如何使用__malloc_hook？

Question

在GNU C Library Reference Manual中,有一个示例程序(p.65),但我不知道这三个句子__malloc_hook = old_malloc_hook; old_malloc_hook = __malloc_hook; __malloc_hook = my_malloc_hook; 是什么 意思.特别是第二个,谁能为我解释一下？谢谢.

static void *
my_malloc_hook (size_t size, const void *caller)
{
    void *result;
    /* Restore all old hooks */
    __malloc_hook = old_malloc_hook;
    __free_hook = old_free_hook;
    /* Call recursively */
    result = malloc (size);
    /* Save underlying hooks */
    old_malloc_hook = __malloc_hook;
    old_free_hook = __free_hook;
    /* printf might call malloc, so protect it too. */
    printf ("malloc (%u) returns %p\n", (unsigned int) size, result);
    /* Restore our own hooks */
    __malloc_hook = my_malloc_hook;
    __free_hook = my_free_hook;
    return result;
}

我写了一个小程序来测试它:

#include <stdio.h>
#include <malloc.h>

/* Prototypes for our hooks.  */
static void my_init_hook(void);
static void *my_malloc_hook(size_t, const void *);

/* Variables to save original hooks. */
static void *(*old_malloc_hook) (size_t, const void *);

/* Override initializing hook from the C library. */
void (*__malloc_initialize_hook) (void) = my_init_hook;

static void my_init_hook(void)
{
    old_malloc_hook = __malloc_hook;
    __malloc_hook = my_malloc_hook;
}

static void *my_malloc_hook(size_t size, const void *caller)
{
    void *result;

    /* Restore all old hooks */
    __malloc_hook = old_malloc_hook;

        printf("1: __malloc_hook = %x  old_malloc_hook = %x\n", __malloc_hook, old_malloc_hook);
    /* Call recursively */
    result = malloc(size);

        printf("2: __malloc_hook = %x  old_malloc_hook = %x\n", __malloc_hook, old_malloc_hook);


    /* Save underlying hooks */
    old_malloc_hook = __malloc_hook;

        printf("3: __malloc_hook = %x  old_malloc_hook = %x\n", __malloc_hook, old_malloc_hook);
    /* printf() might call malloc(), so protect it too. */
    printf("malloc(%u) called from %p returns %p\n",
           (unsigned int)size, caller, result);

    /* Restore our own hooks */
    __malloc_hook = my_malloc_hook;
        printf("4: __malloc_hook = %x  old_malloc_hook = %x\n", __malloc_hook, old_malloc_hook);

    return result;
}

int main(void)
{
        char *p;
        p = malloc(10);
        free(p);
        return 0;
}

该计划的结果是:

1: __malloc_hook = 0  old_malloc_hook = 0
2: __malloc_hook = 0  old_malloc_hook = 0
3: __malloc_hook = 0  old_malloc_hook = 0
malloc(10) called from 0xb7797f38 returns 0x932c008
4: __malloc_hook = 804849d  old_malloc_hook = 0

但现在我有更多的问题,为什么old_malloc_hook都是0,1,2,3,为什么__malloc_hook是0？我真的很困惑.帮我.

Answer 1

据我所知,一切都按预期完成,输出正常.

变量,, __malloc_hook是0(或null)可能是因为系统的默认值是没有 malloc挂钩.

正如David Schwartz上面提到的那样,保存原件__malloc_hook非常重要,这样才能在malloc()调用原件之前恢复原件.这就是评论下方的界限/* Restore all old hooks */.我猜测在这种特殊情况下,它是不必要的,因为原来的malloc钩子是null,但为了安全起见应该这样做.

请放心,此代码正如您所希望的那样运行.现在,我只想让它炖一段时间,也许一个灯泡会熄灭,有一天,你会完全理解它.(对不起,但这是我今天能做的最好的事情.)