Thank you very much. 在 2019年7月31日星期三 UTC+8下午9:49:04，Ian Lance Taylor写道： > > On Wed, Jul 31, 2019 at 4:49 AM jin wang <wangjin...@gmail.com > <javascript:>> wrote: > > > > so mstart just put on stack for call traceback , but actually will not > return at mstart? > > Yes. As you can see, the systemstack function changes the stack > pointer back to its original value before it returns to its caller. > It does not return on the stack where mstart is stored. > > Ian > > > > 在 2019年7月31日星期三 UTC+8下午12:08:51，Ian Lance Taylor写道： > >> > >> On Tue, Jul 30, 2019 at 8:58 PM <wangjin...@gmail.com> wrote: > >> > > >> > I read go runtime code，but I'm poor in assembly language, > >> > i can't understand the following code > >> > // switch to g0 > >> > MOVQ DX, g(CX) > >> > MOVQ (g_sched+gobuf_sp)(DX), BX > >> > // make it look like mstart called systemstack on g0, to stop > traceback > >> > SUBQ $8, BX > >> > MOVQ $runtime·mstart(SB), DX > >> > MOVQ DX, 0(BX) > >> > MOVQ BX, SP > >> > > >> > > >> > what does copy runtime·mstart to stack means,could someone help me > >> > >> It does exactly what the comment says: when doing a stack traceback, > >> it makes it appear that mstart called systemstack. When a function is > >> called, the return address is placed on the stack. In this case the > >> assembly code is deliberately putting mstart on the stack, so that > >> that is what a traceback will see. > >> > >> Ian > > > > -- > > You received this message because you are subscribed to the Google > Groups "golang-nuts" group. > > To unsubscribe from this group and stop receiving emails from it, send > an email to golan...@googlegroups.com <javascript:>. > > To view this discussion on the web visit > https://groups.google.com/d/msgid/golang-nuts/c8587ec0-a09f-4215-8989-6133d307c7f8%40googlegroups.com. > > >
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