drogon982007
2019-07-31 02:08
浏览 16
已采纳

为什么函数能够返回局部变量? [关闭]

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
}
  • 点赞
  • 写回答
  • 关注问题
  • 收藏
  • 邀请回答

1条回答 默认 最新

  • douxi7219 2019-07-31 02:26
    已采纳

    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

    点赞 打赏 评论

相关推荐 更多相似问题