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x86/efi: Remove the always true EFI_DEBUG symbol
This symbol is always set which makes it useless. Additionally we have a kernel command-line switch, efi=debug, which actually controls the printing of the memory map. Reported-by: Robert Elliott <elliott@hpe.com> Signed-off-by: Matt Fleming <matt@codeblueprint.co.uk> Acked-by: Borislav Petkov <bp@suse.de> Cc: Ard Biesheuvel <ard.biesheuvel@linaro.org> Cc: Borislav Petkov <bp@alien8.de> Cc: Peter Zijlstra <peterz@infradead.org> Cc: Thomas Gleixner <tglx@linutronix.de> Cc: linux-efi@vger.kernel.org Link: http://lkml.kernel.org/r/1461614832-17633-16-git-send-email-matt@codeblueprint.co.uk Signed-off-by: Ingo Molnar <mingo@kernel.org>
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@ -54,8 +54,6 @@
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#include <asm/rtc.h>
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#include <asm/uv/uv.h>
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#define EFI_DEBUG
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static struct efi efi_phys __initdata;
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static efi_system_table_t efi_systab __initdata;
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@ -222,7 +220,6 @@ int __init efi_memblock_x86_reserve_range(void)
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void __init efi_print_memmap(void)
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{
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#ifdef EFI_DEBUG
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efi_memory_desc_t *md;
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int i = 0;
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@ -235,7 +232,6 @@ void __init efi_print_memmap(void)
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md->phys_addr + (md->num_pages << EFI_PAGE_SHIFT) - 1,
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(md->num_pages >> (20 - EFI_PAGE_SHIFT)));
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}
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#endif /* EFI_DEBUG */
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}
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void __init efi_unmap_memmap(void)
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