lib: list_sort_test(): simplify and harden cleanup
There is no reason to maintain the list structure while freeing the debug elements. Aside from the redundant pointer manipulations, it is also inefficient from a locality-of-reference viewpoint, since they are visited in a random order (wrt. the order they were allocated). Furthermore, if we jumped to exit: after detecting list corruption, it is actually dangerous. So just free the elements in the order they were allocated, using the backing array elts. Allocate that using kcalloc(), so that if allocation of one of the debug element fails, we just end up calling kfree(NULL) for the trailing elements. Minor details: Use sizeof(*elts) instead of sizeof(void *), and return err immediately when allocation of elts fails, to avoid introducing another label just before the final return statement. Signed-off-by: Rasmus Villemoes <linux@rasmusvillemoes.dk> Cc: Artem Bityutskiy <artem.bityutskiy@linux.intel.com> Cc: Don Mullis <don.mullis@gmail.com> Cc: Dave Chinner <david@fromorbit.com> Signed-off-by: Andrew Morton <akpm@linux-foundation.org> Signed-off-by: Linus Torvalds <torvalds@linux-foundation.org>
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@ -209,16 +209,16 @@ static int __init list_sort_test(void)
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{
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int i, count = 1, err = -ENOMEM;
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struct debug_el *el;
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struct list_head *cur, *tmp;
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struct list_head *cur;
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LIST_HEAD(head);
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printk(KERN_DEBUG "list_sort_test: start testing list_sort()\n");
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elts = kmalloc(sizeof(void *) * TEST_LIST_LEN, GFP_KERNEL);
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elts = kcalloc(TEST_LIST_LEN, sizeof(*elts), GFP_KERNEL);
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if (!elts) {
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printk(KERN_ERR "list_sort_test: error: cannot allocate "
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"memory\n");
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goto exit;
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return err;
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}
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for (i = 0; i < TEST_LIST_LEN; i++) {
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@ -286,11 +286,9 @@ static int __init list_sort_test(void)
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err = 0;
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exit:
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for (i = 0; i < TEST_LIST_LEN; i++)
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kfree(elts[i]);
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kfree(elts);
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list_for_each_safe(cur, tmp, &head) {
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list_del(cur);
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kfree(container_of(cur, struct debug_el, list));
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}
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return err;
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}
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module_init(list_sort_test);
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