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authorJoel Brobecker <brobecker@gnat.com>2011-07-01 18:26:50 +0000
committerJoel Brobecker <brobecker@gnat.com>2011-07-01 18:26:50 +0000
commit8f17729f21e54827ba8de614be382c4fbd6b690f (patch)
tree92add7a0018deb53f1793f0ea2d22f4fc38fd14d /gdb/ada-lang.c
parentf5aa6869dee6874a9d1a3a386b1d5893f0ac46d4 (diff)
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treat identical enum types as the same type
This is to avoid an unnecessary multiple-choice menu for an expression involving an enumeral declared in two types, when the second type is an identical copy of the first type. This happens in the following situation: type Color is (Black, Red, Green, Blue, White); type RGB_Color is new Color range Red .. Blue; In that case, an implict type is created, and is used as the base type for type RGB_Color. This base type is a copy of type Color. We've added some extensive comments explaining the situation and our approach further. gdb/ChangeLog: * ada-lang.c (ada_identical_enum_types_p): New function. (symbols_are_identical_enums): New function. (remove_extra_symbols): Do nothing if NSYMS < 2. Use symbols_are_identical_enums. gdb/testsuite/ChangeLog: * gdb.ada/same_enum: New testcase.
Diffstat (limited to 'gdb/ada-lang.c')
-rw-r--r--gdb/ada-lang.c124
1 files changed, 124 insertions, 0 deletions
diff --git a/gdb/ada-lang.c b/gdb/ada-lang.c
index dd77852..766bfc8 100644
--- a/gdb/ada-lang.c
+++ b/gdb/ada-lang.c
@@ -4365,6 +4365,108 @@ is_nondebugging_type (struct type *type)
return (name != NULL && strcmp (name, "<variable, no debug info>") == 0);
}
+/* Return nonzero if TYPE1 and TYPE2 are two enumeration types
+ that are deemed "identical" for practical purposes.
+
+ This function assumes that TYPE1 and TYPE2 are both TYPE_CODE_ENUM
+ types and that their number of enumerals is identical (in other
+ words, TYPE_NFIELDS (type1) == TYPE_NFIELDS (type2)). */
+
+static int
+ada_identical_enum_types_p (struct type *type1, struct type *type2)
+{
+ int i;
+
+ /* The heuristic we use here is fairly conservative. We consider
+ that 2 enumerate types are identical if they have the same
+ number of enumerals and that all enumerals have the same
+ underlying value and name. */
+
+ /* All enums in the type should have an identical underlying value. */
+ for (i = 0; i < TYPE_NFIELDS (type1); i++)
+ if (TYPE_FIELD_BITPOS (type1, i) != TYPE_FIELD_BITPOS (type2, i))
+ return 0;
+
+ /* All enumerals should also have the same name (modulo any numerical
+ suffix). */
+ for (i = 0; i < TYPE_NFIELDS (type1); i++)
+ {
+ char *name_1 = TYPE_FIELD_NAME (type1, i);
+ char *name_2 = TYPE_FIELD_NAME (type2, i);
+ int len_1 = strlen (name_1);
+ int len_2 = strlen (name_2);
+
+ ada_remove_trailing_digits (TYPE_FIELD_NAME (type1, i), &len_1);
+ ada_remove_trailing_digits (TYPE_FIELD_NAME (type2, i), &len_2);
+ if (len_1 != len_2
+ || strncmp (TYPE_FIELD_NAME (type1, i),
+ TYPE_FIELD_NAME (type2, i),
+ len_1) != 0)
+ return 0;
+ }
+
+ return 1;
+}
+
+/* Return nonzero if all the symbols in SYMS are all enumeral symbols
+ that are deemed "identical" for practical purposes. Sometimes,
+ enumerals are not strictly identical, but their types are so similar
+ that they can be considered identical.
+
+ For instance, consider the following code:
+
+ type Color is (Black, Red, Green, Blue, White);
+ type RGB_Color is new Color range Red .. Blue;
+
+ Type RGB_Color is a subrange of an implicit type which is a copy
+ of type Color. If we call that implicit type RGB_ColorB ("B" is
+ for "Base Type"), then type RGB_ColorB is a copy of type Color.
+ As a result, when an expression references any of the enumeral
+ by name (Eg. "print green"), the expression is technically
+ ambiguous and the user should be asked to disambiguate. But
+ doing so would only hinder the user, since it wouldn't matter
+ what choice he makes, the outcome would always be the same.
+ So, for practical purposes, we consider them as the same. */
+
+static int
+symbols_are_identical_enums (struct ada_symbol_info *syms, int nsyms)
+{
+ int i;
+
+ /* Before performing a thorough comparison check of each type,
+ we perform a series of inexpensive checks. We expect that these
+ checks will quickly fail in the vast majority of cases, and thus
+ help prevent the unnecessary use of a more expensive comparison.
+ Said comparison also expects us to make some of these checks
+ (see ada_identical_enum_types_p). */
+
+ /* Quick check: All symbols should have an enum type. */
+ for (i = 0; i < nsyms; i++)
+ if (TYPE_CODE (SYMBOL_TYPE (syms[i].sym)) != TYPE_CODE_ENUM)
+ return 0;
+
+ /* Quick check: They should all have the same value. */
+ for (i = 1; i < nsyms; i++)
+ if (SYMBOL_VALUE (syms[i].sym) != SYMBOL_VALUE (syms[0].sym))
+ return 0;
+
+ /* Quick check: They should all have the same number of enumerals. */
+ for (i = 1; i < nsyms; i++)
+ if (TYPE_NFIELDS (SYMBOL_TYPE (syms[i].sym))
+ != TYPE_NFIELDS (SYMBOL_TYPE (syms[0].sym)))
+ return 0;
+
+ /* All the sanity checks passed, so we might have a set of
+ identical enumeration types. Perform a more complete
+ comparison of the type of each symbol. */
+ for (i = 1; i < nsyms; i++)
+ if (!ada_identical_enum_types_p (SYMBOL_TYPE (syms[i].sym),
+ SYMBOL_TYPE (syms[0].sym)))
+ return 0;
+
+ return 1;
+}
+
/* Remove any non-debugging symbols in SYMS[0 .. NSYMS-1] that definitely
duplicate other symbols in the list (The only case I know of where
this happens is when object files containing stabs-in-ecoff are
@@ -4377,6 +4479,12 @@ remove_extra_symbols (struct ada_symbol_info *syms, int nsyms)
{
int i, j;
+ /* We should never be called with less than 2 symbols, as there
+ cannot be any extra symbol in that case. But it's easy to
+ handle, since we have nothing to do in that case. */
+ if (nsyms < 2)
+ return nsyms;
+
i = 0;
while (i < nsyms)
{
@@ -4428,6 +4536,22 @@ remove_extra_symbols (struct ada_symbol_info *syms, int nsyms)
i += 1;
}
+
+ /* If all the remaining symbols are identical enumerals, then
+ just keep the first one and discard the rest.
+
+ Unlike what we did previously, we do not discard any entry
+ unless they are ALL identical. This is because the symbol
+ comparison is not a strict comparison, but rather a practical
+ comparison. If all symbols are considered identical, then
+ we can just go ahead and use the first one and discard the rest.
+ But if we cannot reduce the list to a single element, we have
+ to ask the user to disambiguate anyways. And if we have to
+ present a multiple-choice menu, it's less confusing if the list
+ isn't missing some choices that were identical and yet distinct. */
+ if (symbols_are_identical_enums (syms, nsyms))
+ nsyms = 1;
+
return nsyms;
}