C struct - lldb expression fails

Question

Why can't lldb's expression understand my C Struct unless I declare a default variable?

struct YD_MENU {
    char menu_name[10];
    int menu_option;
};

int main() {

    return 0;
}

Adding a breakpoint inside of main...

(lldb) exp struct YD_MENU $b
error: variable has incomplete type 'struct YD_MENU'
forward declaration of 'YD_MENU'

If I change it to the following...

struct YD_MENU {
    char menu_name[10];
    int menu_option;
} default_menu;

(lldb) exp struct YD_MENU $a works fine.

I think this is related to Why can't LLDB evaluate this expression? but the proposed answers don't work.

(lldb) version
lldb-1000.0.29

score 3 · Answer 1 · answered Aug 27 '18 at 17:17

clang (and gcc as well) are fairly aggressive about not emitting type information for "unused types". There are so many types floating around when you do something like import <Cocoa/Cocoa.h> that it has to do this to keep debug information size reasonable. That's why when you define a structure but don't use it, lldb can't see the type. The information about the type is in fact not there for lldb to see.

C struct - lldb expression fails

1 Answers1