Pointers (GoLang)

user_5333957 0
Viewed 144

This is a rookie question, please bear with me. So the doubt is why f() function points different address. My understanding is the variable v must overwrite the old value.

package main
import "fmt"

var p = f()

func f() *int {
    v := 1
    return &v
}

func main() {
    fmt.Println(f())
    fmt.Println(f())
    fmt.Println(f())
    fmt.Println(p)
}

//0xc0000140b0
//0xc0000140b8
//0xc0000140e0
//0xc000014098
go pointers
2 Answers

The compiler detects that v escapes the function f, so it is allocated on the heap. Each call to f returns a new instance of v, that's why you see a different address for each call.

user_119239990

To give a simple answer to this

Go looks for variables that outlive the current stack frame and then heap-allocates them

Basically, the variable v escapes the function f stack frame and gets allocated in heap which is why you see different addresses printed everytime.

Read this nice introduction to escape analysis. https://medium.com/a-journey-with-go/go-introduction-to-the-escape-analysis-f7610174e890

Try running escape analysis to see all the variables that got escaped.

go build -gcflags="-m" main.go:
./main.go:7:2: moved to heap: v   //points to v := 1
./main.go:12:15: moved to heap: v //points to fmt.Println(f())
./main.go:13:15: moved to heap: v //points to fmt.Println(f())
./main.go:14:15: moved to heap: v //points to fmt.Println(f())

Note that the last fmt.Println(f()) statement is not considered for escaping as value passed to Println is p which is a global variable so its already in heap and thus doesn't need to escape.

user_28518980
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