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12 changes: 12 additions & 0 deletions _test/issue-1653.go
Original file line number Diff line number Diff line change
@@ -0,0 +1,12 @@
package main

func f(b uint) uint {
return uint(1) + (0x1 >> b)
}

func main() {
println(f(1))
}

// Output:
// 1
17 changes: 17 additions & 0 deletions interp/cfg.go
Original file line number Diff line number Diff line change
Expand Up @@ -984,6 +984,9 @@ func (interp *Interpreter) cfg(root *node, sc *scope, importPath, pkgName string
// Allocate a new location in frame, and store the result here.
n.findex = sc.add(n.typ)
}
if n.typ != nil && !n.typ.untyped {
fixUntyped(n, sc)
}

case indexExpr:
if isBlank(n.child[0]) {
Expand Down Expand Up @@ -2302,6 +2305,20 @@ func (interp *Interpreter) cfg(root *node, sc *scope, importPath, pkgName string
return initNodes, err
}

// fixUntyped propagates implicit type conversions for untyped binary expressions.
func fixUntyped(nod *node, sc *scope) {
nod.Walk(func(n *node) bool {
if n == nod || (n.kind != binaryExpr && n.kind != parenExpr) || !n.typ.untyped {
return true
}
n.typ = nod.typ
if n.findex >= 0 {
sc.types[n.findex] = nod.typ.frameType()
}
return true
Comment on lines +2310 to +2318
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Does it ever implicitly return false?

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A callback returning false terminates the tree walk. In our case, we just want to skip the node, but continue to process the rest of the expression subtree, looking for untyped expressions, so the callback always returns true.

}, nil)
}

func compDefineX(sc *scope, n *node) error {
l := len(n.child) - 1
types := []*itype{}
Expand Down