我有高通RB5主板。
它是SOC规格表内容如下:
CPU: Kryo 585 CPU, 64-bit, up to 2.84 GHz
ISP: Qualcomm Spectra 480 ISP with Dual 14-bit image signal processing
GPU Adreno 650 GPU w/ support for Open GL ES & Open CL
lscpu 输出如下:
$ lscpu
Architecture: aarch64
CPU op-mode(s): 32-bit, 64-bit
Byte Order: Little Endian
CPU(s): 8
On-line CPU(s) list: 0-7
Thread(s) per core: 1
Core(s) per socket: 2
Socket(s): 3
Vendor ID: Qualcomm
Model: 14
Stepping: 0xd
CPU max MHz: 2841.6001
CPU min MHz: 300.0000
BogoMIPS: 38.40
Vulnerability Itlb multihit: Not affected
Vulnerability L1tf: Not affected
Vulnerability Mds: Not affected
Vulnerability Meltdown: Vulnerable
Vulnerability Mmio stale data: Not affected
Vulnerability Retbleed: Not affected
Vulnerability Spec store bypass: Mitigation; Speculative Store Bypass disabled v
ia prctl
Vulnerability Spectre v1: Mitigation; __user pointer sanitization
Vulnerability Spectre v2: Mitigation; CSV2, BHB
Vulnerability Srbds: Not affected
Vulnerability Tsx async abort: Not affected
Flags: fp asimd evtstrm aes pmull sha1 sha2 crc32 atom
ics fphp asimdhp cpuid asimdrdm lrcpc dcpop asi
mddp
我想了解以下内容的含义:
CPU(s): 8
On-line CPU(s) list: 0-7
Thread(s) per core: 1
Core(s) per socket: 2
Socket(s): 3
我知道这是 8 核处理器。但我有以下疑问:
- 那么
CPU(s): 8
,这是否意味着它有 8 个核心? - 插槽 * 每插槽核心数 = 3 * 2 = 6。为什么它与 8 不同?
- 这是否意味着逻辑核心的总数 = 3 * 2 * 8 = 48?
答案1
CPU 有 8 个内核(4 个“大”高性能内核,4 个“小”低功耗内核),编号为 0 到 7。所有内核都是单线程的,没有 SMT - 逻辑内核 = 物理内核。每插槽核心数和套接字值是无意义的,它只是一个插座。
我猜虚拟插槽是用来区分大核和小核的。为什么标示 3*2 = 6 个核而不是 8 个核,我实在搞不懂。可能有 2 个插槽,每个插槽有 2 个(大核+小核)核,还有 1 个插槽用于 DSP。