Why does GCC for Risc-V generate nop instructions after call
Asked Answered
S

2

7

GCC for Risc-V produces nop instructions after call instructions by default:

$ cat test.c
void g();
void f() {
        g();
}
$ riscv64-unknown-elf-gcc -S test.c -o -    
[...]
f:
        addi    sp,sp,-16
        sd      ra,8(sp)
        sd      s0,0(sp)
        addi    s0,sp,16
        call    g
        nop #### <-----------here
        ld      ra,8(sp)
        ld      s0,0(sp)
        addi    sp,sp,16
        jr      ra
        .size   f, .-f
        .ident  "GCC: (GNU) 8.3.0"

I'd expect that when targeting an architecture that has branch delay slot, but my understanding is that Risc-V is not such an architecture. Actually, the nop disappears when compiling with -O1 or higher.

Is it just a "bug" in GCC that emits nop as a leftover from architectures that have delay slots, or is there an actual reason for this nop instruction?

Systematize answered 28/8, 2019 at 14:47 Comment(0)
S
8

Not a complete answer, but at least an attempt at diving into the reason why the nop appears. I strongly believe it's a bug/leftover from architectures that have delay slots (as it gets added in the very first RTL pass - expand).

Onto the investigation, GCC has 2 types of passes, Tree and RTL. To see them in action, prep 2 folders as there will be many files, noopt and opt and use the -fdump-tree-all-raw -fdump-rtl-all to see the intermediate results. The last stage of the Tree pass gives (noopt case):

$ cat noopt/test.c.232t.optimized

;; Function f (f, funcdef_no=0, decl_uid=1549, cgraph_uid=0, symbol_order=0)

f ()
{
  <bb 2> :
  gimple_call <g, NULL>
  gimple_return <NULL NULL>

}

The opt case (-O1) differs negligibly:

$ diff -u noopt/test.c.232t.optimized opt/test.c.232t.optimized 
--- noopt/test.c.232t.optimized 2019-09-03 14:48:02.874071927 +0200
+++ opt/test.c.232t.optimized   2019-09-03 14:48:29.550278667 +0200
@@ -3,7 +3,7 @@

 f ()
 {
-  <bb 2> :
+  <bb 2> [local count: 1073741825]:
   gimple_call <g, NULL>
   gimple_return <NULL NULL>

The first stage of the RTL passes (expand) is the one that differs:

$ cat noopt/test.c.234r.expand

;; Function f (f, funcdef_no=0, decl_uid=1549, cgraph_uid=0, symbol_order=0)


;; Generating RTL for gimple basic block 2


try_optimize_cfg iteration 1

Merging block 3 into block 2...
Merged blocks 2 and 3.
Merged 2 and 3 without moving.
Merging block 4 into block 2...
Merged blocks 2 and 4.
Merged 2 and 4 without moving.


try_optimize_cfg iteration 2



;;
;; Full RTL generated for this function:
;;
(note 1 0 3 NOTE_INSN_DELETED)
(note 3 1 2 2 [bb 2] NOTE_INSN_BASIC_BLOCK)
(note 2 3 5 2 NOTE_INSN_FUNCTION_BEG)
(call_insn 5 2 8 2 (parallel [
            (call (mem:SI (symbol_ref:DI ("g") [flags 0x41] <function_decl 0x7fbc2827a400 g>) [0 g S4 A32])
                (const_int 0 [0]))
            (clobber (reg:SI 1 ra))
        ]) "../test.c":3 -1
     (nil)
    (nil))
(insn 8 5 0 2 (const_int 0 [0]) "../test.c":4 -1
     (nil))

The difference with -O1 is just removing that const_int 0 [0], which will ultimately lead to the nop:

$ diff -u noopt/test.c.234r.expand opt/test.c.234r.expand 
--- noopt/test.c.234r.expand    2019-09-03 14:48:02.874071927 +0200
+++ opt/test.c.234r.expand  2019-09-03 14:48:29.550278667 +0200
@@ -25,12 +25,10 @@
 (note 1 0 3 NOTE_INSN_DELETED)
 (note 3 1 2 2 [bb 2] NOTE_INSN_BASIC_BLOCK)
 (note 2 3 5 2 NOTE_INSN_FUNCTION_BEG)
-(call_insn 5 2 8 2 (parallel [
-            (call (mem:SI (symbol_ref:DI ("g") [flags 0x41] <function_decl 0x7fbc2827a400 g>) [0 g S4 A32])
+(call_insn 5 2 0 2 (parallel [
+            (call (mem:SI (symbol_ref:DI ("g") [flags 0x41] <function_decl 0x7f0bdec1f400 g>) [0 g S4 A32])
                 (const_int 0 [0]))
             (clobber (reg:SI 1 ra))
         ]) "../test.c":3 -1
      (nil)
     (nil))
-(insn 8 5 0 2 (const_int 0 [0]) "../test.c":4 -1
-     (nil))
Substantialism answered 3/9, 2019 at 13:10 Comment(0)
N
-1

GCC will probably do this on any architecture. I think the nop instruction is related to the void result, it's where a non-void function would set up the return value. Try compiling this:

int g();
int f() {
        g();
        return 1;
}

For a void function, there is nothing to do to generate the result, hence the nop.

Needlecraft answered 3/10, 2019 at 17:30 Comment(0)

© 2022 - 2024 — McMap. All rights reserved.