I am sorry to ask this trivial question but I could not find a definitive answer: if I have explicit static initialization to zero, is it zero initialization or initialization with constant expression? Say if I have
a.hpp:
class A { ... static int x; }
a.cpp;
int A::x = 0;
How many times will 0 be assigned to x? Once during zero initialization or twice during both zero initialization and initialization with constant expression?
The value of the variable will be 0 before any of your code is executed.
How it gets that way depends largely on the system; one typical approach
is to read an image of the date from disk, when loading the program.
Formally, you have zero initialization, followed by static
initialization, but there's no way a conforming implementation can tell,
and I've never heard of an implementation that separates the two.
Under Unix, at least in its older and more traditional versions,
uninitialized static variables were placed in the bs segment,
statically initialized variables in the data segment. The executable
file on the disk contained an image of the data segment, which was
copied into memory; all of the bytes in the bs segment were set to 0.
On a modern machine, with paged virtual memory, I would expect similar
behavior, with the difference that the initialization be deferred until
the page was first accessed.
I would be very surprised if Windows handled this differently (except for the names of the segments).
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;
In C++ I know static and global objects are constructed before the main function. But as you know, in C, there is no such kind initialization procedure before main.
For example, in my code:
int global_int1 = 5;
int global_int2;
static int static_int1 = 4;
static int static_int2;
When are these four variables initialized?
Where values for initialization like 5 and 4 are stored during compilation? How to manage them when initialization?
EDIT:
Clarification of 2nd question.
In my code I use 5 to initialize global_int1, so how can the compiler assign 5 to global_int? For example, maybe the compiler first store the 5 value at somewhere (i.e. a table), and get this value when initialization begins.
As to "How to manage them when initialization?", it is realy vague and I myself does not how to interpret yet. Sometimes, it is not easy to explain a question. Overlook it since I have not mastered the question fully yet.
By static and global objects, I presume you mean objects with
static lifetime defined at namespace scope. When such objects
are defined with local scope, the rules are slightly different.
Formally, C++ initializes such variables in three phases:
1. Zero initialization
2. Static initialization
3. Dynamic initialization
The language also distinguishes between variables which require
dynamic initialization, and those which require static
initialization: all static objects (objects with static
lifetime) are first zero initialized, then objects with static
initialization are initialized, and then dynamic initialization
occurs.
As a simple first approximation, dynamic initialization means
that some code must be executed; typically, static
initialization doesn't. Thus:
extern int f();
int g1 = 42; // static initialization
int g2 = f(); // dynamic initialization
Another approximization would be that static initialization is
what C supports (for variables with static lifetime), dynamic
everything else.
How the compiler does this depends, of course, on the
initialization, but on disk based systems, where the executable
is loaded into memory from disk, the values for static
initialization are part of the image on disk, and loaded
directly by the system from the disk. On a classical Unix
system, global variables would be divided into three "segments":
text:
The code, loaded into a write protected area. Static
variables with `const` types would also be placed here.
data:
Static variables with static initializers.
bss:
Static variables with no-initializer (C and C++) or with dynamic
initialization (C++). The executable contains no image for this
segment, and the system simply sets it all to `0` before
starting your code.
I suspect that a lot of modern systems still use something
similar.
EDIT:
One additional remark: the above refers to C++03. For existing
programs, C++11 probably doesn't change anything, but it does
add constexpr (which means that some user defined functions
can still be static initialization) and thread local variables,
which opens up a whole new can of worms.
Preface: The word "static" has a vast number of different meanings in C++. Don't get confused.
All your objects have static storage duration. That is because they are neither automatic nor dynamic. (Nor thread-local, though thread-local is a bit like static.)
In C++, Static objects are initialized in two phases: static initialization, and dynamic initialization.
Dynamic initialization requires actual code to execute, so this happens for objects that start with a constructor call, or where the initializer is an expression that can only be evaluated at runtime.
Static initialization is when the initializer is known statically and no constructor needs to run. (Static initialization is either zero-initialization or constant-initialization.) This is the case for your int variables with constant initializer, and you are guaranteed that those are indeed initialized in the static phase.
(Static-storage variables with dynamic initialization are also zero-initialzed statically before anything else happens.)
The crucial point is that the static initialization phase doens't "run" at all. The data is there right from the start. That means that there is no "ordering" or any other such dynamic property that concerns static initialization. The initial values are hard-coded into your program binary, if you will.
When are these four variables initialized?
As you say, this happens before program startup, i.e. before main begins. C does not specify it further; in C++, these happen during the static initialisation phase before objects with more complicated constructors or initialisers.
Where values for initialization like 5 and 4 are stored during compilation?
Typically, the non-zero values are stored in a data segment in the program file, while the zero values are in a bss segment which just reserves enough memory for the variables. When the program starts, the data segment is loaded into memory and the bss segment is set to zero. (Of course, the language standard doesn't specify this, so a compiler could do something else, like generate code to initialise each variables before running main).
Paraphrased from the standard:
All variables which do not have dynamic storage duration, do not have thread local storage duration, and are not local, have static storage duration. In other words, all globals have static storage duration.
Static objects with dynamic initialization are not necessarily created before the first statement in the main function. It is implementation defined as to whether these objects are created before the first statement in main, or before the first use of any function or variable defined in the same translation unit as the static variable to be initialized.
So, in your code, global_int1 and static_int1 are definitely initialized before the first statement in main because they are statically initialized. However, global_int2 and static_int2 are dynamically initialized, so their initialization is implementation defined according to the rule I mentioned above.
As for your second point, I'm not sure I understand what you mean. Could you clarify?
I was reading about SIOF from a book and it gave an example :
//file1.cpp
extern int y;
int x=y+1;
//file2.cpp
extern int x;
int y=x+1;
Now My question is :
In above code, will following things happen ?
while compiling file1.cpp, compiler leaves y as it is i.e doesn't allocate storage for it.
compiler allocates storage for x, but doesn't initialize it.
While compiling file2.cpp, compiler leaves x as it is i.e doesn't allocate storage for it.
compiler allocates storage for y, but doesn't initialize it.
While linking file1.o and file2.o, now let file2.o is initialized first, so now:
Does x gets initial value of 0? or doesn't get initialized?
The initialization steps are given in 3.6.2 "Initialization of non-local objects" of the C++ standard:
Step 1: x and y are zero-initialized before any other initialization takes place.
Step 2: x or y is dynamically initialized - which one is unspecified by the standard. That variable will get the value 1 since the other variable will have been zero-initialized.
Step 3: the other variable will be dynamically initialized, getting the value 2.
SIOF is very much a runtime artifact, the compiler and linker don't have much to do with it. Consider the atexit() function, it registers functions to be called at program exit. Many CRT implementations have something similar for program initialization, let's call it atinit().
Initializing these global variables requires executing code, the value cannot be determined by the compiler. So the compiler generates snippets of machine code that execute the expression and assigns the value. These snippets need to be executed before main() runs.
That's where atinit() comes into play. A common CRT implementation walks a list of atinit function pointers and execute the initialization snippets, in order. The problem is the order in which the functions are registered in the atinit() list. While atexit() has a well defined LIFO order, and it is implicitly determined by the order in which the code calls atexit(), such is not the case for atinit functions. The language specification doesn't require an order, there is nothing you could do in your code to specify an order. SIOF is the result.
One possible implementation is the compiler emitting function pointers in a separate section. The linker merges them, producing the atinit list. If your compiler does that then the initialization order will be determined by the order in which you link the object files. Look at the map file, you should see the atinit section if your compiler does this. It won't be called atinit, but some kind of name with "init" is likely. Taking a look at the CRT source code that calls main() should give insight as well.
It is compiler dependent and may be runtime dependent. A compiler may decide to lazily initialize static variables when the first variable in a file is accessed, or as each variable is accessed. Otherwise it will initialize all static variables by file at launch time, with the order usually depending on the link order of files. The file order could change based on dependencies or other, compiler dependent influences.
Static variables are usually initialized to zero unless they have a constant initializer. Again, this is compiler dependent. So one of these variables will probably be zero when the other is initialized. However, since both have initializers some compilers might leave the values undefined.
I think the most likely scenario would be:
Space is allocated for the variables, and both have the value 0.
One variable, say x, is initialized and set to the value 1.
The other, say y, is initialized and set to the value 2.
You could always run it and see. It could be that some compilers would generate code that goes into an infinite loop.
The whole point (and the reason it's called a "fiasco") is that it's impossible to say with any certainty what will happen in a case like this. Essentially, you're asking for something impossible (that two variables each be one greater than the other). Since they can't do that, what they will do is open to some question -- they might produce 0/1, or 1/0, or 1/2, or 2/1, or possibly (best case) just an error message.
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.
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