How come Go doesn't have stackoverflows
I read in this presentation http://golang.org/doc/ExpressivenessOfGo.pdf page 42:
Safe
- no stack overflows
How is this possible? and/or how does Go works to avoid this?
It's a feature called "segmented stacks": every goroutine has its own stack, allocated on the heap.
In the simplest case, programming language implementations use a single stack per process/address space, commonly managed with special processor instructions called push and pop (or something like that) and implemented as a dynamic array of stack frames starting at a fixed address (commonly, the top of virtual memory).
That is (or used to be) fast, but is not particularly safe. It causes trouble when lots of code is executing concurrently in the same address space (threads). Now each needs its own stack. But then, all the stacks (except perhaps one) must be fixed-size, lest they overlap with each other or with the heap.
Any programming language that uses a stack can, however, also be implemented by managing the stack in a different way: by using a list data structure or similar that holds the stack frames, but is actually allocated on the heap. There's no stack overflow until the heap is filled.
it uses a segmented stack. Which basically means it uses a linked list instead of a fixed size array as it's stack. When it runs out of space it makes the stack a little bigger.
edit:
Here is some more information: http://golang.org/doc/go_faq.html#goroutines
The reason this is so great is not because it'll never overflow (that's a nice side-effect), it's that you can create threads with a really small memory footprint, meaning you can have lots of them.
I don't think they can "totally" avoid stack overflows. They provide a way to prevent the most typical programming-related errors to produce a stack overflow.
When the memory finishes there is no way to prevent a stack overflow.
链接地址: http://www.djcxy.com/p/80444.html上一篇: 操作系统如何检测堆栈溢出？
下一篇: Go怎么没有stackoverflows