C 内存分配的调试断言错误

当这段代码编译时，我得到一个“调试断言错误”，表达式为_CrtIsValidHeapPointer（block），有人能解释为什么会发生这种情况，以及如何解决吗

typedef struct _POOL
{
    int size;
    void* memory;

} Pool;

Pool * allocatePool(int n)
{

    int *p = (int*)malloc(sizeof(int) * n);

    Pool pool = { sizeof(*p), *p };

    return &pool;

}

void freePool(Pool* pool)
{
    free(pool);
}

你需要这个：

Pool * allocatePool(int n)
{
    int *p = malloc(sizeof(int) * n);
    Pool *pool = malloc(sizeof(Pool));
    pool->memory = p;
    pool->size = n;
    return pool;
}

---

顺便说一句：

malloc

的类型转换是不必要的。

您的方法中有两个错误：

返回指向具有自动存储的变量的指针

在不指向动态内存的指针上调用

free（）

这意味着您还应该为

池

对象动态分配空间。在

freePool

中，必须释放

void*memory

指向的内存和

Pool*

本身

Pool *allocatePool(int n)
{
    Pool *pool = malloc(sizeof *pool);

    if (!pool) {
        /* malloc failed, bail out */
        return NULL;
    }

    /* Caller is responsible for passing non negative value as n.
       Negative values will have suprising effects. */
    int *p = malloc(sizeof *p * n);

    /* Failed malloc, or n was 0 and current implementation returns
       NULL. */
    if (!p) {
        /* Clean up already allocated memory */
        free(pool);
        /* Many dislike multiple return points in a function, but
           I think for this little example it's ok. That, or gotos. */
        return NULL;
    }

    /* Storing the size of single item seems dubious. Size of the whole
       allocated area in objects is much more useful. */
    pool->size = n;
    pool->memory = p;
    return pool;    
}

及

这两个功能

Pool * allocatePool(int n)
{

    int *p = (int*)malloc(sizeof(int) * n);

    Pool pool = { sizeof(*p), *p };

    return &pool;

}

void freePool(Pool* pool)
{
    free(pool);
}

没有道理

例如，第一个函数返回指向

Pool

类型的本地对象的指针，该对象在退出函数后将不活动。此外，此初始化无效

Pool pool = { sizeof(*p), *p };

我想你的意思至少是

Pool pool = { n, p };

或许

Pool pool = { n * sizeof( int ), p };

这些函数可以按以下方式显示

Pool allocatePool( int n )
{
    Pool pool = { 0, NULL };
    void *p;

    if ( n > 0 && ( p = malloc( n * sizeof( int ) ) ) != NULL )
    {
        pool.size = n;
        pool.memory = p;
    }

    return pool;
}

void freePool( Pool *pool )
{
    free( pool->memory );
    pool->size = 0;
    pool->memory = NULL;
}

---

戴维他们需要的是C++，而不是C。

Pool allocatePool( int n )
{
    Pool pool = { 0, NULL };
    void *p;

    if ( n > 0 && ( p = malloc( n * sizeof( int ) ) ) != NULL )
    {
        pool.size = n;
        pool.memory = p;
    }

    return pool;
}

void freePool( Pool *pool )
{
    free( pool->memory );
    pool->size = 0;
    pool->memory = NULL;
}