Does const mean thread-safe in C++11? - c++

Question

I hear that const means thread-safe in C++11. Is that true?
Does that mean const is now the equivalent of Java's synchronized?
Are they running out of keywords?

Answer 1

I hear that const means thread-safe in C++11. Is that true?
It is somewhat true...
This is what the Standard Language has to say on thread-safety:
[1.10/4]
Two expression evaluations conflict if one of them modifies a memory location (1.7) and the other one accesses or modifies the same memory location.
[1.10/21]
The execution of a program contains a data race if it contains two conflicting actions in different threads, at least one of which is not atomic, and neither happens before the other. Any such data race results in undefined behavior.
which is nothing else than the sufficient condition for a data race to occur:
There are two or more actions being performed at the same time on a given thing; and
At least one of them is a write.
The Standard Library builds on that, going a bit further:
[17.6.5.9/1]
This section specifies requirements that implementations shall meet to prevent data races (1.10). Every standard library function shall meet each requirement unless otherwise specified. Implementations may prevent data races in cases other than those specified below.
[17.6.5.9/3]
A C++ standard library function shall not directly or indirectly modify objects (1.10) accessible by threads other than the current thread unless the objects are accessed directly or indirectly via the function’s non-const arguments, including this.
which in simple words says that it expects operations on const objects to be thread-safe. This means that the Standard Library won't introduce a data race as long as operations on const objects of your own types either
Consist entirely of reads --that is, there are no writes--; or
Internally synchronizes writes.
If this expectation does not hold for one of your types, then using it directly or indirectly together with any component of the Standard Library may result in a data race. In conclusion, const does mean thread-safe from the Standard Library point of view. It is important to note that this is merely a contract and it won't be enforced by the compiler, if you break it you get undefined behavior and you are on your own. Whether const is present or not will not affect code generation --at least not in respect to data races--.
Does that mean const is now the equivalent of Java's synchronized?
No. Not at all...
Consider the following overly simplified class representing a rectangle:
class rect {
int width = 0, height = 0;
public:
/*...*/
void set_size( int new_width, int new_height ) {
width = new_width;
height = new_height;
}
int area() const {
return width * height;
}
};
The member-function area is thread-safe; not because its const, but because it consist entirely of read operations. There are no writes involved, and at least one write involved is necessary for a data race to occur. That means that you can call area from as many threads as you want and you will get correct results all the time.
Note that this doesn't mean that rect is thread-safe. In fact, its easy to see how if a call to area were to happen at the same time that a call to set_size on a given rect, then area could end up computing its result based on an old width and a new height (or even on garbled values).
But that is alright, rect isn't const so its not even expected to be thread-safe after all. An object declared const rect, on the other hand, would be thread-safe since no writes are possible (and if you are considering const_cast-ing something originally declared const then you get undefined-behavior and that's it).
So what does it mean then?
Let's assume --for the sake of argument-- that multiplication operations are extremely costly and we better avoid them when possible. We could compute the area only if it is requested, and then cache it in case it is requested again in the future:
class rect {
int width = 0, height = 0;
mutable int cached_area = 0;
mutable bool cached_area_valid = true;
public:
/*...*/
void set_size( int new_width, int new_height ) {
cached_area_valid = ( width == new_width && height == new_height );
width = new_width;
height = new_height;
}
int area() const {
if( !cached_area_valid ) {
cached_area = width;
cached_area *= height;
cached_area_valid = true;
}
return cached_area;
}
};
[If this example seems too artificial, you could mentally replace int by a very large dynamically allocated integer which is inherently non thread-safe and for which multiplications are extremely costly.]
The member-function area is no longer thread-safe, it is doing writes now and is not internally synchronized. Is it a problem? The call to area may happen as part of a copy-constructor of another object, such constructor could have been called by some operation on a standard container, and at that point the standard library expects this operation to behave as a read in regard to data races. But we are doing writes!
As soon as we put a rect in a standard container --directly or indirectly-- we are entering a contract with the Standard Library. To keep doing writes in a const function while still honoring that contract, we need to internally synchronize those writes:
class rect {
int width = 0, height = 0;
mutable std::mutex cache_mutex;
mutable int cached_area = 0;
mutable bool cached_area_valid = true;
public:
/*...*/
void set_size( int new_width, int new_height ) {
if( new_width != width || new_height != height )
{
std::lock_guard< std::mutex > guard( cache_mutex );
cached_area_valid = false;
}
width = new_width;
height = new_height;
}
int area() const {
std::lock_guard< std::mutex > guard( cache_mutex );
if( !cached_area_valid ) {
cached_area = width;
cached_area *= height;
cached_area_valid = true;
}
return cached_area;
}
};
Note that we made the area function thread-safe, but the rect still isn't thread-safe. A call to area happening at the same time that a call to set_size may still end up computing the wrong value, since the assignments to width and height are not protected by the mutex.
If we really wanted a thread-safe rect, we would use a synchronization primitive to protect the non-thread-safe rect.
Are they running out of keywords?
Yes, they are. They have been running out of keywords since day one.
Source: You don't know const and mutable - Herb Sutter

Answer 2

This is an addition to K-ballo's answer.
The term thread-safe is abused in this context. The correct wording is: a const function implies thread-safe bitwise const or internally synchronised, as stated by Herb Sutter (29:43) himself
It should be thread-safe to call a const function from multiple threads simultaneously, without calling a non-const function at the same time in another thread.
So, a const function should not (and will not most of the time) be really thread-safe, as it may read memory (without internal synchronisation) that could be changed by another non-const function. In general, this is not thread-safe as a data race occurs even if only one thread is writing (and another reading the data).
See also my answer to the related question What is the definition of a thread safe function according to the C++11 (Language/Library) Standard?.

Answer 3

No! Counterexample:
#include <memory>
#include <thread>
class C
{
std::shared_ptr<int> refs = std::make_shared<int>();
public:
C() = default;
C(C const &other) : refs(other.refs)
{ ++*this->refs; }
};
int main()
{
C const c;
std::thread t1([&]() { C const dummy(c); });
std::thread t2([&]() { C const dummy(c); });
}
The copy-constructor of C is perfectly legitimate, but it is not thread-safe despite C being const.