[PATCH] dmesg: print only 2 hex digits for each hex-escaped byte

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As buf is passed as a signed char buffer in fwrite_hex, fprintf will
print every byte from 0x80 as a signed-extended int causing each of
these bytes to be printed as "\xffffff80" and such, which can be pretty
confusing. Force fprintf to use the argument as a char to make it print
only 2 digits, e.g. "\x80".

Signed-off-by: Ivan Delalande <colona@xxxxxxxxxx>
---
 sys-utils/dmesg.c | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/sys-utils/dmesg.c b/sys-utils/dmesg.c
index b83cfb1bb..821d8bbb2 100644
--- a/sys-utils/dmesg.c
+++ b/sys-utils/dmesg.c
@@ -613,7 +613,7 @@ static int fwrite_hex(const char *buf, size_t size, FILE *out)
 	size_t i;
 
 	for (i = 0; i < size; i++) {
-		int rc = fprintf(out, "\\x%02x", buf[i]);
+		int rc = fprintf(out, "\\x%02hhx", buf[i]);
 		if (rc < 0)
 			return rc;
 	}
-- 
2.13.1
--
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