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Is the Go compiler optimizing the code?
In this code:
package main import "time" func main() { i := 1 go func() { for { i++ } }() <-time.After(1 * time.Second) println(i) }
The output is always 1. However, it is surprising that 1s is enough for the for loop to go over many times. The reason for this is because the Go compiler is optimizing the code.
The Go memory model specifies the conditions under which reads of a variable in one goroutine can be guaranteed to observe values produced by writes to the same variable in a different goroutine. The assignment to i, via increment i (i = i 1), is not followed by any synchronization event, so it is not guaranteed to be observed by any other goroutine. In fact, an aggressive compiler might delete the entire i statement.
For example, in this code:
package main import "time" func main() { i := 1 go func() { for { i++ } }() <-time.After(1 * time.Millisecond) println(i) }
The output is 1. The goroutine is reduced to:
"".main.func1 STEXT nosplit size=2 args=0x8 locals=0x0 0x0000 00000 (elide.go:7) TEXT "".main.func1(SB), NOSPLIT, <pre class="brush:php;toolbar:false">for { i++ }-8 0x0000 00000 (elide.go:7) FUNCDATA
package main import "time" func main() { i := 1 go func() { for { i++ println("+1") } }() <-time.After(1 * time.Millisecond) println(i) }, gclocals·2a5305abe05176240e61b8620e19a815(SB) 0x0000 00000 (elide.go:7) FUNCDATA , gclocals·33cdeccccebe80329f1fdbee7f5874cb(SB) 0x0000 00000 (elide.go:9) JMP 0
To the compiler, the for loop can be implemented by incrementing a register forever, essentially a no-op for loop:
+1 +1 << SNIP >> +1 +1 432
After inserting a print statement,
"".main.func1 STEXT size=81 args=0x8 locals=0x18 0x0000 00000 (elide.go:7) TEXT "".main.func1(SB), -8 0x0000 00000 (elide.go:7) MOVQ (TLS), CX 0x0009 00009 (elide.go:7) CMPQ SP, 16(CX) 0x000d 00013 (elide.go:7) JLS 74 0x000f 00015 (elide.go:7) SUBQ , SP 0x0013 00019 (elide.go:7) MOVQ BP, 16(SP) 0x0018 00024 (elide.go:7) LEAQ 16(SP), BP 0x001d 00029 (elide.go:7) FUNCDATA <pre class="brush:php;toolbar:false">================== WARNING: DATA RACE Read at 0x00c420094000 by main goroutine: main.main() /home/peter/gopath/src/lucky.go:14 +0xac Previous write at 0x00c420094000 by goroutine 5: main.main.func1() /home/peter/gopath/src/lucky.go:9 +0x4e Goroutine 5 (running) created at: main.main() /home/peter/gopath/src/lucky.go:7 +0x7a ==================, gclocals·a36216b97439c93dafebe03e7f0808b5(SB) 0x001d 00029 (elide.go:7) FUNCDATA , gclocals·33cdeccccebe80329f1fdbee7f5874cb(SB) 0x001d 00029 (elide.go:8) MOVQ "".&i+32(SP), AX 0x0022 00034 (elide.go:9) INCQ (AX) 0x0025 00037 (elide.go:10) PCDATA
package main import ( "sync" "time" ) func main() { mx := new(sync.Mutex) i := 1 go func() { for { mx.Lock() i++ mx.Unlock() } }() <-time.After(1 * time.Second) mx.Lock() println(i) mx.Unlock() },
418078380x0025 00037 (elide.go:10) CALL runtime.printlock(SB) 0x002a 00042 (elide.go:10) LEAQ go.string."+1\n"(SB), AX 0x0031 00049 (elide.go:10) MOVQ AX, (SP) 0x0035 00053 (elide.go:10) MOVQ , 8(SP) 0x003e 00062 (elide.go:10) PCDATA , 0x003e 00062 (elide.go:10) CALL runtime.printstring(SB) 0x0043 00067 (elide.go:10) PCDATA , 0x0043 00067 (elide.go:10) CALL runtime.printunlock(SB) 0x0048 00072 (elide.go:9) JMP 29 0x004a 00074 (elide.go:9) NOP 0x004a 00074 (elide.go:7) PCDATA , $-1 0x004a 00074 (elide.go:7) CALL runtime.morestack_noctxt(SB) 0x004f 00079 (elide.go:7) JMP 0
The output is:
The goroutine expands to:
The increased complexity of the goroutine means that the compiler no longer considers dedicating a register to the value of i. The in-memory value of i is incremented, which makes the updates visible, with a data race, to the main goroutine.
To get the expected result, add some synchronization:
Output:
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