How GDB handles SIGSEGV

Question

When debugging a C++ program emit a SIGSEGV with gdb,it is possible to handle the signal and asked to nostop.

How gdb handles this kind of scenario ??

Have searched gdb source code and couldn't find a starting point.

score 1 · Answer 1 · edited May 23 '17 at 12:15

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You cannot automatically ignore SIGSEGV. I also wouldn't recommend doing that anyway. Although you can make gdb ignore the signal and not pass it to the program, the kernel will attempt to re-run the offending instruction once the signal handler returns and results in an infinite loop. See this answer for more information.

One way to work around it is to the skip the instruction or change register values so that it does not segfault. The link shows an example of setting a register. You can also use the jump command to skip over an instruction.

edited May 23 '17 at 12:15

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answered Feb 17 '16 at 05:38

Kurt Stutsman

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Yes understood handling signal anyway and the question is how gdb does it self – Rimash Mohomed Feb 17 '16 at 06:19

score 1 · Answer 2 · answered Feb 20 '16 at 03:14

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is possible to handle the signal and asked to nostop.

It's unclear whether you want GDB to handle the signal, or the program itself.

If the latter, gdb handle SIGSEGV nostop noprint pass will do exactly that.

answered Feb 20 '16 at 03:14

Employed Russian

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score 0 · Answer 3 · answered Feb 17 '16 at 05:26

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This is actually something the OS does. On Windows, if a program has a debugger attached and an exception is thrown, Windows will ask the debugger if it wants to handle it. If/when it declines, it passes it to the program. If the program doesn't handle it, Windows passes it to the debugger again.

answered Feb 17 '16 at 05:26

1201ProgramAlarm

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Yes understood handling signal anyway and the question is how gdb does it self – Rimash Mohomed Feb 17 '16 at 06:19

How GDB handles SIGSEGV

3 Answers3