While benchmarking, I noticed a surprising heap memory allocation. After reducing the repro, I ended up with the following:
// --- Repro file ---
func memAllocRepro(values []int) *[]int {
for {
break
}
return &values
}
// --- Benchmark file ---
func BenchmarkMemAlloc(b *testing.B) {
values := []int{1, 2, 3, 4}
for i := 0; i < b.N; i++ {
memAllocRepro(values)
}
}
And here is the benchmark output:
BenchmarkMemAlloc-4 50000000 40.2 ns/op 32 B/op 1 allocs/op
PASS
ok memalloc_debugging 2.113s
Success: Benchmarks passed.
Now the funny this is, if I remove the for loop, or if I return the slice directly instead of a slice pointer, there are no more heap alloc:
// --- Repro file ---
func noAlloc1(values []int) *[]int {
return &values // No alloc!
}
func noAlloc2(values []int) []int {
for {
break
}
return values // No alloc!
}
// --- Benchmark file ---
func BenchmarkNoAlloc(b *testing.B) {
values := []int{1, 2, 3, 4}
for i := 0; i < b.N; i++ {
noAlloc1(values)
noAlloc2(values)
}
Benchmark result:
BenchmarkNoAlloc-4 300000000 4.20 ns/op 0 B/op 0 allocs/op
PASS
ok memalloc_debugging 1.756s
Success: Benchmarks passed.
I found that very confusing and confirmed with Delve that the disassembly does has an allocation at the start of the memAllocRepro function:
(dlv) disassemble
TEXT main.memAllocRepro(SB) memalloc_debugging/main.go
main.go:10 0x44ce10 65488b0c2528000000 mov rcx, qword ptr gs:[0x28]
main.go:10 0x44ce19 488b8900000000 mov rcx, qword ptr [rcx]
main.go:10 0x44ce20 483b6110 cmp rsp, qword ptr [rcx+0x10]
main.go:10 0x44ce24 7662 jbe 0x44ce88
main.go:10 0x44ce26 4883ec18 sub rsp, 0x18
main.go:10 0x44ce2a 48896c2410 mov qword ptr [rsp+0x10], rbp
main.go:10 0x44ce2f 488d6c2410 lea rbp, ptr [rsp+0x10]
main.go:10 0x44ce34 488d0525880000 lea rax, ptr [rip+0x8825]
main.go:10 0x44ce3b 48890424 mov qword ptr [rsp], rax
=> main.go:10 0x44ce3f* e8bcebfbff call 0x40ba00 runtime.newobject
I must say though, once I hit that point, I couldn't easily dig further. I'm pretty sure it would be possible to know at least which type is allocated by looking at the structure pointed to by the RAX register, but I wasn't very successful doing so. It's been a long time since I've read disassembly like this.
(dlv) regs
Rip = 0x000000000044ce3f
Rsp = 0x000000c042039f30
Rax = 0x0000000000455660
(...)
All that being said, I have 2 questions: * Anyone can tell why is there a heap allocation there and if it's "expected"? * How could I have gone further in my debugging session? Dumping memory to hex has a different address layout and go tool objdump will output disassembly, which mangle the content at the address location
Full function dump with go tool objdump:
TEXT main.memAllocRepro(SB) memalloc_debugging/main.go
main.go:10 0x44ce10 65488b0c2528000000 MOVQ GS:0x28, CX
main.go:10 0x44ce19 488b8900000000 MOVQ 0(CX), CX
main.go:10 0x44ce20 483b6110 CMPQ 0x10(CX), SP
main.go:10 0x44ce24 7662 JBE 0x44ce88
main.go:10 0x44ce26 4883ec18 SUBQ $0x18, SP
main.go:10 0x44ce2a 48896c2410 MOVQ BP, 0x10(SP)
main.go:10 0x44ce2f 488d6c2410 LEAQ 0x10(SP), BP
main.go:10 0x44ce34 488d0525880000 LEAQ runtime.types+34656(SB), AX
main.go:10 0x44ce3b 48890424 MOVQ AX, 0(SP)
main.go:10 0x44ce3f e8bcebfbff CALL runtime.newobject(SB)
main.go:10 0x44ce44 488b7c2408 MOVQ 0x8(SP), DI
main.go:10 0x44ce49 488b442428 MOVQ 0x28(SP), AX
main.go:10 0x44ce4e 48894708 MOVQ AX, 0x8(DI)
main.go:10 0x44ce52 488b442430 MOVQ 0x30(SP), AX
main.go:10 0x44ce57 48894710 MOVQ AX, 0x10(DI)
main.go:10 0x44ce5b 8b052ff60600 MOVL runtime.writeBarrier(SB), AX
main.go:10 0x44ce61 85c0 TESTL AX, AX
main.go:10 0x44ce63 7517 JNE 0x44ce7c
main.go:10 0x44ce65 488b442420 MOVQ 0x20(SP), AX
main.go:10 0x44ce6a 488907 MOVQ AX, 0(DI)
main.go:16 0x44ce6d 48897c2438 MOVQ DI, 0x38(SP)
main.go:16 0x44ce72 488b6c2410 MOVQ 0x10(SP), BP
main.go:16 0x44ce77 4883c418 ADDQ $0x18, SP
main.go:16 0x44ce7b c3 RET
main.go:16 0x44ce7c 488b442420 MOVQ 0x20(SP), AX
main.go:10 0x44ce81 e86aaaffff CALL runtime.gcWriteBarrier(SB)
main.go:10 0x44ce86 ebe5 JMP 0x44ce6d
main.go:10 0x44ce88 e85385ffff CALL runtime.morestack_noctxt(SB)
main.go:10 0x44ce8d eb81 JMP main.memAllocRepro(SB)
:-1 0x44ce8f cc INT $0x3
Disassemble of the memory pointed to by the RAX register:
(dlv) disassemble -a 0x0000000000455660 0x0000000000455860
.:0 0x455660 1800 sbb byte ptr [rax], al
.:0 0x455662 0000 add byte ptr [rax], al
.:0 0x455664 0000 add byte ptr [rax], al
.:0 0x455666 0000 add byte ptr [rax], al
.:0 0x455668 0800 or byte ptr [rax], al
.:0 0x45566a 0000 add byte ptr [rax], al
.:0 0x45566c 0000 add byte ptr [rax], al
.:0 0x45566e 0000 add byte ptr [rax], al
.:0 0x455670 8e66f9 mov fs, word ptr [rsi-0x7]
.:0 0x455673 1b02 sbb eax, dword ptr [rdx]
.:0 0x455675 0808 or byte ptr [rax], cl
.:0 0x455677 17 ?
.:0 0x455678 60 ?
.:0 0x455679 0d4a000000 or eax, 0x4a
.:0 0x45567e 0000 add byte ptr [rax], al
.:0 0x455680 c01f47 rcr byte ptr [rdi], 0x47
.:0 0x455683 0000 add byte ptr [rax], al
.:0 0x455685 0000 add byte ptr [rax], al
.:0 0x455687 0000 add byte ptr [rax], al
.:0 0x455689 0c00 or al, 0x0
.:0 0x45568b 004062 add byte ptr [rax+0x62], al
.:0 0x45568e 0000 add byte ptr [rax], al
.:0 0x455690 c0684500 shr byte ptr [rax+0x45], 0x0
Escape analysis determines whether any references to a value escape the function in which the value is declared.
In Go, arguments are passed by value, typically on the stack; the stack is reclaimed at the end of the function. However, returning the reference &values from the memAllocRepro function gives the values parameter declared in memAllocRepro a lifetime beyond the end of the function. The values variable is moved to the heap.
memAllocRepro: &values: Alloc
./escape.go:3:6: cannot inline memAllocRepro: unhandled op FOR
./escape.go:7:9: &values escapes to heap
./escape.go:7:9: from ~r1 (return) at ./escape.go:7:2
./escape.go:3:37: moved to heap: values
The noAlloc1 function is inlined in the main function. The values argument, if necessary, is declared in and does not escape from the main function.
noAlloc1: &values: No Alloc
./escape.go:10:6: can inline noAlloc1 as: func([]int)*[]int{return &values}
./escape.go:23:10: inlining call to noAlloc1 func([]int)*[]int{return &values}
The noAlloc2 function values argument is returned as values. values is returned on the stack. There is no reference to values in the noAlloc2 function and so no escape.
noAlloc2: values: No Alloc
package main
func memAllocRepro(values []int) *[]int {
for {
break
}
return &values
}
func noAlloc1(values []int) *[]int {
return &values
}
func noAlloc2(values []int) []int {
for {
break
}
return values
}
func main() {
memAllocRepro(nil)
noAlloc1(nil)
noAlloc2(nil)
}
Output:
$ go build -a -gcflags='-m -m' escape.go
# command-line-arguments
./escape.go:3:6: cannot inline memAllocRepro: unhandled op FOR
./escape.go:10:6: can inline noAlloc1 as: func([]int) *[]int { return &values }
./escape.go:14:6: cannot inline noAlloc2: unhandled op FOR
./escape.go:21:6: cannot inline main: non-leaf function
./escape.go:23:10: inlining call to noAlloc1 func([]int) *[]int { return &values }
./escape.go:7:9: &values escapes to heap
./escape.go:7:9: from ~r1 (return) at ./escape.go:7:2
./escape.go:3:37: moved to heap: values
./escape.go:11:9: &values escapes to heap
./escape.go:11:9: from ~r1 (return) at ./escape.go:11:2
./escape.go:10:32: moved to heap: values
./escape.go:14:31: leaking param: values to result ~r1 level=0
./escape.go:14:31: from ~r1 (return) at ./escape.go:18:2
./escape.go:23:10: main &values does not escape
$