在 Java 中将八进制字符串设置为 0

Question

我正在使用以下 snmp set 命令将零值发送到硬件中。

snmpset -v 3 -u <Username> -a SHA -A "password" -x AES -X "password" -l AuthPriv udp6:[<ipv6>%bond0] <oid> x 0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

有 1024 个零，现在我想使用代码执行相同的操作。

但是我将如何获得带有给定值的八位字节字符串。

我尝试使用：

final List<Integer> list = new ArrayList<>();
list.add(0);
final OctetString externalVlanIdsBitMapOctetString = BitMapUtils.createLeftBitmap(list);

但是我得到 snmpset 失败，错误代码 = 5。

谁能帮我？

Answer 1

我不确定是字节 0，还是字符"0"的字节。后一个版本我注释掉了。

byte[] bytes = new byte[1024];
//Arrays.fill(bytes, (byte) '0');
final OctetString externalVlanIdsBitMapOctetString = new OctetString(bytes);

Answer 2

如果要

创建一个包含 1024 个零的字符串：

Sring zeros = String.format("%01024d", 0);