Re: Is this a bug in gdb

On 2/17/20 2:50 PM, Luis Machado wrote:
> On 2/17/20 10:21 AM, Volker Weißmann wrote:
>> On 2/17/20 1:37 PM, Luis Machado wrote:
>>
>>> On 2/16/20 7:40 PM, Volker Weißmann wrote:
>>>> Hello,
>>>>
>>>> The help text of the watch command claims that the -l option
>>>> watches the
>>>> memory of the variable. When I tried this, I was surprised by the
>>>> outcome (reproducible):
>>>>
>>>>
>>>> (gdb) watch this->v_
>>>> Hardware watchpoint 2: this->v_
>>>
>>> This command tells GDB to watch the value of this->v_, whatever
>>> address &(this->v_) points to. That, of course, can change across the
>>> execution of the program.
>>>
>>>> (gdb) watch -l this->v_
>>>
>>> This command tells GDB to watch for changes in a particular location.
>>> Since this->v_ is a value rather than a location, the error is thrown.
>>>
>>> The correct invocation would be ...
>>>> A syntax error in expression, near `restrict *) 0x00007fffffffd398'.
>>>> (gdb)
>>>>
>>>> Note: Using print &(this->v_) and  watch (char[8])
>>>> *outputoftheprintcommand worked.
>>>>
>>>
>>> ... the above. It points to an address that will be watched.
>>>
>>>>
>>>> I am asking you whether this is a bug in gdb or not, because if it
>>>> is a
>>>> bug in gdb, I will try to make a minimal example an file a bug report.
>>>
>>> I don't think it is a bug. But maybe the documentation isn't doing a
>>> good enough job of making it clear how to invoke these commands?
>>
>> The reason why I thought that this might be a bug, is that I wrote the
>> following toy program to test it:
>>
>> class MyClass {
>> public:
>>      int var = 0;
>>      void member() {
>>          var = 1;
>>      }
>> };
>> int main() {
>>      MyClass obj;
>>      obj.member();
>>      obj.var = 2;
>>      return obj.var;
>> }
>>
>> Lets say I want to know why my program returns 2 and not 1. I can do
>> this like this:
>
> I went to actually read the documentation (by habit i only use watch
> with the address itself) and i was mistaken. The -location option
> should take care of grabbing the address of whatever expression you
> passed to it. So ...
>
>>
>> (gdb) break main.cpp:5
>> Breakpoint 1 at 0x118c: file main.cpp, line 5.
>> (gdb) run
>> Starting program: /home/volker/Sync/DatenVolker/bugreports/gdb/a.out
>>
>> Breakpoint 1, MyClass::member (this=0x7fffffffe344) at main.cpp:5
>> 5               var = 1;
>> (gdb) watch -l this->var
>> Hardware watchpoint 2: -location this->var
>> (gdb) continue
>> Continuing.
>>
>> Hardware watchpoint 2: -location this->var
>>
>> Old value = 0
>> New value = 1
>> MyClass::member (this=0x7fffffffe344) at main.cpp:6
>> 6           }
>> (gdb) continue
>> Continuing.
>>
>> Hardware watchpoint 2: -location this->var
>>
>> Old value = 1
>> New value = 2
>> main () at main.cpp:12
>> 12          return obj.var;
>> (gdb)
>
> ... this indeed is supposed to work.
>
>>
>> In this toy program, watch -l this->var works, but in my large program
>> watch -l this->v_ did not. And I do not understand the difference.
>>
>>
>
> I'm guessing GDB got confused while trying to grab the address of your
> this->v_ expression. If you have a short testcase for that, it would
> be great to have this reported (if it isn't already).
Thank you for your help. I only have a large testcase for it, but I will
try to reduce it to a short testcase. After that, I will open a bug report.