Why calling `free(malloc(8))`?

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The Objective-C runtime's hashtable2.mm file contains the following code:

static void bootstrap (void) {
    free(malloc(8));
    prototypes = ALLOCTABLE (DEFAULT_ZONE);
    prototypes->prototype = &protoPrototype; 
    prototypes->count = 1;
    prototypes->nbBuckets = 1; /* has to be 1 so that the right bucket is 0 */
    prototypes->buckets = ALLOCBUCKETS(DEFAULT_ZONE, 1);
    prototypes->info = NULL;
    ((HashBucket *) prototypes->buckets)[0].count = 1;
    ((HashBucket *) prototypes->buckets)[0].elements.one = &protoPrototype;
};

Why does it allocate and release the 8-bytes space immediately?

Another source of confusion is this method from objc-os.h:

static __inline void *malloc_zone_malloc(malloc_zone_t z, size_t size) { return malloc(size); }

While it uses only one parameter, does the signature ask for two?

malloc objective-c c
1 Answers

For the first question I can only assume. My bet it was done to avoid/reduce memory churn, or segment the memory for some other reason. You can briefly find where it's discussed in the Changelog of bmalloc (which is not quite relevant, but i could not find a better reference):

2017-06-02 Geoffrey Garen <ggaren@apple.com>

...

Updated for new APIs. Note that we cache one free chunk per page class. This avoids churn in the large allocator when you free(malloc(X))

It's unclear however, if the memory churn is caused by this technique or it was supposed to address it.

For the second question, Objective-C runtime used to work with "zones" in order to destroy all allocated variables by just destroying the said zone, but it proved being error prone and later it was agreed to not use it anymore. The API, however still uses it for historical reasons (backward compatibility, i assume), but says that zones are ignored:

Zones are ignored on iOS and 64-bit runtime on OS X. You should not use zones in current development.

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