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reflect: include the alignment of zero-sized types in stack offsets
This change modifies the reflect ABI assignment algorithm to catch zero-sized types at the top level of each argument and faux-stack-assign them. It doesn't actually generate an ABI step, which is unnecessary, but it ensures that the offsets of further stack-assigned arguments are aligned to the alignment of that zero-sized argument. This change is necessary to have the register ABI assignment algorithm gracefully degrade to ABI0 when no registers are present in the ABI. Fixes #44377. Change-Id: Ia95571688a61259302bb3c6d5fb33fbb6b5e8db8 Reviewed-on: https://go-review.googlesource.com/c/go/+/293789 Run-TryBot: Michael Knyszek <mknyszek@google.com> TryBot-Result: Go Bot <gobot@golang.org> Reviewed-by: Than McIntosh <thanm@google.com> Reviewed-by: Cherry Zhang <cherryyz@google.com> Trust: Than McIntosh <thanm@google.com>
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@ -123,6 +123,24 @@ func (a *abiSeq) stepsForValue(i int) []abiStep {
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func (a *abiSeq) addArg(t *rtype) *abiStep {
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pStart := len(a.steps)
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a.valueStart = append(a.valueStart, pStart)
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if t.size == 0 {
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// If the size of the argument type is zero, then
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// in order to degrade gracefully into ABI0, we need
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// to stack-assign this type. The reason is that
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// although zero-sized types take up no space on the
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// stack, they do cause the next argument to be aligned.
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// So just do that here, but don't bother actually
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// generating a new ABI step for it (there's nothing to
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// actually copy).
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//
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// We cannot handle this in the recursive case of
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// regAssign because zero-sized *fields* of a
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// non-zero-sized struct do not cause it to be
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// stack-assigned. So we need a special case here
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// at the top.
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a.stackBytes = align(a.stackBytes, uintptr(t.align))
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return nil
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}
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if !a.regAssign(t, 0) {
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a.steps = a.steps[:pStart]
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a.stackAssign(t.size, uintptr(t.align))
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