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parse: fix loadw when assigned to l temporary

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The documentation states that loadw is syntactic sugar for loadsw,
but it actually got parsed as Oload. If the result is an l temporary,
Oload behaves like Oloadl, not Oloadsw.

To fix this, parse Tloadw as Oloadsw explicitly.
This commit is contained in:
Michael Forney 2021-09-14 13:49:26 -07:00 • committed by Quentin Carbonneaux
parent 52c8eb48ee
commit bb16529b34
1 changed files with 3 additions and 1 deletions
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4
parse.c
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@ -633,7 +633,9 @@ DoOp:
arg[1] = R;
goto Ins;
}
if (op >= Tloadw && op <= Tloadd)
if (op == Tloadw)
op = Oloadsw;
if (op >= Tloadl && op <= Tloadd)
op = Oload;
if (op == Talloc1 || op == Talloc2)
op = Oalloc;