-right + 0 x))
s := x sum.
Dim s As Integer = x.Sum()