At the bottom of a Go func with defer, the assembly looks like this:
0x000000000047e1a0 <+256>: callq *%rsi
0x000000000047e1a2 <+258>: movb $0x0,0x7(%rsp)
0x000000000047e1a7 <+263>: mov 0x748(%rsp),%rdx
0x000000000047e1af <+271>: mov (%rdx),%rbx
0x000000000047e1b2 <+274>: callq *%rbx
0x000000000047e1b4 <+276>: mov 0x750(%rsp),%rbp
0x000000000047e1bc <+284>: add $0x758,%rsp
0x000000000047e1c3 <+291>: retq
0x000000000047e1c4 <+292>: callq 0x42ea40 <runtime.deferreturn>
0x000000000047e1c9 <+297>: mov 0x750(%rsp),%rbp
0x000000000047e1d1 <+305>: add $0x758,%rsp
0x000000000047e1d8 <+312>: retq
0x000000000047e1d9 <+313>: callq 0x458900 <runtime.morestack>
As we can see, there is a retq
before call runtime.deferreturn
. So, it seems that the deferreturn
will never be called. Do I understand this right?
As we can see, there is a retq before call runtime.deferreturn. So, it seems that the deferreturn will never be called. Do I understand this right?
No. The code may be used by the Go runtime, for example, while panicking.