Get line number for lambda using GDB

Question

We've a backtrace for a segfault that quotes a compiler-generated name for a lambda:

(gdb) bt
#0  std::_Function_handler<std::function<bool()>(), bold::AdHocOptionTreeBuilder::buildTree(bold::Agent*)::__lambda59>::_M_invoke(const std::_Any_data &) (__functor=...) at /usr/include/c++/4.8/functional:2057
#1  0x08146d2c in operator() (this=<optimized out>) at /usr/include/c++/4.8/functional:2464
...

The assigned name is bold::AdHocOptionTreeBuilder::buildTree(bold::Agent*)::__lambda59. However as you can tell that file has a lot of lambda in it! Is there a way to map that generated function's name to a line number in the source code? We have line numbers for other functions, however here it's only quoted as a type param for std::_Function_handler<>.

Compile the file, introducing errors in the lambda bodies, see if the compiler spits out a name? It may number them sequentially which coukd make searching fast. — Yakk - Adam Nevraumont
– Yakk - Adam Nevraumont, Commented Jan 30, 2014 at 12:48

MSalters · Accepted Answer · 2014-01-30 14:59:54Z

1

The linker option -Map mapfile should give you the information showing where each function originated, including lambda's. nm --line-numbers might work too, if the program was compiled with debug info -g.

Also, I think you can use set print symbol-filename on in GDB, and then evaluate &bold::AdHocOptionTreeBuilder::buildTree(bold::Agent*)::__lambda59

edited Jan 30, 2014 at 14:59

answered Jan 30, 2014 at 14:51

MSalters

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3 Comments

Drew Noakes Over a year ago

Thanks very much for this. I'll be sure to try this when I next reproduce the error.

Drew Noakes Over a year ago

Testing this out today I find that nm doesn't list any lambdas. The set command works, though does not change the information in the backtrace. Is there a specific command to use for evaluating the full type name? print &bold::AdHocOptionTreeBuilder::buildTree(bold::Agent*) works, but print &bold::AdHocOptionTreeBuilder::buildTree(bold::Agent*)::__lambda59 gives A syntax error in expression, near `__lambda58'.

MSalters Over a year ago

@DrewNoakes: Tricky. When we say "lambda", we usually mean an unnamed object of an unnamed type, which always has an operator() member. nm shows functions, not types, but I guess the std::_Function_handler template parameter in your example is a type, not a function.

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