is (local) static initialization allowed to happen prematurley? - c++

Lets say foo is called once in a program, and
void foo()
{
if(sometimes_false())
{
static int xx = func_with_sideeffect();
}
}
the condition wasn't met, is the side-effect
1. Allowed to have happened
2. Mandated to have happened (I'm guessing not if my compiler is conforming)
3. Mandated to not have happened

The non-constant initialization of all variable with local scope and static storage duration is from the point it is encountered till the end of the program. So, if the variable is not encountered because of a condition, it would not be initialized and the side effect won't happen.
The following quote from the standard supports the answer (particularly the part in bold)
6.7 Declaration statement [stmt.dcl]
The zero-initialization (8.5) of all block-scope variables with static
storage duration (3.7.1) or thread storage duration (3.7.2) is
performed before any other initialization takes place. Constant
initialization (3.6.2) of ablock-scope entity with static storage
duration, if applicable, is performed before its block is first
entered.An implementation is permitted to perform early initialization
of other block-scope variables with static orthread storage duration
under the same conditions that an implementation is permitted to
statically initializea variable with static or thread storage duration
in namespace scope (3.6.2). Otherwise such a variable is initialized
the first time control passes through its declaration; such a
variable is considered initialized upon the completion of its
initialization. If the initialization exits by throwing an exception,
the initialization is not complete, so it will be tried again the next
time control enters the declaration. If control enters the declaration
concurrently while the variable is being initialized, the concurrent
execution shall wait for completion of the initialization.88 If control
re-enters the declaration recursively while the variable is
being initialized, the behavior is undefined.

The answer is 3 but you need to think about threading.
The static int xx will be initialised by the first thread encountering that variable, and C++11 will block all other threads at that point until func_with_sideeffect() returns and the result has been assigned to xx. (That's not the case with older standards: a mutex would have been required).
It's more difficult to predict the destruction of xx if it was an instance of an object with a non-trivial destructor.

Related

Understanding static initialization in standard 6.7/4

From the Standard 6.7/4:
The zero-initialization (8.5) of all block-scope variables with static
storage duration (3.7.1) or thread storage duration (3.7.2) is
performed before any other initialization takes place.
It's not clear if the zero-initialization's being performed during the program-startup or during the control first executes the block?
I will report a longer snippet. That's your secction
The zero-initialization (8.5) of all block-scope variables with static
storage duration (3.7.1) or thread storage duration (3.7.2) is
performed before any other initialization takes place. Constant
initialization (3.6.2) of a block-scope entity with static storage
duration, if applicable, is performed before its block is first
entered.
That's the follow up, and I understand that this means it can do both. That's the meaning I give to this section (the continuation of the same paragraph)
An implementation is permitted to perform early
initialization of other block-scope variables with static or thread
storage duration under the same conditions that an implementation is
permitted to statically initialize a variable with static or thread
storage duration in namespace scope (3.6.2).
All we can do now is wait a native english speaker to confirm/confute my legalese decryption...

Does C++11 core language takes care of Singleton Dead Reference?

I have recently read Andrei Alexandrescu's Modern C++ Design. After reading 6. chapter, I begin to worry about our singletons at company. Since our experienced team leader writes core helper libraries like singletons etc... . I asked him if the way he handles singleton takes care of on dead reference problem ? If he used at_exit function call which is given by C core language?
He told me C++11 has singleton support and will execute CTORs and DTORs in a row that they will not be any dead reference problem. User will not have to cope with synchronization.
Even it sounds awesome I couldn't find any information which confirms him on internet. So please tell me if C++11 takes care of Dead Reference Problem for singletons and if so please explain a little what dark magic going behind ?
Presumably your team leader is talking about singletons implemented as follows:
T &get_value() {
static T val;
return val;
}
In this case, the standard gives two guarantees. The first is that the val objects will be constructed exactly once, the first time that the flow of program execution passes the declaration of the local static variable, even if that happens simultaneously on several threads 6.7/4:
An implementation is permitted to perform early initialization of other block-scope variables with static or thread storage duration under the same conditions that an implementation is permitted to statically initialize a variable with static or thread storage duration in namespace scope (3.6.2). Otherwise such a variable is initialized the first time control passes through its declaration; such a variable is considered initialized upon the completion of its initialization. If the initialization exits by throwing an exception, the initialization is not complete, so it will be tried again the next time control enters the declaration. If control enters the declaration concurrently while the variable is being initialized, the concurrent execution shall wait for completion of the initialization.
Static initialisation is only allowed in the case of constants, so as long as T does not have a constexpr constructor you shouldn't have to worry (but read 3.6.2 for the full rules, in case there is some edge case that is relevant to your code).
The second guarantee is that all variables with static storage duration will be destructed in the reverse order of their construction 3.6.3/1:
Destructors (12.4) for initialized objects (that is, objects whose lifetime (3.8) has begun) with static storage duration are called as a result of returning from main and as a result of calling std::exit (18.5). Destructors for initialized objects with thread storage duration within a given thread are called as a result of returning from the initial function of that thread and as a result of that thread calling std::exit. The completions of the destructors for all initialized objects with thread storage duration within that thread are sequenced before the initiation of the destructors of any object with static storage duration. If the completion of the constructor or dynamic initialization of an object with thread storage duration is sequenced before that of another, the completion of the destructor of the second is sequenced before the initiation of the destructor of the first. If the completion of the constructor or dynamic initialization of an object with static storage duration is sequenced before that of another, the completion of the destructor of the second is sequenced before the initiation of the destructor of the first. [Note: This definition permits concurrent destruction. — end note ] If an object is initialized statically, the object is destroyed in the same order as if the object was dynamically initialized. For an object of array or class type, all subobjects of that object are destroyed before any block-scope object with static storage duration initialized during the construction of the subobjects is destroyed. If the destruction of an object with static or thread storage duration exits via an exception, std::terminate is called (15.5.1).
While this paragraph gives a lot of scope for concurrent destruction of static objects when their construction was concurrent, the main thing to take away from it is that destruction happens in the reverse order of construction.
Together, these mean means that if T val depends on some U val in another of these singleton functions, the U val will always be constructed before T val and destructed after the T val, so overall, this is a safe way of implementing singletons (unless you're doing something very crazy).

Thread safe singleton in C++ after gcc 4?

I heard that after some version of gcc using simply something like:
static A* a = new A();
return a;
Is thread-safe for a singleton and one wouldn't need something adapted from say http://locklessinc.com/articles/singleton_pattern/ anymore...
Does anyone have a specific reference or link to where I can read about this?
Section 6.7 of draft standard (n3337.pdf), point 4:
The zero-initialization (8.5) of all block-scope variables with static storage duration (3.7.1) or thread storage
duration (3.7.2) is performed before any other initialization takes place. Constant initialization (3.6.2) of a
block-scope entity with static storage duration, if applicable, is performed before its block is first entered.
An implementation is permitted to perform early initialization of other block-scope variables with static or
thread storage duration under the same conditions that an implementation is permitted to statically initialize
a variable with static or thread storage duration in namespace scope (3.6.2). Otherwise such a variable is
initialized the first time control passes through its declaration; such a variable is considered initialized upon
the completion of its initialization. If the initialization exits by throwing an exception, the initialization
is not complete, so it will be tried again the next time control enters the declaration. If control enters
the declaration concurrently while the variable is being initialized, the concurrent execution shall wait for
completion of the initialization.88 If control re-enters the declaration recursively while the variable is being
initialized, the behavior is undefined.
GCC follows the cross-vendor Itanium C++ ABI. The relevant sections covering thread-safe initialization of function-local statics are 2.8 Initialization guard variables and 3.3.2 One-time Construction API, which says:
An implementation that does not anticipate supporting multi-threading may simply check the first byte (i.e., the byte with lowest address) of that guard variable, initializing if and only if its value is zero, and then setting it to a non-zero value.
However, an implementation intending to support automatically thread-safe, one-time initialization (as opposed to requiring explicit user control for thread safety) may make use of the following API functions:
...
There were some bugs in the early GCC implementation of that API, I think they are all fixed and it works correctly from GCC version 4.3 (possibly earlier, I don't recall and can't find a reference right now.)
However, Singleton is a bad, bad pattern, do not use it!

Thread-safe initialisation of local statics: MSVC [duplicate]

This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Is static init thread-safe with VC2010?
I know that gcc and llvm-clang emit code to initialise local static variables in a threadsafe manner (which allows one to escape the static order initialisation fiasco by wrapping global statics in functions).
This msdn blog post, however, is the best documentation I can find of vcc's behaviour in these circumstances, and purports that static initialisation cannot ever be threadsafe, because the initialiser for a local static could recursively call into the same scope.
I don't buy this argument - it is clearly a programming error if the initialiser relies on its own result.
So, given that this article is from 2004, that gcc and clang can do it, and that the current msvc documentation is ambiguous (stating that 'assigning' to a local static isn't threadsafe, but nothing more):
Is the initialisation of local statics now threadsafe in MSVC?
If not, why not, since it is clearly possible for gcc to do this, but very difficult for the programmer to add in afterwards.
The C++0x Standard says:
§6.7 Declaration statement [stmt.dcl]
4/ The zero-initialization (8.5) of all block-scope variables with static storage duration (3.7.1) or thread storage duration (3.7.2) is performed before any other initialization takes place. Constant initialization (3.6.2) of a block-scope entity with static storage duration, if applicable, is performed before its block is first entered.
An implementation is permitted to perform early initialization of other block-scope variables with static or thread storage duration under the same conditions that an implementation is permitted to statically initialize a variable with static or thread storage duration in namespace scope (3.6.2). Otherwise such a variable is initialized the first time control passes through its declaration; such a variable is considered initialized upon the completion of its initialization.
If the initialization exits by throwing an exception, the initialization is not complete, so it will be tried again the next time control enters the declaration.
If control enters the declaration concurrently while the variable is being initialized, the concurrent execution shall wait for completion of the initialization.88
If control re-enters the declaration recursively while the variable is being initialized, the behavior is undefined.
[ Example:
int foo(int i) {
static int s = foo(2*i); // recursive call - undefined
return i+1;
}
—end example ]
88) The implementation must not introduce any deadlock around execution of the initializer.
As expected, it is quite complete.
However the fact is that even older versions of gcc already complied with this, and in fact do even better: in case of recursive initialization, an exception is thrown.
Finally, regarding a programmer adding it afterward: you can normally do it if you have something like Compare And Swap available, and use a sufficiently small variable, relying on zero-initialization of the variable to mark its non-computed state. However I do agree it's much easier if it's baked in.
I am afraid I stopped followed VC++ progresses though, so I don't know where it stands now. My only advice would be... look it up at assembly level.
I heard it is already implemented in vs2010, but can not find any reference. Anyway in c++0x standard such initializations are explicitly required to be thread-safe, so sooner or later ms would comply I guess.

C++ destructor & function call order

Suppose I have the following snipplet:
Foo foo;
....
return bar();
Now, does the C++ standard guarantees me that bar() will be called before foo::~Foo() ? Or is this the compiler/implementation's choice?
Thanks!
It is guaranteed behaviour. The actual execution is unrolled as follows:
0: enter block (scope)
1: Foo::Foo()
2. evaluation of bar(); as expression in return statement
3. save result of the expression as value returned from function
4. finalize return statement to leave function to its caller (request exit from current scope)
5: exit block (scope) with call to Foo::~Foo()
Here are some references from the standard:
What program execution guarantees, generally
1.9 Program execution
10 An instance of each object with automatic storage duration (3.7.2) is
associated with each entry into its
block.
The foo is of automatic storage duration and:
3.7.2 Automatic storage duration
1 Local objects explicitly declared auto or register or not explicitly declared
static or extern have automatic storage duration. The storage for
these objects lasts until the block in which they are created exits.
What is actual effect of return statement
6.6.3 The return statement
2 (...) the value of the expression is returned to the caller of the function
and
6.6 Jump statements (return belongs to jump statements)
2 On exit from a scope (however accomplished), destructors (12.4) are called for all
constructed objects with automatic storage duration (3.7.2)
What guarantees that the effect occurs
6.7 Declaration statement
2 Variables with automatic storage duration declared in the block are
destroyed on exit from the block
and
12.4 Destructors
10 Destructors are invoked implicitly (1) for a constructed
object with static storage duration
(3.7.1) at program termination
(3.6.3), (2) for a constructed object
with automatic storage duration
(3.7.2) when the block in which the
object is created exits (6.7)
It is not easy to grasp single idea form details scattered around all the C++ standard. Hopefully, quick overview will help you to make such analysis yourself too.
Yes, bar() will be called before the destructor of foo.
The standard says:
6.6: "On exit from a scope (however accomplished), destructors (12.4) are
called for all constructed objects with automatic storage duration
(3.7.2) (named objects or temporaries) that are declared in that scope,
in the reverse order of their declaration."
The scope isn't left until the return statement is completed.
The result of calling bar() must be evaluated before the stack frame containing Foo can be cleaned up, so yes, bar() will be called before Foo::~Foo().
Objects destruct when leaving the scope.
return leaves the scope, but it can't return until it has executed bar(). Ergo, bar() is called.
Just think, what if it was return bar(foo);? That just has to work, and it'd be silly if the destruction order was different depending on whether you pass that as an argument or not.