How do you find the physical memory free on the machine in Ruby?

I would like to know how much physical memory is available on the sys开发者_StackOverflowtem, excluding any swap. Is there a method to get this information in Ruby?

---

If you are using linux,You usually use a "free" command to find physical memory ie RAM details on the system

 output = %x(free)

output will look slightly like the following string

" total used free shared buffers cached\nMem: 251308 201500 49808 0 3456 48508\n-/+ buffers/cache: 149536 101772\nSwap: 524284 88612 435672\n"

You can extract the information you need using simple string manipulations like

output.split(" ")[7] will give total memory output.split(" ")[8] will give used memory output.split(" ")[9] will give free memory

---

Slightly slicker version of AndrewKS's answer:

total_memory_usage_in_k = `ps -Ao rss=`.split.map(&:to_i).inject(&:+)

---

Well, the Unix command "top" doesn't seem to work in Ruby, so try this:

# In Kilobytes
memory_usages = `ps -A -o rss=`.split("\n")
total_mem_usage = memory_usages.inject { |a, e| a.to_i + e.strip.to_i }

This "seems" correct. I don't guarantee it. Also, this takes a lot more time than the system will so by the time it's finished the physical memory would have changed.

---

You can use this gem to get various system info http://threez.github.com/ruby-vmstat/

---

This answers for both, Ruby and Bash and probably also for Python and the rest:

#!/usr/bin/env bash 
# This file is in public domain. 
# Initial author: martin.vahi@softf1.com

#export S_MEMINFO_FIELD="Inactive"; \
export S_MEMINFO_FIELD="MemTotal"; \
ruby -e "s=%x(cat /proc/meminfo | grep $S_MEMINFO_FIELD: | \
gawk '{gsub(/MemTotal:/,\"\");print}' | \
gawk '{gsub(/kB/,\"*1024\");print}' | \
gawk '{gsub(/KB/,\"*1024\");print}' | \
gawk '{gsub(/KiB/,\"*1024\");print}' | \
gawk '{gsub(/MB/,\"*1048576\");print}' | \
gawk '{gsub(/MiB/,\"*1048576\");print}' | \
gawk '{gsub(/GB/,\"*1073741824\");print}' | \
gawk '{gsub(/GiB/,\"*1073741824\");print}' | \
gawk '{gsub(/TB/,\"*1099511627776\");print}' | \
gawk '{gsub(/TiB/,\"*1099511627776\");print}' | \
gawk '{gsub(/B/,\"*1\");print}' | \
gawk '{gsub(/[^1234567890*]/,\"\");print}' \
); \
s_prod=s.gsub(/[\\s\\n\\r]/,\"\")+\"*1\";\
ar=s_prod.scan(/[\\d]+/);\
i_prod=1;\
ar.each{|s_x| i_prod=i_prod*s_x.to_i};\
print(i_prod.to_s+\" B\")"

The thing to notice here is that the "\" at the ends of the lines are Bash line continuations. Basically it should all be a single-liner, with the exception of the out-commented lines, which must be deleted. The colon at the end of the

grep $S_MEMINFO_FIELD:

is important, because

cat /proc/meminfo | grep Inactive

prints multiple lines, but the rest of the script works with an assumption that the grep outputs only a single line. According to

https://unix.stackexchange.com/questions/263881/convert-meminfo-kb-to-bytes

the /proc/meminfo uses 1024 regardless of whether the unit is "kB" or "KB". I did not have any info about the GiB and GB and TB parts, but I assume that they follow the same style and 1MB=1024*1024KiB.

I created a Linux specific Bash script that takes /proc/meminfo field name as its 1. command line argument and prints out the value in Bytes:

http://longterm.softf1.com/2016/comments/stackoverflow_com/2016_03_07_mmmv_proc_meminfo_filter_t1.bash

(archival copy: https://archive.is/vjcNf )

Thank You for reading my comment. I hope that it helps. :-)

---

You may use total gem (I'm the author):

require 'total'
puts Total::Mem.new.bytes