linux-0.11调试教程，update_line()函数分析，和update_line()函数的数据模型
如何由old变成new的？答案在最后！
例子一：
[/root]# c --> [/root]# ac 按下a键， length为5
具体过程为：[/root]# c --> [/root]# a --> [/root]# ac
static
update_line (old, new, current_line)
register char *old, *new;
int current_line;
{ ....
else
{
/* At the end of a line the characters do not have to
be "inserted". They can just be placed on the screen. */
output_some_chars (nfd, lendiff);
last_c_pos += lendiff;
}
/* Copy (new) chars to screen from first diff to last match. */
if (((nls - nfd) - lendiff) > 0)
{
output_some_chars (&nfd[lendiff], ((nls - nfd) - lendiff));
last_c_pos += ((nls - nfd) - lendiff);
}
....
}
output_some_chars (nfd, lendiff);时：
output_some_chars (&nfd[lendiff], ((nls - nfd) - lendiff));时：
可见nfd为0x59c15,lendiff为1，nls为0x59c17,ne为0x59c17
(ofd为0x63415，ols和oe都为0x63416)
/* if (len (new) > len (old)) */
lendiff = (nls - nfd) - (ols - ofd); /* lendiff = 2-1=1
例子2：下面的对应下面的图，退格键时的6个位置
else /* Delete characters from line. */
{
/* If possible and inexpensive to use terminal deletion, then do so. */
if (term_dc && (2 * (ne - nfd)) >= (-lendiff))
{
if (lendiff)
delete_chars (-lendiff); /* delete (diff) characters /*产生字符0x50315b1b*/
/* Copy (new) chars to screen from first diff to last match */
if ((nls - nfd) > 0)
{
output_some_chars (nfd, (nls - nfd)); /*产生字符0x63*/
last_c_pos += (nls - nfd);
}
}
[/root]# abc --> [/root]# ac 按下退格键，length为b
delete_chars (-lendiff);时
output_some_chars()时
lendiff = (nls - nfd) - (ols - ofd);
lendiff = 10 - 15 = -5;
先删除5个字符，delete_chars (-lendiff);/*产生字符0x50355b1b*/，会调用内核的csi_P()函数。
然后再打印出10个字符buggy say，把le-buggy i覆盖掉！！！
既调用output_some_chars (nfd, (nls - nfd));。
static void delete_char(int currcons)
{
int i;
unsigned short * p = (unsigned short *) pos;
if (x>=video_num_columns)
return;
i = x;
while (++i < video_num_columns) {
*p = *(p+1);
p++;
}
*p = video_erase_char;
}
static void csi_P(int currcons, unsigned int nr)
{
if (nr > video_num_columns)
nr = video_num_columns;
else if (!nr)
nr = 1;
while (nr--)
delete_char(currcons);
}