How can I access synthetic children whose names have special characters in LLDB?

Question

Consider this program.cpp:

struct Foo {
  int field1;
  int field2;
};

int main() {
  Foo foo{.field1 = -1, .field2 = -2};
  return 0;
}

I've implemented a custom LLDB data formatter for the Foo type:

class Foo:
    def __init__(self, valobj, internal_dict):
        self.valobj = valobj

    def num_children(self):
        return 2

    def get_child_index(self, name):
        match name:
            case "bar":
                return 0
            case "[baz]":
                return 1

    def get_child_at_index(self, idx):
        match idx:
            case 0:
                return self.bar
            case 1:
                return self.baz

    def update(self):
        self.bar = self.valobj.GetChildMemberWithName("field1").Clone("bar")
        self.baz = self.valobj.GetChildMemberWithName("field2").Clone("[baz]")

    def has_children(self):
        return True


def __lldb_init_module(debugger, internal_dict):
    debugger.HandleCommand("type synthetic add Foo -l example.Foo")

As you can see, one of these children is named bar which has no special characters, and the other is named [baz] which does have special characters.

I can access bar just fine:

$ clang++ -g program.cpp -o program && lldb ./program
(lldb) command script import example.py
(lldb) b program.cpp:8
(lldb) run
(lldb) v 
(Foo) foo = (bar = -1, [baz] = -2)
(lldb) v foo.bar
(int) foo.bar = -1

But I can't figure out how to access [baz]. Here's what I've tried:

(lldb) v foo.[baz]
error: incomplete expression path after "foo" in "foo.[baz]"
(lldb) v foo.'[baz]'
error: incomplete expression path after "foo" in "foo.[baz]"
(lldb) v foo."[baz]"
error: incomplete expression path after "foo" in "foo.[baz]"
(lldb) v foo.\[baz\]
error: "\" is not a member of "(Foo) foo"

Using LLDB, how can I access a synthetic child whose name contains special characters?

Addendum regarding pointers: a hacky workaround is to just refer to the children by their indices of their names:

(lldb) v foo[0]
(int) foo[0] = -1
(lldb) v foo[1]
(int) foo[1] = -2

But then I don't know what to do if foo is a pointer:

struct Foo {
  int field1;
  int field2;
};

int main() {
  Foo *foo = new Foo{.field1 = -1, .field2 = -2};
  return 0;
}

(lldb) v foo->bar
(int) foo->bar = -1
(lldb) foo->[1]
error: 'foo-' is not a valid command.
(lldb) v foo->[1]
error: incomplete expression path after "foo" in "foo->[1]"
(lldb) v (*foo)[1]
error: no variable named '(*foo)' found in this frame
(lldb) v *foo[1]
error: not a pointer or reference type: (Foo) *(foo)

Is there a particular reason you need this? Choosing names that aren't valid C++ identifiers sounds like a recipe for extreme confusion even if it can be made to work. — Nate Eldredge
– Nate Eldredge, Commented Jul 24 at 20:20
@NateEldredge Here's a link with some context in case it's helpful: the codebase I'm working in already had Natvis files which used square brackets in item names, as well as an LLDB file for some basic types. For instance, there's a List type, and its children become listed in LLDB as [0], [1], etc; and for richer types, the Natvis file uses fancier names like [operand0] : "getDynObject" to make things visually easier to read in an IDE. Also, the LLDB docs do this with examples like deque. — Sam Estep
– Sam Estep, Commented Jul 24 at 20:58

Jim Ingham · Accepted Answer · 2025-07-25 23:06:09Z

0

lldb treats [] specially in parsing variable path expressions, so that you can make an "array-like" presentation that you access as foo[0] not foo.[0] which would be more awkward. So you can't use a name of the form [something] - that's reserved for this purpose.

answered Jul 25 at 23:06

Jim Ingham

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2 Comments

Sam Estep Jul 26 at 18:16

Right, I understand that. What I'm asking is, is there any way to escape those characters?

Jim Ingham Jul 28 at 16:50

Not using the "variable path expression" parser, no.

Sam Estep · Accepted Answer · 2025-07-28 13:57:48Z

0

It's not the most convenient, but you can always use Python:

(lldb) script
Python Interactive Interpreter. To exit, type 'quit()', 'exit()' or Ctrl-D.
>>> lldb.frame.FindVariable("foo").GetChildMemberWithName("bar")
(int) bar = -1
>>> lldb.frame.FindVariable("foo").GetChildMemberWithName("[baz]")
(int) [baz] = -2
>>> ^D
now exiting InteractiveConsole...
(lldb)

answered Jul 28 at 13:57

Sam Estep

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