Before replacing the instruction with INT 3, the debugger keeps a note of what instruction was at that point in the code. When the CPU encounters INT 3, it hands control to the debugger.
When the debugging operations are done, the debugger replaces the INT 3 with the original instruction and makes the instruction pointer go back one step, thereby ensuring that the original instruction is executed.
The debug version you compile doesn’t affect the code; it just stores more information about symbols. The whole shtick about the debugger replacing instructions with INT3 still happens.
You can validate that the code isn’t affected yourself by running objdump on two binaries, one compiled with debug symbols and one without. Otherwise if you’re lazy (like me 😄):
Hey thank you!
Not what I thought it was for sure 😃
How does it work if an instruction gets replaced by the INT3 though?
Excellent question!
Before replacing the instruction with INT 3, the debugger keeps a note of what instruction was at that point in the code. When the CPU encounters INT 3, it hands control to the debugger.
When the debugging operations are done, the debugger replaces the INT 3 with the original instruction and makes the instruction pointer go back one step, thereby ensuring that the original instruction is executed.
Whoo that seems complicated, I mean you akready compile a debug version.
Thanks for the explanation!
The debug version you compile doesn’t affect the code; it just stores more information about symbols. The whole shtick about the debugger replacing instructions with INT3 still happens.
You can validate that the code isn’t affected yourself by running objdump on two binaries, one compiled with debug symbols and one without. Otherwise if you’re lazy (like me 😄):
https://stackoverflow.com/a/8676610
And for completeness: https://gcc.gnu.org/onlinedocs/gcc-14.1.0/gcc/Debugging-Options.html