[gdb][qemu]: Unable to create debug configuration without "program".

[dibalavs]

Type: Debugger

Hello all.
I use vscode to debug programs, launched in QEMU.
As you may know, qemu has built-in gdbserver stub which is available at localhost:1234 .
I am debugging application, launched in qemu:

My build system generates .gdbinit file at build directory:

set architecture i386:x86-64:intel
add-symbol-file /path/to/program1 0xaddr1
add-symbol-file /path/to/program2 0xaddr2
add-symbol-file /path/to/program3 0xaddr3
target remote :1234 

So, to start debugging, I just need to launch gdb in build directory without any parameters.
But I cannot do it, because parameter "program" is necessary.
Please make it optional.

Describe the bug

• OS and Version: Ubuntu 18.04
• VS Code Version: 1.28.2
• C/C++ Extension Version: 0.19.0

To Reproduce
Steps to reproduce the behavior:

1. Go to debug view
2. Click on gear icon (open launch.json)
3. Remove "program" property
4. See error message "missing property program" on hover mouse over { symbol

[pieandcakes]

@dibalavs Hi, this seems like a very specific case where program is not required. Unfortunately for most other users it will be a required parameter. If you can propose a way where we can tell it is in your specific case, we may be able to add a check to ignore the program field without disrupting other users.

[michelleangela]

Closing this issue due to long inactivity and by design.