SNMP UCD dskTotalLow + dskTotalHigh

Question

我对我的一个linux机器做了snmpwalk，下面是输出。

UCD SNMP-MIB：：dskTotalLow.1= Gauge32: 276240608

UCD SNMP-MIB：：dskTotalLow.2= Gauge32: 0

UCD SNMP-MIB：：dskTotalLow.3= Gauge32: 0

UCD SNMP-MIB：：dskTotalLow.4= Gauge32: 0

UCD SNMP-MIB：：dskTotalLow.5= Gauge32: 74346640

UCD-SNMP-MIB::dskTotalLow.6 = Gauge32: 4031680

UCD SNMP-MIB：：dskTotalLow.7= Gauge32: 1677158400

UCD-SNMP-MIB::dskTotalLow.8 = Gauge32: 1071652864

UCD SNMP-MIB：：dskTotalLow.9= Gauge32: 0

UCD-SNMP-MIB::dskTotalLow.10 = Gauge32: 0

UCD SNMP-MIB：：dskTotalHigh.1= Gauge32: 0

UCD SNMP-MIB：：dskTotalHigh.2= Gauge32: 0

UCD SNMP-MIB：：dskTotalHigh.3= Gauge32: 0

UCD SNMP-MIB：：dskTotalHigh.4= Gauge32: 0

UCD SNMP-MIB：：dskTotalHigh.5= Gauge32: 0

UCD-SNMP-MIB::dskTotalHigh.6 = Gauge32: 0

UCD SNMP-MIB：：dskTotalHigh.7= Gauge32: 0

UCD-SNMP-MIB::dskTotalHigh.8 = Gauge32: 11

UCD SNMP-MIB：：dskTotalHigh.9= Gauge32: 0

UCD-SNMP-MIB::dskTotalHigh.10 = Gauge32: 0

磁盘索引8的总大小约为45Tb。在UCD MIB文档中

dskTotalLow：“磁盘/分区(kBytes)的总大小。与dskTotalHigh一起构成64位数字。”

dskTotalHigh：“磁盘/分区(kBytes)的总大小。与dskTotalLow一起构成64位数字。”

如何从dskTotalLow和dskTotalHigh中获取总大小。请提前帮我一下。

谢谢

Answer 1

对于索引8，得到高部分0x0B (= 11)，低部分0x3FE0 2000 (= 1,071,652,864)。所以把它们放在一起，我们得到0x0B 3FE0 2000 (= 48,316,293,120)。这与45 TB匹配。

或者，您可以通过计算获得该值

11 * 4,294,967,296 + 1,071,652,864 = 48,316,293,120

where 4,294,967,296 = 0x1 0000 0000