Calling a lambda function in GDB - c++

Question

//gdb-call-lambda.cpp
#include <iostream>
void do_something(void) {
std::cout << "blah blah" << std::endl;
auto lambda_func = [](void){
std::cout << "in lambda" << std::endl;
return;
};
lambda_func();
std::cout << "..." << std::endl;
return;
}
int main(int argc, char **argv) {
do_something();
return 0;
}
In this example program, if you compile (g++ gdb-call-lambda.cpp --std=c++11 -g) and then run it in gdb (gdb ./a.out), you can have GDB call any "normal" function. Example:
(gdb) break main
Breakpoint 1 at 0x4008e7: file gdb-call-lambda.cpp, line 20.
(gdb) r
Starting program: /home/keithb/dev/mytest/gdb-call-lambda/a.out
Breakpoint 1, main (argc=1, argv=0x7fffffffdfb8) at gdb-call-lambda.cpp:20
20 do_something();
(gdb) call do_something()
blah blah
in lambda
...
However, if you then try to call the lambda:
(gdb) break do_something
Breakpoint 2 at 0x400891: file gdb-call-lambda.cpp, line 5.
(gdb) c
Continuing.
Breakpoint 2, do_something () at gdb-call-lambda.cpp:5
5 std::cout << "blah blah" << std::endl;
(gdb) n
blah blah
12 lambda_func();
(gdb) n
in lambda
14 std::cout << "..." << std::endl;
(gdb) call lambda_func()
Invalid data type for function to be called
GDB kinda freaks out. So my question is thus: how do you call a lambda in GDB? Asking GDB what it expects reveals nothing of interest when compared to a normal function:
(gdb) whatis lambda_func
type = __lambda0
(gdb) whatis do_something
type = void (void)
I went to see if lambda_func has any special members, eg a function pointer to call, akin to std::function and/or std::bind:
(gdb) print lambda_func
$1 = {<No data fields>}
No special members? Okay maybe it's just a glorified function pointer?
(gdb) call ((void (void)) lambda_func)()
Program received signal SIGSEGV, Segmentation fault.
0x00007fffffffdeaf in ?? ()
The program being debugged was signaled while in a function called from GDB.
GDB remains in the frame where the signal was received.
To change this behavior use "set unwindonsignal on".
Evaluation of the expression containing the function
(at 0x0x7fffffffdeaf) will be abandoned.
When the function is done executing, GDB will silently stop.
So I'm not even 100% sure what order to pass any arguments or especially captured types.
I tried additionally call lambda_func.operator()(), call lambda_func::operator(), call lambda_func::operator()(), call __lambda0, call __lambda0(), call __lambda0::operator(), call __lambda0::operator()(), all to no avail.
A search on google reveals things about setting breakpoints in lambdas, but nothing on how to call those lambdas from the debugger.
For what it's worth, this is on Ubuntu 14.04 64-bit using g++ 4.8.2-19ubuntu1 and gdb 7.7-0ubuntu3.1

Answer 1

I was expecting call __lambdaX::operator()() works but it doesn't. I think it is related to GCC implementation. I am not sure if there is a better way but this is my workaround solution when I need to call lambda in GDB.
Briefly, GDB has disassemble command and it gives __lambda0::operator()() const as debug information at the call instruction line. Then, convert that address into a function pointer and call it.
Example explains better.
$ g++ -g -std=c++0x lambda.cpp
$ ./a.out
blah blah
in lambda
...
GDB:
$ gdb ./a.out
GNU gdb (GDB) Fedora 7.7.1-13.fc20
Copyright (C) 2014 Free Software Foundation, Inc.
License GPLv3+: GNU GPL version 3 or later <http://gnu.org/licenses/gpl.html>
This is free software: you are free to change and redistribute it.
There is NO WARRANTY, to the extent permitted by law. Type "show copying"
and "show warranty" for details.
This GDB was configured as "x86_64-redhat-linux-gnu".
Type "show configuration" for configuration details.
For bug reporting instructions, please see:
<http://www.gnu.org/software/gdb/bugs/>.
Find the GDB manual and other documentation resources online at:
<http://www.gnu.org/software/gdb/documentation/>.
For help, type "help".
Type "apropos word" to search for commands related to "word"...
Reading symbols from ./a.out...done.
(gdb) b do_something()
Breakpoint 1 at 0x4008a3: file lambda.cpp, line 4.
(gdb) run
Starting program: /home/alper/cplusplus/a.out
Breakpoint 1, do_something () at lambda.cpp:4
4 std::cout << "blah blah" << std::endl;
Missing separate debuginfos, use:
(gdb) n
blah blah
11 lambda_func();
Disassemble do_something
(gdb) disassemble do_something
Dump of assembler code for function do_something():
0x40089b <+0>: push %rbp
0x40089c <+1>: mov %rsp,%rbp
0x40089f <+4>: sub $0x10,%rsp
=> 0x4008a3 <+8>: mov $0x4009fb,%esi
0x4008a8 <+13>: mov $0x601060,%edi
0x4008ad <+18>: callq 0x400750 <_ZStlsISt11char_traitsIcEERSt13basic_ostreamIcT_ES5_PKc#plt>
0x4008b2 <+23>: mov $0x400770,%esi
0x4008b7 <+28>: mov %rax,%rdi
0x4008ba <+31>: callq 0x400760 <_ZNSolsEPFRSoS_E#plt>
0x4008bf <+36>: lea -0x1(%rbp),%rax
0x4008c3 <+40>: mov %rax,%rdi
0x4008c6 <+43>: callq 0x400870 <__lambda0::operator()() const>
0x4008cb <+48>: mov $0x400a05,%esi
0x4008d0 <+53>: mov $0x601060,%edi
0x4008d5 <+58>: callq 0x400750 <_ZStlsISt11char_traitsIcEERSt13basic_ostreamIcT_ES5_PKc#plt>
0x4008da <+63>: mov $0x400770,%esi
0x4008df <+68>: mov %rax,%rdi
0x4008e2 <+71>: callq 0x400760 <_ZNSolsEPFRSoS_E#plt>
0x4008e7 <+76>: nop
0x4008e8 <+77>: leaveq
0x4008e9 <+78>: retq
GDB outputs line callq 0x400870 <__lambda0::operator()() const> so convert 0x400870 into a function pointer and call it.
(gdb) call ((void (*)()) 0x400870)()
in lambda
(gdb) call ((void (*)()) 0x400870)()
in lambda
(gdb) call ((void (*)()) 0x400870)()
in lambda
Note: If GCC inlines the lambda, there is nothing to call. For example, if the example above is compiled with optimisation switch -O3, there is no line with __lambda0::operator()() const in GDB disassemble output.

Answer 2

In a less synthetic use case I had success in calling the lambda by calling it as
call myLambda.operator()(param1,param2)
But in the test case in the OP GDB appears to think the function is inlined — even though it's not. I've reported this as bug 28137.

Answer 3

Note: these methods are not guaranteed to work with C++14 generic lambda.
Case 1. When modifying the source code is allowed
Requires modifying the source code for each individual lambda.
Method 1.1. std::function
Wrap the lambda in a std::function, and explicitly instantiate it.
(it does have an extra performance hit for dynamically allocate memory, but it doesn't matter because this is only the debug build)
If the lambda itself
does not take any local data types as input or return one,
it's possible to explicitly instantiate the whole class
Otherwise, operator() can be instantiated by using it in the program.
#include<iostream>
#include<functional>
template class std::function<int(int)>; // explicit instantiation
int main(){
auto a=[](int x){
std::cout<<x<<'\n';
return x+1;
};
std::function a_function{a};
a_function(1); // implicit instantiation
__builtin_trap();
}
Call in gdb as a_function.operator()(1).
Reference:
C++, STL, GDB: Cannot evaluate function maybe inlined - Stack Overflow
c++ - gdb stl functions still show as inlined after disabling optimizations - Stack Overflow
Method 1.2: store the address of the lambda
Program:
#include<iostream>
int main(){
auto a=[](int x){
std::cout<<x<<'\n';
return x+1;
};
auto a_operator_call=&decltype(a)::operator();
__builtin_trap();
}
Call like this: (requires appropriate compilation option so that value of a_operator_call is not optimized out)
(gdb) print (a.*a_operator_call)(1)
1
$1 = 2
Reference:
c++ - Possible to call inline functions in gdb and/or emit them using GCC? - Stack Overflow
Case 2. When modifying the source code is not allowed
Case 2.1. Global lambda
#include<iostream>
auto a=[](int x){
std::cout<<x<<'\n';
return x+1;
};
int main(){
__builtin_trap();
}
You need to disable xmethod (at least in my current gdb version. I think it's a bug)
and call with the syntax a.operator()(...):
(gdb) print a
$1 = {<No data fields>}
(gdb) print a(1)
Invalid data type for function to be called.
(gdb) print a.operator()(1)
Python Exception <class 'TypeError'> expected string or bytes-like object:
Error while looking for matching xmethod workers defined in Python.
(gdb) disable xmethod
(gdb) print a.operator()(1)
1
$2 = 2
Reference:
#836820 - gdb: Python Exception <class 'TypeError'> expected string or bytes-like object: - Debian Bug report logs
Gentoo Forums :: View topic - [solved] gdb can't print types anymore
Case 2.2. Local lambda
Note: for this method it may be required to compile with -fkeep-inline-functions.
Consider a simple program
#include<iostream>
int main(){
auto a=[](int x){
std::cout<<x<<'\n';
return x+1;
};
__builtin_trap();
}
And see what symbols are generated in the executable:
(gdb) shell nm --demangle ./a.out |grep lambda
000000000001fe4c t main::{lambda(int)#1}::operator()(int) const
000000000001ff5c t main::{lambda(int)#1}::operator int (*)(int)() const
000000000001ff40 t main::{lambda(int)#1}::_FUN(int)
(gdb) print 'main::{lambda(int)#1}::operator()(int) const'
$1 = {int (const struct {...} * const, int)} 0x555555573e4c <operator()(int) const>
(gdb) print 'main::{lambda(int)#1}::operator()(int) const' (a, 1)
No symbol "(null)" in current context.
(gdb) set $c='main::{lambda(int)#1}::operator()(int) const'
(gdb) print (a.*$c)(1)
Non-pointer-to-member value used in pointer-to-member construct
(gdb) print $c(&a, 1)
1
$2 = 2
You can see that the symbol has the type of a function pointer, not a pointer-to-member-function,
so it must be called with normal syntax (although as can be seen in method 1.2 above gdb does support function-to-member dereference)
It's possible (for example with a Python program) to
automatically parse the symbols with lambda in its demangled name
get its type with maintenance print type command
on lambda invocation: get the lambda object's actual type with the maintenance print type command
(ptype alone is insufficient as some different types have identical string representation and considered equal by gdb, see example below)
then match the types against each other to determine the correct symbol.
Failed attempts
Use gdb:
(gdb) print a
$1 = {<No data fields>}
(gdb) ptype a
type = struct {
}
(gdb) ptype $c
type = int (*)(const struct {...} * const, int)
Use Python API:
Unfortunately, for some reason it consider different types with identical definition equal.
For example in this program:
#include<iostream>
int main(){
auto a=[](int x){ std::cout<<x<<'\n'; return x+1; };
auto b=[](int x){ std::cout<<x<<'\n'; return x+2; };
struct{ int x; } c;
struct{ int x; } d;
struct{ int y; } e;
__builtin_trap();
}
in an IPython interactive shell embedded in gdb:
In [1]: gdb.parse_and_eval("a").type == gdb.parse_and_eval("b").type
Out[38]: True
In [39]: gdb.parse_and_eval("c").type == gdb.parse_and_eval("d").type
Out[39]: True
In [40]: gdb.parse_and_eval("c").type == gdb.parse_and_eval("e").type
Out[40]: False
Things that can be tried out (although not necessary in this particular case)
-fno-inline-functions
-fkeep-inline-functions
-gdwarf-5
-g3
__attribute__((used))
-fno-eliminate-unused-debug-types
-fkeep-static-functions
-fkeep-static-consts
Reference:
c++ - Cannot evaluate function -- may be inlined - Stack Overflow
c++ - Function to mangle/demangle functions - Stack Overflow
c++ - Getting mangled name from demangled name - Stack Overflow
compiler construction - What is Linux utility to mangle a C++ symbol name? - Stack Overflow
Debugging Options (Using the GNU Compiler Collection (GCC))
How to make gdb show the original non-mangling function name on disassembly model? - Stack Overflow
https://dwarfstd.org/doc/Debugging%20using%20DWARF.pdf