Linux是否在较低的堆栈末端下方提供了保证的不可访问的内存区域? [英] Does linux provide a guaranteed inaccessible memory area below the lower stack end?

问题描述

Linux是否在堆栈下端下方提供了无法访问的内存区域，该区域具有保证的最小大小?如果存在这样保证的最小尺寸，那是什么?

Does Linux provide an inaccessible memory area below the lower stack end that has a guaranteed minimum size? And if such a guaranteed minimum size exists, what is it?

或者换句话说，我什么时候应该开始担心alloca()或将指针提供给有效的非堆栈内存?

Or in other words, when should I start to worry about alloca() or so giving me pointers into valid, non-stack memory?

推荐答案

作为

如果不能扩展堆栈帧，则没有错误指示. (但是，分配失败后，该程序很可能会收到 如果尝试访问未分配的空间，则为SIGSEGV信号.)

There is no error indication if the stack frame cannot be extended. (However, after a failed allocation, the program is likely to receive a SIGSEGV signal if it attempts to access the unallocated space.)

所以根本没有任何指示，它还说:

So there is no indication at all and it also says:

如果分配导致堆栈溢出，则程序行为未定义.

If the allocation causes stack overflow, program behavior is undefined.

堆栈溢出问题是递归的一个普遍问题，并不是alloca的真正问题，或者说可变长度数组.通常，您要么需要找到一种方法来限制递归的深度，要么将其重构为迭代解决方案，要么使用自己的动态堆栈(可能不适用于这种情况).

The stack overflow problem is a general issue with recursion and not really particular to alloca or let's say variable length arrays. Typically you either need to find a way to limit the depth of the recursion, refactor to an iterative solution or use your own dynamic stack(probably does not apply to this case).

更新

OP发现 Linux确实会通过生成用于解决问题第一部分的SIGBUS信号，在堆栈溢出后使用保护页来提供事后指示.

As the OP discovered Linux does provide an after the fact indication using a guard page after the stack of stack overflow by generating a SIGBUS signal, which addresses the first part of the question.

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