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lib/vsprintf: no_hash_pointers prints all addresses as unhashed
If the no_hash_pointers command line parameter is set, then printk("%p") will print pointers as unhashed, which is useful for debugging purposes. This change applies to any function that uses vsprintf, such as print_hex_dump() and seq_buf_printf(). A large warning message is displayed if this option is enabled. Unhashed pointers expose kernel addresses, which can be a security risk. Also update test_printf to skip the hashed pointer tests if the command-line option is set. Signed-off-by: Timur Tabi <timur@kernel.org> Acked-by: Petr Mladek <pmladek@suse.com> Acked-by: Randy Dunlap <rdunlap@infradead.org> Acked-by: Sergey Senozhatsky <sergey.senozhatsky@gmail.com> Acked-by: Vlastimil Babka <vbabka@suse.cz> Acked-by: Marco Elver <elver@google.com> Signed-off-by: Petr Mladek <pmladek@suse.com> Link: https://lore.kernel.org/r/20210214161348.369023-4-timur@kernel.org
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@@ -35,6 +35,8 @@ KSTM_MODULE_GLOBALS();
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static char *test_buffer __initdata;
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static char *alloced_buffer __initdata;
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extern bool no_hash_pointers;
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static int __printf(4, 0) __init
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do_test(int bufsize, const char *expect, int elen,
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const char *fmt, va_list ap)
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@@ -301,6 +303,12 @@ plain(void)
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{
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int err;
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if (no_hash_pointers) {
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pr_warn("skipping plain 'p' tests");
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skipped_tests += 2;
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return;
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}
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err = plain_hash();
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if (err) {
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pr_warn("plain 'p' does not appear to be hashed\n");
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