douji1877 2010-11-19 16:18
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Go怎么没有堆栈溢出

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?

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  • dpd7122 2010-11-19 16:29
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    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.

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