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Fix "b f(std::string)" when current language is C
If you try to set a breakpoint at a function such as "b
f(std::string)", and the current language is C, the breakpoint fails
to be set, like so:
(gdb) set language c
break f(std::string)
Function "f(std::string)" not defined.
Make breakpoint pending on future shared library load? (y or [n]) n
(gdb)
The problem is that the code in GDB that expands the std::string
typedef hits this in c-typeprint.c:
/* If we have "typedef struct foo {. . .} bar;" do we want to
print it as "struct foo" or as "bar"? Pick the latter for
C++, because C++ folk tend to expect things like "class5
*foo" rather than "struct class5 *foo". We rather
arbitrarily choose to make language_minimal work in a C-like
way. */
if (language == language_c || language == language_minimal)
{
if (type->code () == TYPE_CODE_UNION)
gdb_printf (stream, "union ");
else if (type->code () == TYPE_CODE_STRUCT)
{
if (type->is_declared_class ())
gdb_printf (stream, "class ");
else
gdb_printf (stream, "struct ");
}
else if (type->code () == TYPE_CODE_ENUM)
gdb_printf (stream, "enum ");
}
I.e., std::string is expanded to "class std::..." instead of just
"std::...", and then the "f(class std::..." symbol doesn't exist.
Fix this by making cp-support.c:inspect_type print the expanded
typedef type using the language of the symbol whose type we're
expanding the typedefs for -- in the example in question, the
"std::string" typedef symbol, which is a C++ symbol.
Use type_print_raw_options as it seems to me that in this scenario we
always want raw types, to match the real symbol names.
Adjust the gdb.cp/break-f-std-string.exp testcase to try setting a
breakpoint at "f(std::string)" in both C and C++.
Change-Id: Ib54fab4cf0fd307bfd55bf1dd5056830096a653b
This commit is contained in:
@@ -41,6 +41,7 @@
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#include <atomic>
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#include "event-top.h"
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#include "run-on-main-thread.h"
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#include "typeprint.h"
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#define d_left(dc) (dc)->u.s_binary.left
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#define d_right(dc) (dc)->u.s_binary.right
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@@ -229,7 +230,14 @@ inspect_type (struct demangle_parse_info *info,
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string_file buf;
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try
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{
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type_print (type, "", &buf, -1);
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/* Avoid using the current language. If the language is
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C, and TYPE is a struct/class, the printed type is
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prefixed with "struct " or "class ", which we don't
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want when we're expanding a C++ typedef. Print using
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the type symbol's language to expand a C++ typedef
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the C++ way even if the current language is C. */
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const language_defn *lang = language_def (sym->language ());
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lang->print_type (type, "", &buf, -1, 0, &type_print_raw_options);
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}
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/* If type_print threw an exception, there is little point
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in continuing, so just bow out gracefully. */
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