Normalize round() result to an exact int (#8511)

int.__round__ returned the receiver unchanged on the no-rounding path, so
round() on an int subclass handed back an instance of that subclass:
round(True) was True, not 1. CPython routes both no-op paths (ndigits is
None, and ndigits >= 0) through long_long(), which reuses the reference
only when the value is an exact int and otherwise copies it into a fresh
one.

Return zelf.__int__(vm).into_pyref(), which goes through into_exact_or()
and is the same normalization the neighbouring __trunc__, __floor__ and
__ceil__ already perform.

Assisted-by: Claude Code:claude-opus-5

crates/vm/src/builtins/int.rs

@@ -487,7 +487,9 @@ impl PyInt {
                 return vm.ctx.new_int(rounded);
             }
         }
-        zelf
+        // No rounding to do, but an int subclass must still be normalized to an
+        // exact int, the way CPython's long_long() does.
+        zelf.__int__(vm).into_pyref()
     }
 
     #[pymethod]

extra_tests/snippets/builtin_round.py

@@ -93,3 +93,21 @@ assert math.copysign(1.0, round(0.0, 2)) == 1.0
 assert round(1.0, 1000) == 1.0
 assert round(1.0, -1000) == 0.0
 assert round(1.7976931348623157e308, 0) == 1.7976931348623157e308
+
+
+# round() normalizes an int subclass to an exact int, like CPython's long_long().
+assert round(True) == 1
+assert type(round(True)) is int
+assert type(round(True, 0)) is int
+assert type(round(False)) is int
+
+
+class MyInt(int):
+    pass
+
+
+assert round(MyInt(5)) == 5
+assert type(round(MyInt(5))) is int
+assert type(round(MyInt(5), 2)) is int
+# A negative ndigits already produced a fresh exact int.
+assert type(round(MyInt(15), -1)) is int