Fix translating MOD to R

`x MOD y` should be `x - y * trunc(x / y)`.

Author: gaborcsardi

crates/data-dict/src/emit/mod.rs

@@ -476,6 +476,16 @@ mod r_tests {
     fn modulo_takes_its_sign_from_the_dividend() {
         // R's `%%` follows the divisor, so the language's rule is arithmetic.
         assert_eq!(r("MOD(n, 3) = 0"), "n - 3L * trunc(n / 3L) == 0L");
+        // A compound dividend keeps its brackets inside the division, or `/`
+        // would take only the last term.
+        assert_eq!(
+            r("MOD(n + 1, 3) = 0"),
+            "n + 1L - 3L * trunc((n + 1L) / 3L) == 0L"
+        );
+        assert_eq!(
+            r("MOD(n, qty + 1) = 0"),
+            "n - (qty + 1L) * trunc(n / (qty + 1L)) == 0L"
+        );
     }
 
     #[test]

crates/data-dict/src/emit/r_tidyverse.rs

@@ -398,10 +398,13 @@ fn write_func(cx: &mut Ctx, op: Op, args: &[TypedExpr]) -> Result<(), Unsupporte
             cx.child(p::ADD, Side::Left, x)?;
             cx.push(" - ");
             cx.child(p::MUL, Side::Right, y)?;
+            // The division inside `trunc` is an operator position, not a
+            // delimited one: a compound dividend has to keep its brackets or
+            // `/` steals its last term.
             cx.push(" * trunc(");
-            cx.free(x)?;
+            cx.child(p::MUL, Side::Left, x)?;
             cx.push(" / ");
-            cx.free(y)?;
+            cx.child(p::MUL, Side::Right, y)?;
             cx.push(")");
         }
         Op::Min | Op::Max | Op::Sum | Op::Avg => {