[Solved] Consider this pseudocode for two threads incrementing a shared counter without locks: Thread 1: temp = counter temp = temp + 1 counter = temp Thread 2:... | LLD

LLD - Design — Parking Lot System

Consider this pseudocode for two threads incrementing a shared counter without locks:

Thread 1: temp = counter
          temp = temp + 1
          counter = temp

Thread 2: temp = counter
          temp = temp + 1
          counter = temp

What is the possible final value of counter if it starts at 0?

A2

BAny negative number

C1

D0

Step-by-Step Solution

Solution:

Step 1: Analyze concurrent increments without locks
Both threads read the same initial value 0, increment it to 1, and write back 1, causing one increment to be lost.
Step 2: Determine final counter value
Because of race condition, the counter may only increase once, resulting in final value 1 instead of 2.
Final Answer:
1 -> Option C
Quick Check:
Race condition causes lost update = 1 [OK]

Quick Trick: Without locks, increments can overwrite each other [OK]

Common Mistakes:

MISTAKES

Assuming both increments always succeed
Ignoring race conditions
Thinking counter can be negative here

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Consider this pseudocode for two threads incrementing a shared counter without locks:

Step 1: Analyze concurrent increments without locks

Step 2: Determine final counter value

Final Answer:

Quick Check:

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