I've been studing 'Concurency in Action' position for some time and I have a problem with understanding following example of code (Listing 5.2):
#include <vector>
#include <atomic>
#include <iostream>
std::vector<int> data;
std::atomic<bool> data_ready(false);
void reader_thread()
{
while(!data_ready.load())
{
std::this_thread::sleep(std::milliseconds(1));
}
std::cout<<”The answer=”<<data[0]<<”\n”;
}
void writer_thread()
{
data.push_back(42); //write of data
data_ready=true; //write to data_ready flag
}
The book explaines:
(...) The write of the data happens-before the write to the data_ready flag (...)
My concern is that the sentence does not cover the out-of-order execution. From my understanding out of order execution may happen when at least two instruction do not have depended operands. Taking this into account:
data_ready=true
does not need anything from
data.push_back(42)
to be executed. As a result of that it is not guaranteed that:
The write of the data happens-before the write to the data_ready flag
Is my understadning correct or there is something in out-of-order execution that I don't understand causing misunderstaning of given example?
EDIT
Thank you for answers, it was helpful. My misunserstanding was a result of not knowing that atomic types not only prevents from partialy channing a variable, but also acts as memory barrier.
For example following code may be reordered in many combinations by either compiler or processor:
d=0;
b=5;
a=10
c=1;
Resulting with following order (one of many possibilities):
b=5;
a=10
c=1;
d=0;
It it is not a problem with single-thread code since none of expressions have depended operands on other, but on multithreaded application may result of undefined behaviour. For example following code (initial values: x=0 and y=0):
Thread 1: Thread 2:
x=10; while(y!=15);
y=15; assert(x==10);
Without reordering of code by compiler or reordering execution by processor we could say: "Since assigement y=15 allways happens after assigement x=10 and assert happens after while loop the assert will never fail" But it's not true. The real execution order may be as below (one of many possible combinations):
Thread 1: Thread 2:
x=10; (4) while(y!=15); (3)
y=15; (1) assert(x==10); (2)
By default an atomic variable ensures sequentionally consistent ordering. If y in example above was atomic with memory_order_seq_cst default parameter following sentences are true:
- what happens before in thread 1 (x=10) it is also visible in thread 2 as happening before.
- what happens after while(y!=15) in thread 2 it is also visible in thread 1 as happening after
As a result of it assert will never fail.
Some of sources that may help with understaning: