Kevin Sopp <baraclese <at> googlemail.com> writes:
What happens is that delete (test_case const*) tu.second; first invokes the test_case destructor which will erase the value referenced by tu from the map, then the delete statement continues but tu references deallocated memory now at which point valgrind complains that you do an invalid read of 4 bytes.
From what I understand order of events is as follows:
1. tu.second resolved (memory accessed) 2. ~test_case() invoked 3. ~test_unit() invoked 4. framework::deregister_test_unit( this ) is invoked 5. map.erase( tu ) 6. free( ptr ) If what you saying is true we perform first step twice. Second time before 6th step. Why would compiler do this? Can you test it? just create temp pointer test_unit* tu_ptr = tu.second; before delete. Gennadiy