with parallel programming and security. (he/him)
Very unexpected!
My best guess is that with a deep recursion, you're forcing a stack growth of current goroutine, ending up with local variables in a new memory location.
val doesn't "escape" (to the heap), yet we can't rely on its address 
#golang #pop #quiz
func main() {
var val int
println(&val)
f(10000)
println(&val)
}
func f(i int) {
if i--; i == 0 {
return
}
f(i)
}
this program prints:
[playground (spoilers) https://play.golang.org/p/JQSqP3yV3Et ]
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Yeah, that's my guess, too. There's a nice article online about the switch from the original segmented stacks to contiguous stacks, which require copying to new stacks for growth:https://blog.cloudflare.com/how-stacks-are-handled-in-go/amp/ …
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Dammmnp ( my brain sounds like Hommer Simpson after click for apparently "obvious" 2 distincts results.
@ChaveDeney We need u!!!! -
The nice thing is that if you save those pointers and compare them they are equal to each other
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Replacing println with fmt.Println makes both print statements print the same address. I don't understand why?
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Because interfaces are always treated as escaping to the heap, thus they are never moved
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Try this one: package main func main() { var val int a := &val println(a) f(10000) b := &val println(b) println(a == b) // true? } func f(i int) { if i--; i == 0 { return } f(i) } https://play.golang.org/
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with some experimenting it only gets two diff pointers when i > 65 which i wasn't quite expecting
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What did you expect?
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Čini se da učitavanje traje već neko vrijeme.
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