[qzzi]

No, the debugger does not get involved in preprocessing. When you write "a = __LINE__;", it expands to "a = 10;" (or whatever number) and is compiled, and the debugger has no knowledge of it. Debugging information, including the mapping of positions in the code to positions in the source, is generated by the compiler and embedded directly into the generated binary or an external file, from which the debugger reads it.

The __LINE__ macro is passed to the debugger only if the program itself outputs its value, and the "debugger" is a human reading that output :)