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When converting an expression to type type, retain the resulting instruction as well as the TypeId. (#4355)

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The `TypeId` is lossy, as it represents only the canonical type, and not
the specific computation that produced it.
This commit is contained in:
Richard Smith
2024-10-01 01:49:39 +00:00
committed by GitHub
parent 9d5ec52232
commit 4ca711c175
15 changed files with 48 additions and 33 deletions
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toolchain/check/handle_struct.cpp
+1 -1
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@@ -58,7 +58,7 @@ auto HandleParseNode(Context& context, Parse::StructFieldId node_id) -> bool {
auto HandleParseNode(Context& context, Parse::StructTypeFieldId node_id)
-> bool {
auto [type_node, type_id] = context.node_stack().PopExprWithNodeId();
SemIR::TypeId cast_type_id = ExprAsType(context, type_node, type_id);
SemIR::TypeId cast_type_id = ExprAsType(context, type_node, type_id).type_id;
auto [name_node, name_id] = context.node_stack().PopNameWithNodeId();