为什么函数能够返回局部变量？ [关闭]

Question

This is just something that caught my curiosity.

While we know that in C/C++ returning a local non-pointer type variable declared in a function is illegal, this is perfectly legal in Golang. Why is that so? Does the compiler determines whether to allocate the variable to stack/heap during compile time based on the usage of the variable?

for example

func getVal() *int {
    x := 1
    return &x
}

Answer 1

Yes, exactly. The compiler performs something called "escape analysis" to determine if the variable's scope exceeds the function's, in which case it automatically gets placed on the heap instead. See http://www.agardner.me/golang/garbage/collection/gc/escape/analysis/2015/10/18/go-escape-analysis.html for more details, although I believe more modern versions of go are a bit smarter now than what's described there.

This is also touched on (a bit confusingly in my opinion) in the official Go FAQ: https://golang.org/doc/faq#stack_or_heap