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Make diverging procedure types different from ones without a return type

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This commit is contained in:
gingerBill
2018-09-09 13:48:33 +01:00
parent f5549f6bde
commit 12902821d6
6 changed files with 17 additions and 16 deletions
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9
src/types.cpp
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@@ -157,7 +157,7 @@ struct TypeUnion {
bool are_offsets_set; \
}) \
TYPE_KIND(Proc, struct { \
Ast *node; \
Ast *node; \
Scope * scope; \
Type * params; /* Type_Tuple */ \
Type * results; /* Type_Tuple */ \
@@ -174,7 +174,7 @@ struct TypeUnion {
bool is_poly_specialized; \
bool has_proc_default_values; \
bool has_named_results; \
bool no_return; \
bool diverging; /* no return */ \
isize specialization_count; \
ProcCallingConvention calling_convention; \
}) \
@@ -1405,8 +1405,9 @@ bool are_types_identical(Type *x, Type *y) {
case Type_Proc:
if (y->kind == Type_Proc) {
return x->Proc.calling_convention == y->Proc.calling_convention &&
x->Proc.c_vararg == y->Proc.c_vararg &&
x->Proc.variadic == y->Proc.variadic &&
x->Proc.c_vararg == y->Proc.c_vararg &&
x->Proc.variadic == y->Proc.variadic &&
x->Proc.diverging == y->Proc.diverging &&
are_types_identical(x->Proc.params, y->Proc.params) &&
are_types_identical(x->Proc.results, y->Proc.results);
}