# XO0178: a line starts with a binary operator; end the previous line with it

A newline ends a statement when the line's last token is an identifier, literal, `)`, `]`, `}`, `?`, or `return`, `break`, `continue` (core 1.3, Go's rule). So an expression continued on the next line must leave its operator at the end of the first line: a line that starts with `||`, `&&`, `+`, `==`, `??`, or another binary operator would begin a new statement. Rust, Swift, Kotlin, and some Go style guides put the operator first; Xo cannot, because `-x`, `!x`, and `..n` start expressions and a leading `||` reads as a closure. The machine fix moves the operator to the end of the previous line, and the parser reads the expression as continued, so no other error follows. A line starting with `.` continues a method chain and is allowed. See decision 0136.

## Example

```xo
fn ready(a: Int, b: Int) -> Bool {
  a > 0
    || b > 0
}
```

## Fix

```xo
fn ready(a: Int, b: Int) -> Bool {
  a > 0 ||
    b > 0
}
```

Run `xo explain XO0178` for this text in a terminal, or see
[section 10.6 of the specification](../../spec/core/#106-diagnostic-codes).