提交 167bfa71 编写于 作者: A Andy Shevchenko 提交者: David S. Miller

ppp: reuse print_hex_dump_bytes

There is a native function to dump hex buffers.
Signed-off-by: NAndy Shevchenko <andriy.shevchenko@linux.intel.com>
Signed-off-by: NDavid S. Miller <davem@davemloft.net>
上级 5a048e3b
......@@ -105,64 +105,15 @@ static const struct ppp_channel_ops sync_ops = {
};
/*
* Utility procedures to print a buffer in hex/ascii
* Utility procedure to print a buffer in hex/ascii
*/
static void
ppp_print_hex (register __u8 * out, const __u8 * in, int count)
{
register __u8 next_ch;
static const char hex[] = "0123456789ABCDEF";
while (count-- > 0) {
next_ch = *in++;
*out++ = hex[(next_ch >> 4) & 0x0F];
*out++ = hex[next_ch & 0x0F];
++out;
}
}
static void
ppp_print_char (register __u8 * out, const __u8 * in, int count)
{
register __u8 next_ch;
while (count-- > 0) {
next_ch = *in++;
if (next_ch < 0x20 || next_ch > 0x7e)
*out++ = '.';
else {
*out++ = next_ch;
if (next_ch == '%') /* printk/syslogd has a bug !! */
*out++ = '%';
}
}
*out = '\0';
}
static void
ppp_print_buffer (const char *name, const __u8 *buf, int count)
{
__u8 line[44];
if (name != NULL)
printk(KERN_DEBUG "ppp_synctty: %s, count = %d\n", name, count);
while (count > 8) {
memset (line, 32, 44);
ppp_print_hex (line, buf, 8);
ppp_print_char (&line[8 * 3], buf, 8);
printk(KERN_DEBUG "%s\n", line);
count -= 8;
buf += 8;
}
if (count > 0) {
memset (line, 32, 44);
ppp_print_hex (line, buf, count);
ppp_print_char (&line[8 * 3], buf, count);
printk(KERN_DEBUG "%s\n", line);
}
print_hex_dump_bytes("", DUMP_PREFIX_NONE, buf, count);
}
......
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