The book Object oriented programming in c++ by Robert Lafore says,
A static local variable has the visibility of an automatic local
variable (that is, inside the function containing it). However, its
lifetime is the same as that of a global variable, except that it
doesn’t come into existence until the first call to the function
containing it. Thereafter it remains in existence for the life of the
program
What does coming into existence after first call of function mean? The storage for static local is allocated at the time program is loaded in the memory.
The storage is allocated before main is entered, but (for example) if the static object has a ctor with side effects, those side effects might be delayed until just before the first time the function is called.
Note, however, that this is not necessarily the case. Constant initialization is only required to happen before that block is entered (not necessarily just as execution "crosses" that definition). Likewise, implementations are allowed to initialize other block-scope static variables earlier than required under some circumstances (if you want to get into the gory details of the circumstances, you can look at [basic.start.init] and [stmt.dcl], but it basically comes down to: as long as it doesn't affect the value with which it's initialized. For example, if you had something like:
int i;
std::cin >> i;
{
static int x = i;
...the implementation wouldn't be able to initialize x until the block was entered, because the value with which it was being initialized wouldn't be known until them. On the other hand, if you had:
{
static int i = 0;
...the implementation could carry out the initialization as early as it wished (and most would/will basically carry out such an initialization at compile time, so it won't involve executing any instructions at run-time at all). Even for less trivial cases, however, earlier initialization is allowed when logically possible (e.g., the value isn't coming from previous execution).
In C++ storage duration of an object (when raw memory gets allocated for it) and lifetime of an object are two separate concepts. The author was apparently referring to the latter one when he was talking about object's "coming into existence".
In general case it is not enough to allocate storage for an object to make it "come into existence". Lifetime of an object with non-trivial initialization begins once its initialization is complete. For example, an object of a class with a non-trivial constructor does not officially "live" until its constructor has completed execution.
Initialization of a static local object is performed when the control passes over the declaration for the very first time. Before that the object does not officially exist, even if the memory for it is already allocated.
Note that the author is not painstakingly precise in his description. It is not sufficient to just call the function containing the declaration. The control has to pass through the declaration of the object for it to begin its lifetime. If the function contains branching, this does not necessarily happen during the very first call to the function.
For object with trivial initialization (like int objects), there's no difference between storage duration and lifetime. For such objects allocating memory is all that needs to be done. But in general case allocating memory alone is not sufficient.
It means that the static variable inside a function doesn't get initialized (by the constructor or the assignment operator) until the first call for that function.
As soon as the function, which contains a static local variable, is called the static local variable is initialized.
Related
My csapp book says that if global and static variables are initialized, than they are contained in .data section in ELF relocatable object file.
So my question is that if some foo.c code contains
int a;
int main()
{
a = 3;
}`
and example.c contains,
int b = 3;
int main()
{
...
}
is it only b that considered to be initialized? In other words, does initialization mean declaration and definition in same line?
It means exactly what it says. Initialized static storage duration objects will have their init values set before the main function is called. Not initialized will be zeroed. The second part of the statement is actually implementation dependant, and implementation has the full freedom of the way it will be archived.
When you declare the variable without the keyword extern you always define it as well
Both are considered initialized
They get zero initialized or constant initalized (in short: if the right hand side is a compile time constant expression).
If permitted, Constant initialization takes place first (see Constant
initialization for the list of those situations). In practice,
constant initialization is usually performed at compile time, and
pre-calculated object representations are stored as part of the
program image. If the compiler doesn't do that, it still has to
guarantee that this initialization happens before any dynamic
initialization.
For all other non-local static and thread-local variables, Zero
initialization takes place. In practice, variables that are going to
be zero-initialized are placed in the .bss segment of the program
image, which occupies no space on disk, and is zeroed out by the OS
when loading the program.
To sum up, if the implementation cannot constant initialize it, then it must first zero initialize and then initialize it before any dynamic initialization happends.
In the snippet:
int a;
int main()
{
a = 3;
}
a is not initialized; it is assigned. Assignment is a run-time execution of code. For example, should main be called multiple times (which is not, but any user function could), then a is set to 3 each time the function is called.
You second snippet is initializaion of the globalvariable b and it will be placed in the .data segment.
I will answer this question in general and complete way and not with respect to any programming language
There is a hell lot of confusion between declaration, definition, and initialization. Sometimes they all look similar and sometimes completely different.
Before understanding the differences, It is very important to be aware of two things:
The difference between declaration, definition, and initialization
varies from one programming language to other. Each programming has
its own way of doing these three things.The “thing” which you are
defining, declaring or initializing also affects the difference
between the three of them. That “thing” can be a variable, a class or
a function. All of them have different meanings of definitions,
declaration, and initialization. Once we are aware of the above two
things, most of the doubts get cleared and we stop seeking exact
differences because it’s not there.
In general terms ( irrespective of any language or “thing”)
The declaration means we are saying to a computer that this “thing”
(it can be a variable, a function or a class) exists but we don’t know
where. In the future, we may tell but right now it just exists
somewhere. In simple words, we don’t allocate memory while declaring.
We can declare that “thing” many times.
The definition means we are saying to the computer that this “thing” needs memory and it needs to be located somewhere. In simple
words, defining means we have allocated memory for it. We can define
something only once
The initialization means whatever our “thing “ is, we are giving it an initial value. That “thing” must be in some memory location and
if we keep that location empty, it may be a house for bugs and errors.
Initialization is not always necessary but it’s important.
Many people assume that declaration + definition = Initialization .
It's not wrong, but it’s not correct in all places. Its correct only for variables that too in a language like C ++ or maybe C.
In python, there is no concept of the declaration . We don’t need to declare anything in it.
The general meaning of the three is valid everywhere but the way that is performed varies from language to language and the “thing”.
Hope it helps :)
Variables with static storage duration that are initialized to zero end up in .bss.
Variables with static storage duration that are initialized with a non-zero value end up in .data.
NOTE: the C standard guarantees that if the programmer doesn't explicitly initialize a variable with static storage duration, such as static int a;, it is then initialized to zero implicitly1). Therefore a ends up in .bss.
Examples here.
1) C11 6.7.9
If an object that has static or thread storage duration is not initialized
explicitly, then:
if it has arithmetic type, it is initialized to (positive or unsigned) zero;
I have C++ code which declares static-lifetime variables which are initialized by function calls. The called function constructs a vector instance and calls its push_back method. Is the code risking doom via the C++ static initialization order fiasco? If not, why not?
Supplementary information:
What's the "static initialization order fiasco"?
It's explained in C++ FAQ 10.14
Why would I think use of vector could trigger the fiasco?
It's possible that the vector constructor makes use of the value of another static-lifetime variable initialized dynamically. If so, then there is nothing to ensure that vector's variable is initialized before I use vector in my code. Initializing result (see code below) could end up calling the vector constructor before vector's dependencies are fully initialized, leading to access to uninitialized memory.
What does this code look like anyway?
struct QueryEngine {
QueryEngine(const char *query, string *result_ptr)
: query(query), result_ptr(result_ptr) { }
static void AddQuery(const char *query, string *result_ptr) {
if (pending == NULL)
pending = new vector<QueryEngine>;
pending->push_back(QueryEngine(query, result_ptr));
}
const char *query;
string *result_ptr;
static vector<QueryEngine> *pending;
};
vector<QueryEngine> *QueryEngine::pending = NULL;
void Register(const char *query, string *result_ptr) {
QueryEngine::AddQuery(query, result_ptr);
}
string result = Register("query", &result);
Fortunately, static objects are zero-initialised even before any other initialisation is performed (even before the "true" initialisation of the same objects), so you know that the NULL will be set on that pointer long before Register is first invoked.1
Now, in terms of operating on your vector, it appears that (technically) you could run into such a problem:
[C++11: 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.
[C++11: 17.6.5.9/4]: [Note: This means, for example, that implementations can’t use a static object for internal purposes without synchronization because it could cause a data race even in programs that do not explicitly share objects between threads. —end note]
Notice that, although synchronisation is being required in this note, that's been mentioned within a passage that ultimately acknowledges that static implementation details are otherwise allowed.
That being said, it seems like the standard should further state that user code should avoid operating on standard containers during static initialisation, if the intent were that the semantics of such code could not be guaranteed; I'd consider this a defect in the standard, either way. It should be clearer.
1 And it is a NULL pointer, whatever the bit-wise representation of that may be, rather than a blot to all-zero-bits.
vector doesn't depend on anything preventing its use in dynamic initialisation of statics. The only issue with your code is a lack of thread safety - no particular reason to think you should care about that, unless you have statics whose construction spawns threads....
Initializing result (see code below) could end up calling the vector constructor before that class is fully initialized, leading to access to uninitialized memory.
No... initialising result calls AddQuery which checks if (pending == NULL) - the initialisation to NULL will certainly have been done before any dynamic initialisation, per 3.6.2/2:
Constant initialization is performed:
...
— if an object with static or thread storage duration is not initialized by a constructor call and if either the object is value-initialized or every full-expression that appears in its initializer is a constant expression
So even if the result assignment is in a different translation unit it's safe. See 3.6.2/2:
Together, zero-initialization and constant initialization are called static initialization; all other initialization is dynamic initialization. Static initialization shall be performed before any dynamic initialization takes place.
I read several posts on C++ initialization from Google, some of which direct me here on StackOverflow. The concepts I picked from those posts are as follows:
The order of initialization of C++ is:
Zero Initialization;
Static Initialization;
Dynamic Initialization.
Static objects (variables included) are first Zero-initialized, and then Static-initialized.
I have several inquiries as to the initialization issue (storage class issue may be related as well):
Global objects (defined without static keyword) are also static objects, right?
Global objects are also initialized like static objects by two steps like above, right?
What is the Static Initialization? Does it refer to initializing static objects (defined with static keyword)?
I also read that objects defined within block (i.e. in a function) with static keyword is initialized when the execution thread first enters the block! This means that local static objects are not initialized before main function execution. This means they are not initialized as the two steps mentioned above, right?
Dynamic initialization refers to initialization of objects created by new operator, right? It might refer to initialization like myClass obj = myClass(100); or myClass obj = foo();
I have too many inquiries on the initialization and storage class specifier issues. I read the C++2003 Standard document, but cannot find a clear logic since they are scattered throughout the document.
I hope you give me an answer that logically explains the whole map of storage class specifier and initialization. Any reference is welcome!
Code that might explain my question:
class myClass{
public:
int i;
myClass(int j = 10): j(i){}
// other declarations
};
myClass obj1;//global scope
static myClass obj2(2);//file scope
{ //local scope
myClass obj3(3);
static myClass obj4(4);
}
EDIT:
If you think my question is rather tedious, you can help explain your ideas based on the code above.
I read several posts on C++ initialization from Google, some of which direct me here on StackOverflow. The concepts I picked from those posts are as follows:
The order of initialization of C++ is:
Zero Initialization;
Static Initialization;
Dynamic Initialization.
Yes, indeed there are 3 phases (in the Standard). Let us clarify them before continuing:
Zero Initialization: the memory is filled with 0s at the byte level.
Constant Initialization: a pre-computed (compile-time) byte pattern is copied at the memory location of the object
Static Initialization: Zero Initialization followed by Constant Initialization
Dynamic Initialization: a function is executed to initialize the memory
A simple example:
int const i = 5; // constant initialization
int const j = foo(); // dynamic initialization
Static objects (variables included) are first Zero-initialized, and then Static-initialized.
Yes and no.
The Standard mandates that the objects be first zero-initialized and then they are:
constant initialized if possible
dynamically initialized otherwise (the compiler could not compute the memory content at compile-time)
Note: in case of constant initialization, the compiler might omit to first zero-initialized memory following the as-if rule.
I have several inquiries as to the initialization issue (storage class issue may be related as well):
Global objects (defined without static keyword) are also static objects, right?
Yes, at file scope the static object is just about the visibility of the symbol. A global object can be referred to, by name, from another source file whilst a static object name is completely local to the current source file.
The confusion stems from the reuse of the world static in many different situations :(
Global objects are also initialized like static objects by two steps like above, right?
Yes, as are local static objects in fact.
What is the Static Initialization? Does it refer to initializing static objects (defined with static keyword)?
No, as explained above it refers to initializing objects without executing a user-defined function but instead copying a pre-computed byte pattern over the object's memory. Note that in the case of objects that will later be dynamically initialized, this is just zero-ing the memory.
I also read that objects defined within block (i.e. in a function) with static keyword is initialized when the execution thread first enters the block! This means that local static objects are not initialized before main function execution. This means they are not initialized as the two steps mentioned above, right?
They are initialized with the two steps process, though indeed only the first time execution pass through their definition. So the process is the same but the timing is subtly different.
In practice though, if their initialization is static (ie, the memory pattern is a compile-time pattern) and their address is not taken they might be optimized away.
Note that in case of dynamic initialization, if their initialization fails (an exception is thrown by the function supposed to initialize them) it will be re-attempted the next time flow-control passes through their definition.
Dynamic initialization refers to initialization of objects created by new operator, right? It might refer to initialization like myClass obj = myClass(100); or myClass obj = foo();
Not at all, it refers to initialization requiring the execution of a user defined function (note: std::string has a user-defined constructor as far as the C++ language is concerned).
EDIT: My thanks to Zach who pointed to me I erroneously called Static Initialization what the C++11 Standard calls Constant Initialization; this error should now be fixed.
I believe there are three different concepts: initializing the variable, the location of the variable in memory, the time the variable is initialized.
First: Initialization
When a variable is allocated in memory, typical processors leave the memory untouched, so the variable will have the same value that somebody else stored earlier. For security, some compilers add the extra code to initialize all variables they allocate to zero. I think this is what you mean by "Zero Initialization". It happens when you say:
int i; // not all compilers set this to zero
However if you say to the compiler:
int i = 10;
then the compiler instructs the processor to put 10 in the memory rather than leaving it with old values or setting it to zero. I think this is what you mean by "Static Initialization".
Finally, you could say this:
int i;
...
...
i = 11;
then the processor "zero initializes" (or leaves the old value) when executing int i; then when it reaches the line i = 11 it "dynamically initializes" the variable to 11 (which can happen very long after the first initialization.
Second: Location of the variable
There are: stack-based variables (sometimes called static variables), and memory-heap variables (sometimes called dynamic variables).
Variables can be created in the stack segment using this:
int i;
or the memory heap like this:
int *i = new int;
The difference is that the stack segment variable is lost after exiting the function call, while memory-heap variables are left until you say delete i;. You can read an Assembly-language book to understand the difference better.
Third: The time the variable is initialized
A stack-segment variable is "zero-initialized" or statically-initialized" when you enter the function call they are defined within.
A memory-heap variable is "zero-initialized" or statically-initialized" when it is first created by the new operator.
Final Remark
You can think about static int i; as a global variable with a scope limited to the function it is defined in. I think the confusion about static int i; comes because static hear mean another thing (it is not destroyed when you exit the routine, so it retains its value). I am not sure, but I think the trick used for static int i; is to put it in the stack of main() which means it is not destroyed until you exit the whole program (so it retains the first initialization), or it could be that it is stored in the data segment of the application.
I've been reading up on C++ on the Internet, and here's one thing that I haven't been quite able to find an answer to.
I know that static variables used within functions are akin to globals, and that subsequent invocations of that function will have the static variable retain its value between calls.
However, if the function is never called, does the static variable get allocated?
Thanks
If the function is never called, it is likely that your linker will deadstrip both the function and the static variable, preventing it from entering .rodata, .data, or .bss segments (or your executable file format's equivalents).
However, there are various reasons why a linker might not deadstrip (flags telling it not to, an inability to determine what depends on the symbol, etc).
It's worth checking your linker map file (sometimes just a text file!), or using objdump, nm, or dumpbin utilities on the final executable to see if the symbol or related symbols (such as static initializer code) survived.
The C++ Standard, section 6.7 says:
The zero-initialization (8.5) of all
local objects with static storage
duration (3.7.1) is performed before
any other initialization takes place.
A local object of POD type (3.9) with
static storage duration initialized
with constant-expressions is
initialized before its block is first
entered. An implementation is
permitted to per- form early
initialization of other local objects
with static storage duration under the
same conditions that an implementation
is permitted to statically initialize
an object with static storage duration
in namespace scope (3.6.2). Otherwise
such an object is initialized the
first time control passes through its
declaration; such an object is
considered initialized upon the
completion of its initialization.
Which indicates that local static objects are normally initialised the first time the control flow encounters them. However, they may well be allocated before this - the standard is somewhat reticent on what static storage actually is, except with reference to static object lifetimes.
Every object in C++ has two nested time-periods associated with it: storage duration and lifetime. Storage duration is the period for which the raw memory occupied by the object is allocated. Lifetime is the period between construction and destruction of an actual object in that memory. (For objects of POD-types construction-destruction either doesn't matter or not applicable, so their lifetime matches their storage duration).
When someone says "allocated" they usually refer to storage duration. The language doesn't actually specify exactly when the object's storage duration begins. It is sufficient to require that shall begin at some point before the object's lifetime begins.
For this reason, in general case a static object defined inside a function might never begin its lifetime and, theoretically, it's storage duration does not have to begin either. So, in theory, in might not even get "allocated".
In practice though, all objects with static storage duration ("globals", local statics, etc.) are normally treated equally: they are assigned a specific amount of storage early, at the program's startup.
As an additional note, if a local object with static storage duration requires a non-trivial initialization, this initialization is carried out when the control passes over the definition for the very first time. So in this example
void foo() {
static int *p = new int[100];
}
the dynamic array will never be allocated if the function is never called. And it will be allocated only once if the function is called. This doesn't look like what you are asking about, but I mention this just in case.
Im sure that thats going to be up to the implementation. What MSVC does is - static objects are allocated in the automatic data segment of the EXE or DLL. However, the constructor is only executed the first time the function containing the static is executed.
Yes, actual allocation is compiler dependent, although I think that every compiler just reserves the space in the .static segment of the executable (or the equivalent in its executable file format).
The initialization, however takes place only the firs time that the execution flow encounters the static object, and that is required by the standard.
Beware that initialization of global static objects works in a different way, though.
You can get very good answers to almost every question at the C++ FAQ lite site.
I am also fond of Scott Meyers's "Effective C++".
Depends. If you mean, never called, as in, the function is literally never invoked, then your compiler will probably not allocate it, or even put in the function code. If, however, you made it dependent on, say, user input, and that user input just happened to never come up, then it will probably be pre-allocated. However, you're treading in a minefield here, and it's best just to assume that it is always created by the time control enters the function(s) that refer to it.
Static variables defined on classes (members) or functions are not allocated dynamically on stack during function call, like non static ones. They are allocated in another area of generated code reserved for global and static data. So, if you call the function or not, instantiate classes that contain static members or not, a space to their data will be reserved on program data area anyway.
In C/C++, why are globals and static variables initialized to default values?
Why not leave it with just garbage values? Are there any special
reasons for this?
Security: leaving memory alone would leak information from other processes or the kernel.
Efficiency: the values are useless until initialized to something, and it's more efficient to zero them in a block with unrolled loops. The OS can even zero freelist pages when the system is otherwise idle, rather than when some client or user is waiting for the program to start.
Reproducibility: leaving the values alone would make program behavior non-repeatable, making bugs really hard to find.
Elegance: it's cleaner if programs can start from 0 without having to clutter the code with default initializers.
One might then wonder why the auto storage class does start as garbage. The answer is two-fold:
It doesn't, in a sense. The very first stack frame page at each level (i.e., every new page added to the stack) does receive zero values. The "garbage", or "uninitialized" values that subsequent function instances at the same stack level see are really the previous values left by other method instances of your own program and its library.
There might be a quadratic (or whatever) runtime performance penalty associated with initializing auto (function locals) to anything. A function might not use any or all of a large array, say, on any given call, and it could be invoked thousands or millions of times. The initialization of statics and globals, OTOH, only needs to happen once.
Because with the proper cooperation of the OS, 0 initializing statics and globals can be implemented with no runtime overhead.
Section 6.7.8 Initialization of C99 standard (n1256) answers this question:
If an object that has automatic storage duration is not initialized explicitly, its value is indeterminate. If an object that has static storage duration is not initialized explicitly, then:
— if it has pointer type, it is initialized to a null pointer;
— if it has arithmetic type, it is initialized to (positive or unsigned) zero;
— if it is an aggregate, every member is initialized (recursively) according to these rules;
— if it is a union, the first named member is initialized (recursively) according to these rules.
Think about it, in the static realm you can't tell always for sure something is indeed initialized, or that main has started. There's also a static init and a dynamic init phase, the static one first right after the dynamic one where order matters.
If you didn't have zeroing out of statics then you would be completely unable to tell in this phase for sure if anything was initialized AT ALL and in short the C++ world would fly apart and basic things like singletons (or any sort of dynamic static init) would simple cease to work.
The answer with the bulletpoints is enthusiastic but a bit silly. Those could all apply to nonstatic allocation but that isn't done (well, sometimes but not usually).
In C, statically-allocated objects without an explicit initializer are initialized to zero (for arithmetic types) or a null pointer (for pointer types). Implementations of C typically represent zero values and null pointer values using a bit pattern consisting solely of zero-valued bits (though this is not required by the C standard). Hence, the bss section typically includes all uninitialized variables declared at file scope (i.e., outside of any function) as well as uninitialized local variables declared with the static keyword.
Source: Wikipedia