您可以看到lscpu
命令的输出-
jack@042:~$ lscpu
Architecture: x86_64
CPU op-mode(s): 32-bit, 64-bit
Byte Order: Little Endian
CPU(s): 56
On-line CPU(s) list: 0-55
Thread(s) per core: 2
Core(s) per socket: 14
Socket(s): 2
NUMA node(s): 2
Vendor ID: GenuineIntel
CPU family: 6
Model: 79
Model name: Intel(R) Xeon(R) CPU E5-2690 v4 @ 2.60GHz
Stepping: 1
CPU MHz: 2600.000
CPU max MHz: 2600.0000
CPU min MHz: 1200.0000
BogoMIPS: 5201.37
Virtualization: VT-x
Hypervisor vendor: vertical
Virtualization type: full
L1d cache: 32K
L1i cache: 32K …
Run Code Online (Sandbox Code Playgroud) 我的Intel Core2 Duo T9400M当前的 CPU 频率 ( CPU MHz)在高负载时怎么会高于最大 MHz ?
\n\n\xe2\x9e\x9c lscpu\n[...]\nModel name: Intel(R) Core(TM)2 Duo CPU T9400M @ 2.53GHz\nStepping: 10\nCPU MHz: 2606.581\nCPU max MHz: 2534.0000\nCPU min MHz: 800.0000\n[...]\n
Run Code Online (Sandbox Code Playgroud)\n\n这不限于 lscpu:我从/proc/cpuinfo中得到类似的值:
\n\n\xe2\x9e\x9c bat /proc/cpuinfo\n[...]\nmodel name : Intel(R) Core(TM)2 Duo CPU T9400 @ 2.53GHz\ncpu MHz : 2635.237\n[...]\n
Run Code Online (Sandbox Code Playgroud)\n\n我在查看 Node.js 文档时发现了这一点,并发现os.cpus()的当前速度值(即使在文档的示例中)也高于根据模型的最大 CPU 速度:
\n\n[\n {\n model: \'Intel(R) Core(TM) i7 CPU 860 @ 2.80GHz\',\n speed: 2926,\n[...]\n
Run Code Online (Sandbox Code Playgroud)\n 在我的 Linux 机器上,我有 32 个 vcore。
从lscpu
,我可以看到相同的。
对于我的 CPU,“每核线程”是 2。
那么这是否意味着我实际上有 64 个 vcores?
我刚刚检查了我们一台服务器上的 CPU 信息,我有点困惑。英特尔说这个模型有 28 个内核,而lscpu
显示只有 2 个。我错过了什么吗?虚拟化?
Architecture: x86_64
CPU op-mode(s): 32-bit, 64-bit
Byte Order: Little Endian
CPU(s): 2
On-line CPU(s) list: 0,1
Thread(s) per core: 1
Core(s) per socket: 2
Socket(s): 1
NUMA node(s): 1
Vendor ID: GenuineIntel
CPU family: 6
Model: 79
Model name: Intel(R) Xeon(R) Platinum 8180 CPU @ 2.50GHz
Stepping: 0
CPU MHz: 2494.140
BogoMIPS: 4988.28
Hypervisor vendor: VMware (<--- ??)
Virtualization type: full
L1d cache: 32K
L1i cache: 32K
L2 cache: …
Run Code Online (Sandbox Code Playgroud)