ZooKeeper使用ACL进行权限控制C++

C++中对zookeeper节点进行ACL权限控制资料少，让人误解。zhandle_t *zk = zookeeper_init（"127.0.0.1:5555", NULL, 10000, 0, NULL, 0）;
char buffer[512];
char scheme[10]="digest";
char usr_pa[100]="test:V28q/NynI4JI3Rk54h0r8O5kMug=";
struct ACL_vector aclv;
aclv.count=1;
struct ACL acl;
acl.perms = ZOO_PERM_ALL;
acl.id.scheme = scheme;
acl.id.id = usr_pa;
aclv.data = &acl;
zoo_create（zk, "/mynode", "cluster", strlen（"cluster"）, &aclv, 0, buffer, sizeof（buffer）-1）;zoo_add_auth（zk,"digest","test:test",7,NULL,NULL）;memset（buffer,0,sizeof（buffer））;
int buflen= sizeof（buffer）;
struct Stat stat;
int rc = zoo_get（zk, "/mynode", 0, buffer, &buflen, &stat）;
if （rc） {
fprintf（stderr, "zoo_get Error %d for %d ", rc, __LINE__）;
}
printf（"get buffer: %s ", buffer）;rc = zoo_delete（zk, "/mynode", -1）;
if （rc） {
fprintf（stderr, "zoo_delete Error %d for %d ", rc, __LINE__）;
}
sleep（1000000）;其中对于希望使用的用户名、密码分别为user,passwd时： ACL的id中保存的字符串为user:base64encode（sha1（user:passwd））其中编码为：单字符8位，字符集为ASCII 在java中有DigestAuthenticationProvider.java模块插件可以进行此加密：对应加密函数static public String generateDigest（String idPassword）
throws NoSuchAlgorithmException {
String parts[] = idPassword.split（":", 2）;
byte digest[] = MessageDigest.getInstance（"SHA1"）.digest（
idPassword.getBytes（））;
return parts[0] + ":" + base64Encode（digest）;
}在zkCli.sh命令行中：创建节点create -s /test null digest:test:V28q/NynI4JI3Rk54h0r8O5kMug=:rwcda 让zookeeper知道自己的权限addauth digest test:test 获取节点数据get /testZooKeeper 的详细介绍：请点这里
ZooKeeper 的下载地址：请点这里相关阅读：ZooKeeper集群配置 http://www.linuxidc.com/Linux/2013-06/86348.htm使用ZooKeeper实现分布式共享锁 http://www.linuxidc.com/Linux/2013-06/85550.htm分布式服务框架 ZooKeeper -- 管理分布式环境中的数据 http://www.linuxidc.com/Linux/2013-06/85549.htmZooKeeper集群环境搭建实践 http://www.linuxidc.com/Linux/2013-04/83562.htmZooKeeper服务器集群环境配置实测 http://www.linuxidc.com/Linux/2013-04/83559.htmZooKeeper集群安装 http://www.linuxidc.com/Linux/2012-10/72906.htm