[Solved] What will this Ruby code output?count = 0 threads = 2.times.map do Thread.new do 5.times { count += 1 } end end threads.each(&:join) puts count | Ruby

Ruby - Concurrent Programming

What will this Ruby code output?

count = 0
threads = 2.times.map do
  Thread.new do
    5.times { count += 1 }
  end
end
threads.each(&:join)
puts count

A10

B5

CMay vary between 5 and 10

D0

Step-by-Step Solution

Solution:

Step 1: Analyze shared variable access
Two threads increment count 5 times each, but no lock means race conditions.
Step 2: Understand race condition effect
Count may not reach 10 because increments can overlap and overwrite each other.
Final Answer:
May vary between 5 and 10 -> Option C
Quick Check:
Race condition causes variable count to vary [OK]

Quick Trick: Shared variables need locks to avoid race conditions [OK]

Common Mistakes:

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