As far as I know SW-breakpoints are working as follows:

The instruction the BP is set to gets substituted by a int/trap instruction, than the trap is handled in a trap handler, on continue the trap is replaced by the original instruction, the instruction is executed in single step mode, now the PC points to the next instruction and the original instruction is replaced again by a int/trap instruction.

HW Breakpoints work as follows according to my understanding:

The address of the instruction the BP is set to is written in a HW-BP Register. If the instruction is hit respectively the PC matches the address in the HW-BP Register, the CPU raises an interrupt which is also handled by a trap handler. Now if the program returns to the orignial instruction the HW BP is still active and one is caught in an infinite loop.

How is that problem treated?

Is the HW BP disabled before continuing and is the orignal instruction also getting executed in single step mode? Or is the original instruction executed before the trap handler is entered, so that the trap handler returns to the instruction after the original instruction? Or is there an other mechanism?

CodePudding user response：

In case of the Intel 64 and IA-32 ("x64/x86") architectures, this is the task of the Resume Flag (RF), bit 16 in EFLAGS. (Other processor architectures that support hardware breakpoints probably have a similar mechanism.)

See section 18.3.1.1 in the

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