Ruby: puts printing stderr of the result of a command

Question

It's not clear to me how "puts" works in Ruby for the output of a command that throws its output to stderr.

Look at this code:

command="/usr/bin/java -version"
result=`#{command}`

puts result
puts "XX#{result}XX"

The result is:

java version "1.8.0_25"
Java(TM) SE Runtime Environment (build 1.8.0_25-b17)
Java HotSpot(TM) 64-Bit Server VM (build 25.25-b02, mixed mode)

XXXX

java -version sends its output to stderr (I am aware that in order to prevent this I should have used 2>&1 at the end of the command)

My question: The variable "result" in reality is empty, but the first puts prints the stderr output and the second puts the stdout (which is empty). Why? What's going on?

Backticks still run the command and print its output. The *capture*, the return of the backticks, is stdout. — Dave Newton, Aug 14 '15 at 00:25

score 2 · Answer 1 · answered Aug 14 '15 at 00:24

Actually, both puts statements print nothing. If you do this:

puts "YY#{result}YY"
puts "XX#{result}XX"

You'll see that you get YYYY and XXXX.

The reason that you see the output from Java is because Java itself is printing to stderr which isn't being captured by Ruby.

score 0 · Answer 2 · answered Aug 14 '15 at 00:22

0

The first puts result prints only newline. The output come from the following line (by the command itself, printing to stderr)

result=`#{command}`

You can confirm that by removing all puts ... lines.

answered Aug 14 '15 at 00:22

falsetru

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Duh! I was fooled by myself! Thanks! – Mario R. Aug 14 '15 at 02:19

Ruby: puts printing stderr of the result of a command

2 Answers2