Alloca()在堆栈上分配内存,而不是像malloc()那样在堆上分配内存。因此,当我从例程返回时,内存被释放。这实际上解决了动态分配内存的问题。通过malloc()释放分配的内存是一个非常令人头痛的问题,如果以某种方式错过释放会导致各种内存问题。

尽管有上述特性,为什么不鼓励使用alloca() ?


当前回答

老问题了,但是没有人提到它应该被可变长度数组取代。

char arr[size];

而不是

char *arr=alloca(size);

它存在于标准C99中,并作为编译器扩展存在于许多编译器中。

其他回答

alloca() is very useful if you can't use a standard local variable because its size would need to be determined at runtime and you can absolutely guarantee that the pointer you get from alloca() will NEVER be used after this function returns. You can be fairly safe if you do not return the pointer, or anything that contains it. do not store the pointer in any structure allocated on the heap do not let any other thread use the pointer The real danger comes from the chance that someone else will violate these conditions sometime later. With that in mind it's great for passing buffers to functions that format text into them :)

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
    }
}

其他答案都是正确的。但是,如果使用alloca()要分配的对象相当小,我认为这是一种比使用malloc()或其他方法更快、更方便的好技术。

换句话说,alloca(0x00ffffff)是危险的,可能会导致溢出,就像char hugeArray[0x00ffffff];是多少。小心谨慎,通情达理,你会没事的。

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)。

一个问题是,它不是标准的,尽管它得到了广泛的支持。在其他条件相同的情况下,我总是使用标准函数,而不是常用的编译器扩展。