alloca的一个缺陷是longjmp将它倒带。

也就是说，如果你用setjmp保存一个上下文，然后分配一些内存，然后longjmp到上下文，你可能会失去分配的内存。堆栈指针回到原来的位置，因此内存不再保留;如果你调用一个函数或执行另一个分配，你将破坏原来的分配。

为了澄清，我在这里特别提到的是一种情况，即longjmp不返回发生分配的函数!相反，函数使用setjmp保存上下文;然后使用alloca分配内存，最后在该上下文中执行longjmp。该函数的分配内存没有全部释放;就是从setjmp开始分配的所有内存。当然，我说的是观察到的行为;据我所知，任何分配都没有这样的要求。

The focus in the documentation is usually on the concept that alloca memory is associated with a function activation, not with any block; that multiple invocations of alloca just grab more stack memory which is all released when the function terminates. Not so; the memory is actually associated with the procedure context. When the context is restored with longjmp, so is the prior alloca state. It's a consequence of the stack pointer register itself being used for allocation, and also (necessarily) saved and restored in the jmp_buf.

顺便说一句，如果这样工作的话，这提供了一种合理的机制来故意释放使用alloca分配的内存。

我曾经遇到过这种情况，这是一个bug的根本原因。

仍然不鼓励使用分配，为什么?

我没有看到这样的共识。很多强大的专业人士;一些缺点:

C99 provides variable length arrays, which would often be used preferentially as the notation's more consistent with fixed-length arrays and intuitive overall many systems have less overall memory/address-space available for the stack than they do for the heap, which makes the program slightly more susceptible to memory exhaustion (through stack overflow): this may be seen as a good or a bad thing - one of the reasons the stack doesn't automatically grow the way heap does is to prevent out-of-control programs from having as much adverse impact on the entire machine when used in a more local scope (such as a while or for loop) or in several scopes, the memory accumulates per iteration/scope and is not released until the function exits: this contrasts with normal variables defined in the scope of a control structure (e.g. for {int i = 0; i < 2; ++i) { X } would accumulate alloca-ed memory requested at X, but memory for a fixed-sized array would be recycled per iteration). modern compilers typically do not inline functions that call alloca, but if you force them then the alloca will happen in the callers' context (i.e. the stack won't be released until the caller returns) a long time ago alloca transitioned from a non-portable feature/hack to a Standardised extension, but some negative perception may persist the lifetime is bound to the function scope, which may or may not suit the programmer better than malloc's explicit control having to use malloc encourages thinking about the deallocation - if that's managed through a wrapper function (e.g. WonderfulObject_DestructorFree(ptr)), then the function provides a point for implementation clean up operations (like closing file descriptors, freeing internal pointers or doing some logging) without explicit changes to client code: sometimes it's a nice model to adopt consistently in this pseudo-OO style of programming, it's natural to want something like WonderfulObject* p = WonderfulObject_AllocConstructor(); - that's possible when the "constructor" is a function returning malloc-ed memory (as the memory remains allocated after the function returns the value to be stored in p), but not if the "constructor" uses alloca a macro version of WonderfulObject_AllocConstructor could achieve this, but "macros are evil" in that they can conflict with each other and non-macro code and create unintended substitutions and consequent difficult-to-diagnose problems missing free operations can be detected by ValGrind, Purify etc. but missing "destructor" calls can't always be detected at all - one very tenuous benefit in terms of enforcement of intended usage; some alloca() implementations (such as GCC's) use an inlined macro for alloca(), so runtime substitution of a memory-usage diagnostic library isn't possible the way it is for malloc/realloc/free (e.g. electric fence) some implementations have subtle issues: for example, from the Linux manpage:

在许多系统中，alloca()不能在函数调用的参数列表中使用，因为由alloca()保留的堆栈空间将出现在堆栈中用于函数参数的空间中间。

我知道这个问题被标记为C，但作为一名c++程序员，我认为我应该使用c++来说明alloca的潜在效用:下面的代码(以及这里的ideone)创建了一个向量，跟踪不同大小的多态类型，这些类型是堆栈分配的(生命期与函数返回绑定)，而不是堆分配的。

#include <alloca.h>
#include <iostream>
#include <vector>

struct Base
{
    virtual ~Base() { }
    virtual int to_int() const = 0;
};

struct Integer : Base
{
    Integer(int n) : n_(n) { }
    int to_int() const { return n_; }
    int n_;
};

struct Double : Base
{
    Double(double n) : n_(n) { }
    int to_int() const { return -n_; }
    double n_;
};

inline Base* factory(double d) __attribute__((always_inline));

inline Base* factory(double d)
{
    if ((double)(int)d != d)
        return new (alloca(sizeof(Double))) Double(d);
    else
        return new (alloca(sizeof(Integer))) Integer(d);
}

int main()
{
    std::vector<Base*> numbers;
    numbers.push_back(factory(29.3));
    numbers.push_back(factory(29));
    numbers.push_back(factory(7.1));
    numbers.push_back(factory(2));
    numbers.push_back(factory(231.0));
    for (std::vector<Base*>::const_iterator i = numbers.begin();
         i != numbers.end(); ++i)
    {
        std::cout << *i << ' ' << (*i)->to_int() << '\n';
        (*i)->~Base();   // optionally / else Undefined Behaviour iff the
                         // program depends on side effects of destructor
    }
}

原因如下:

char x;
char *y=malloc(1);
char *z=alloca(&x-y);
*z = 1;

并不是说任何人都可以编写这段代码，但是您传递给alloca的size参数几乎肯定来自某种输入，它可能恶意地目的是让您的程序分配一个像这样巨大的值。毕竟，如果大小不是基于输入，或者不可能很大，为什么不声明一个小的、固定大小的本地缓冲区呢?

几乎所有使用alloca和/或C99 vlas的代码都有严重的错误，这些错误会导致崩溃(如果你幸运的话)或特权损害(如果你不那么幸运的话)。

alloca并不比变长数组(VLA)更糟糕，但它比在堆上分配更危险。

在x86上(最常见的是在ARM上)，堆栈向下增长，这带来了一定的风险:如果你不小心写超出了用alloca分配的块(例如由于缓冲区溢出)，那么你将覆盖你的函数的返回地址，因为它位于堆栈的“上面”，即在你分配的块之后。

这样做的后果是双重的:

程序将崩溃的壮观，它将不可能告诉为什么或哪里崩溃(堆栈将最有可能unwind到一个随机地址，由于覆盖的帧指针)。 它使缓冲区溢出的危险增加了许多倍，因为恶意用户可以制作一个特殊的有效负载，将其放在堆栈上，因此最终可以执行。

相反，如果你在堆上写超过一个块，你“只是”得到堆损坏。程序可能会意外终止，但会正确地展开堆栈，从而减少恶意代码执行的机会。

alloca的一个缺陷是longjmp将它倒带。

也就是说，如果你用setjmp保存一个上下文，然后分配一些内存，然后longjmp到上下文，你可能会失去分配的内存。堆栈指针回到原来的位置，因此内存不再保留;如果你调用一个函数或执行另一个分配，你将破坏原来的分配。

为了澄清，我在这里特别提到的是一种情况，即longjmp不返回发生分配的函数!相反，函数使用setjmp保存上下文;然后使用alloca分配内存，最后在该上下文中执行longjmp。该函数的分配内存没有全部释放;就是从setjmp开始分配的所有内存。当然，我说的是观察到的行为;据我所知，任何分配都没有这样的要求。

The focus in the documentation is usually on the concept that alloca memory is associated with a function activation, not with any block; that multiple invocations of alloca just grab more stack memory which is all released when the function terminates. Not so; the memory is actually associated with the procedure context. When the context is restored with longjmp, so is the prior alloca state. It's a consequence of the stack pointer register itself being used for allocation, and also (necessarily) saved and restored in the jmp_buf.

顺便说一句，如果这样工作的话，这提供了一种合理的机制来故意释放使用alloca分配的内存。

我曾经遇到过这种情况，这是一个bug的根本原因。

进程只有有限的堆栈空间可用——远远小于malloc()可用的内存量。

通过使用alloca()，您将极大地增加获得Stack Overflow错误的机会(如果幸运的话，或者如果运气不好，则会出现莫名其妙的崩溃)。