xjdrew / crab Goto Github PK

// construct dictinory tree
static void
_dict_close(Table *t) {
if(t == NULL) {
return;
}
int i = 0;
for(i=0; icapacity; i++) {
TableNode *node = t->node + i;
if(node->flag != 0) {
_dict_close(node->value);
}
}
free(t->node);
free(t);
}

根据rfc3629，每个字符只有一种有效的utf8编码，0只应该编码成0，不能是0xc080;

ref:http://tools.ietf.org/html/rfc3629

1. Determine the number of octets required from the character number
   and the first column of the table above. It is important to note
   that the rows of the table are mutually exclusive, i.e., there is
   only one valid way to encode a given character.