Tutorial 2 QnA

Index

• Index
• Q1: synchronises-with if/while
• Optional Food for thought pop quiz: ambiguous sw

Q1: synchronises-with if/while

Tutorial 2, question 4.1. Just to clarify, if we remove the while loop and replaced it with an if-statement, it will also not synchronize with right?

The snippet in question is:

I’m assuming the question is asking if (b) sw (x) if we change while to if.

If we remove the while loop and replace it with an if-statement, you might get a synchronize-with, based on what (x) actually reads.

If y.load in (x) reads from (b) i.e. 2, then we have sw.
If y.load in (x) reads 0, then we do not have sw.

Why? See the definition of synchronizes-with:

If an atomic store in thread A is a release operation, an atomic load in thread B from the same variable is an acquire operation, and the load in thread B reads a value written by the store in thread A, then the store in thread A synchronizes-with the load in thread B.

For the tutorial examples, the while loop is a way to force a synchronizes-with relationship. In the example with the while, (x) may read 0, but it will always read (b) before breaking out of the spinlock. This guarantees us a sw relationship between (x) and (b).

Optional Food for thought pop quiz: ambiguous sw

while are often used to force sw between threads. But don’t blindly assume!

0
1
2
3
4
5
6
7

// Thread 1
x.store(2, stdmo::release); // (a)

// Thread 2
x.store(2, stdmo::release); // (b)

// Thread 3
while(x.load(stdmo::acquire) != 2); // (x)

^ the FFT here is tricky. I don’t think this sort of puzzles will come out during exams/quizzes. It’s just to verify your understanding of sw.

 0
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12

// Thread 1
x.store(1, stdmo::relaxed); // (a)

// Thread 2
y.store(2, stdmo::release); // (b)

// Thread 3 
z.store(3, stdmo::release); // (c)

// Thread 4
while(x.load(stdmo::acquire) != 1); // (p)
while(y.load(stdmo::relaxed) != 2); // (q)
while(z.load(stdmo::acquire) != 3); // (r)