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Pipe Operator

The |> operator injects the left-hand value as the first argument of the right-hand call. This shifts reading from inside-out (f(g(h(x)))) to the natural execution order (x |> h() |> g() |> f()), making the code self-describing.

The simplest form chains single-argument functions:

Simple and multi-line pipelines; |> with extra arguments — the left-hand value always fills the first slot.

01-pipe-basics.zolo
Playground
// Feature: Pipe operator `|>` — fundamentals
// Syntax: `value |> f()` injects `value` as the 1st argument of `f`.
// When to use: read transformations in natural order (data-first), instead
// of nesting calls. Replaces `f(g(h(x)))` with `x |> h() |> g() |> f()`.

fn double(x: int) -> int {
  return x * 2
}

fn add_one(x: int) -> int {
  return x + 1
}

fn square(x: int) -> int {
  return x * x
}

// Simple pipeline: 3 -> 6 -> 7 -> 49.
let r = 3 |> double() |> add_one() |> square()
print(r)

// expected: 49

// Same computation without pipe — reads inside-out.
let r2 = square(add_one(double(3)))
print(r2)

// expected: 49

// Pipe with extra arguments: the LHS goes as the FIRST arg.
fn add(a: int, b: int) -> int {
  return a + b
}

fn mul(a: int, b: int) -> int {
  return a * b
}

// 5 |> add(10) calls add(5, 10) = 15; then mul(15, 3) = 45.
let calc = 5 |> add(10) |> mul(3)
print(calc)

// expected: 45

// Multi-line pipeline — each step stands out.
let chain = 1
  |> add(2)
  |> mul(10)
  |> add(5)
print(chain)
// expected: 35

Challenge

Add a function negate(x: int) -> int that returns -x and insert it as the fourth step of the chain pipeline. What is the expected result?

See also

enespt-br