commands/i386/pc/sendkey: Fix "writing 1 byte into a region of size 0" build error
Latest GCC may complain in that way:
commands/i386/pc/sendkey.c: In function ‘grub_sendkey_postboot’:
commands/i386/pc/sendkey.c:223:21: error: writing 1 byte into a region of size 0 [-Werror=stringop-overflow=]
223 | *((char *) 0x41a) = 0x1e;
| ~~~~~~~~~~~~~~~~~~^~~~~~
The volatile keyword addition helps and additionally assures us the
compiler will not optimize out fixed assignments.
Signed-off-by: Daniel Kiper <daniel.kiper@oracle.com>
Reviewed-by: Robbie Harwood <rharwood@redhat.com>
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@ -220,8 +220,8 @@ grub_sendkey_postboot (void)
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*flags = oldflags;
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*((char *) 0x41a) = 0x1e;
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*((char *) 0x41c) = 0x1e;
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*((volatile char *) 0x41a) = 0x1e;
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*((volatile char *) 0x41c) = 0x1e;
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return GRUB_ERR_NONE;
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}
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@ -236,8 +236,8 @@ grub_sendkey_preboot (int noret __attribute__ ((unused)))
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oldflags = *flags;
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/* Set the sendkey. */
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*((char *) 0x41a) = 0x1e;
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*((char *) 0x41c) = keylen + 0x1e;
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*((volatile char *) 0x41a) = 0x1e;
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*((volatile char *) 0x41c) = keylen + 0x1e;
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grub_memcpy ((char *) 0x41e, sendkey, 0x20);
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/* Transform "any ctrl" to "right ctrl" flag. */
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