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interp: fix a bug when assigning to an empty interface

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The concrete type was not forwarded propertly in case of a binary
expression involving a valueT. The corresponding part in type.go
has been refactored and the now the multi-assign case should be
handled as well.

Fixes #1094.
This commit is contained in:
Marc Vertes
2021-05-07 16:30:09 +02:00
committed by GitHub
parent cdc6b773c2
commit 33a532ee01
3 changed files with 29 additions and 23 deletions
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32
interp/type.go
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@@ -277,40 +277,26 @@ func nodeType(interp *Interpreter, sc *scope, n *node) (*itype, error) {
}
}
// Because an empty interface concrete type "mutates" as different values are
// assigned to it, we need to make a new itype from scratch everytime a new
// assignment is made, and not let different nodes (of the same variable) share the
// same itype. Otherwise they would overwrite each other.
if n.anc.kind == assignStmt && isInterface(n.anc.child[0].typ) && len(n.anc.child[0].typ.field) == 0 {
// TODO(mpl): do the indexes properly for multiple assignments on the same line.
// Also, maybe we should use nodeType to figure out dt.cat? but isn't it always
// gonna be an interfaceT anyway?
dt := new(itype)
dt.cat = interfaceT
val := new(itype)
val.cat = t.cat
dt.val = val
// TODO(mpl): do the indexes properly for multiple assignments on the same line.
// Also, maybe we should use nodeType to figure out dt.cat? but isn't it always
// gonna be an interfaceT anyway?
n.anc.child[0].typ = dt
// TODO(mpl): not sure yet whether we should do that last step. It doesn't seem
// to change anything either way though.
// t = dt
break
}
// If the node is to be assigned or returned, the node type is the destination type.
dt := t
switch a := n.anc; {
case a.kind == assignStmt && isEmptyInterface(a.child[0].typ):
// Because an empty interface concrete type "mutates" as different values are
// assigned to it, we need to make a new itype from scratch everytime a new
// assignment is made, and not let different nodes (of the same variable) share the
// same itype. Otherwise they would overwrite each other.
a.child[0].typ = &itype{cat: interfaceT, val: dt}
case a.kind == defineStmt && len(a.child) > a.nleft+a.nright:
if dt, err = nodeType(interp, sc, a.child[a.nleft]); err != nil {
return nil, err
}
case a.kind == returnStmt:
dt = sc.def.typ.ret[childPos(n)]
}
if isInterface(dt) {
dt.val = t
}