8d034f544c
/proc/meminfo reports memory in KiloBytes and so needs a multiplier of 1024 instead of 1000. The kernel reports in terms of pages and the proc filesystem is left shifting by 2 for 4KB pages to get KB. Since this is a binary shift, Bytes will need to shift by 10 and so get multiplied by 1024. From the kernel code. PAGE_SHIFT = 12 for 4KB pages "MemTotal: %8lu kB\n", K(i.totalram) Closes #131 |
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| .. | ||
| mem.go | ||
| mem_darwin.go | ||
| mem_freebsd.go | ||
| mem_linux.go | ||
| mem_test.go | ||
| mem_windows.go | ||