Very frequently, I want to test the value of a variable in a function via a breakpoint. In many cases, that variable never actually gets referenced by any code that "does stuff", but that doesn't mean I don't still want to see it. Unfortunately, the optimizer is working against me and simply removing all such code when compiling, so I have to come up with convoluted grossness to fool the compiler into thinking those values actually matter so they don't get optimized away. I don't want to turn the optimizer off, as it's doing important stuff in other places, but just for this one block of code I'd like to temporarily disable it for the sake of debugging.
Code the produces observable behavior fits the requirements by definition. For example, printf("").
Access to volatile variable also formally constitutes observable behavior, although I won't be surprised if some compilers still discarded "unnecessary" volatile variables.
For this reason, a call to an I/O function appears to be the best canditate to me.
You can try "volatile" keyword. Some intro is at http://en.wikipedia.org/wiki/Volatile_variable .
Generally speaking, the volatile keyword is intended to prevent the compiler from applying any optimizations on the code that assume values of variables cannot change "on their own."
Have you tried
#pragma optimize("",off)
?
MSVS specific I think - http://msdn.microsoft.com/en-us/library/chh3fb0k(v=vs.80).aspx
You do not specify your compiler so I will just add here that I have been using the volatile specifier on any variables that I use specifically for debugging. In my compiler (Embarcadero RAD Studio C++ Builder) I have been using this for a couple of years and not once has the variable been optimized out. Maybe you don't use this compiler, but if you do I can say volatile has certainly always worked for me.
Here's an example of a trick I've used:
#include <ctime>
#include <iostream>
int main() {
std::cout << "before\n";
if (std::time(NULL) == 0) {
std::cout << "This should not appear\n";
}
std::cout << "after\n";
}
The time() call should always return a positive value, but the compiler has no way of knowing that.
If the only purpose of the variable is to be viewed while in a breakpoint with a debugger, you can make the variable global. You could, for instance, maintain a global buffer:
#ifdef DEBUG
char dbg_buffer[512];
template <typename T>
void poke_dbg (const T &t) {
memcpy(dbg_buffer, &t, sizeof(T));
}
#else
#define poke_dbg(x)
#endif
Then during debugging, you can inspect the contents of the dbg_buffer (with an appropriate cast if desired).
Related
tl;dr: Can it be ensured somehow (e.g. by writing a unit test) that some things are optimized away, e.g. whole loops?
The usual approach to be sure that something is not included in the production build is wrapping it with #if...#endif. But I prefer to stay with C++ mechanics instead. Even there, instead of complicated template specializations I like to keep implementations simple and argue "hey, the compiler will optimize this out anyway".
Context is embedded SW in automotive (binary size matters) with often poor compilers. They are certified in the sense of safety, but usually not good in optimizations.
Example 1: In a container the destruction of elements is typically a loop:
for(size_t i = 0; i<elements; i++)
buffer[i].~T();
This works also for build-in types such as int, as the standard allows the explicit call of the destructor also for any scalar types (C++11 12.4-15). In such case the loop does nothing and is optimized out. In GCC it is, but in another (Aurix) not, I saw a literally empty loop in the disassembly! So that needed a template specialization to fix it.
Example 2: Code, which is intended for debugging, profiling or fault-injection etc. only:
constexpr bool isDebugging = false; // somehow a global flag
void foo(int arg) {
if( isDebugging ) {
// Albeit 'dead' section, it may not appear in production binary!
// (size, security, safety...)
// 'if constexpr..' not an option (C++11)
std::cout << "Arg was " << arg << std::endl;
}
// normal code here...
}
I can look at the disassembly, sure. But being an upstream platform software it's hard to control all targets, compilers and their options one might use. The fear is big that due to any reason a downstream project has a code bloat or performance issue.
Bottom line: Is it possible to write the software in a way, that certain code is known to be optimized away in a safe manner as a #if would do? Or a unit tests, which give a fail if optimization is not as expected?
[Timing tests come to my mind for the first problem, but being bare-metal I don't have convenient tools yet.]
There may be a more elegant way, and it's not a unit test, but if you're just looking for that particular string, and you can make it unique,
strings $COMPILED_BINARY | grep "Arg was"
should show you if the string is being included
if constexpr is the canonical C++ expression (since C++17) for this kind of test.
constexpr bool DEBUG = /*...*/;
int main() {
if constexpr(DEBUG) {
std::cerr << "We are in debugging mode!" << std::endl;
}
}
If DEBUG is false, then the code to print to the console won't generate at all. So if you have things like log statements that you need for checking the behavior of your code, but which you don't want to interact with in production code, you can hide them inside if constexpr expressions to eliminate the code entirely once the code is moved to production.
Looking at your question, I see several (sub-)questions in it that require an answer. Not all answers might be possible with your bare-metal compilers as hardware vendors don't care that much about C++.
The first question is: How do I write code in a way that I'm sure it gets optimized. The obvious answer here is to put everything in a single compilation unit so the caller can see the implementation.
The second question is: How can I force a compiler to optimize. Here constexpr is a bless. Depending on whether you have support for C++11, C++14, C++17 or even the upcoming C++20, you'll get different feature sets of what you can do in a constexpr function. For the usage:
constexpr char c = std::string_view{"my_very_long_string"}[7];
With the code above, c is defined as a constexpr variable. Because you apply it to the variable, you require some things:
Your compiler should optimize the code so the value of c is known at compile time. This even holds true for -O0 builds!
All functions used for calculate c are constexpr and available. (and by result, enforce the behaviour of the first question)
No undefined behaviour is allowed to be triggered in the calculation of c. (For the given value)
The negative about this is: Your input needs to be known at compile time.
C++17 also provides if constexpr which has similar requirements: condition needs to be calculated at compile time. The result is that 1 branch of the code ain't allowed to be compiled (as it even can contain elements that don't work on the type you are using).
Which than brings us to the question: How do I ensure sufficient optimizations for my program to run fast enough, even if my compiler ain't well behaving. Here the only relevant answer is: create benchmarks and compare the results. Take the effort to setup a CI job that automates this for you. (And yes, you can even use external hardware although not being that easy) In the end, you have some requirements: handling A should take less than X seconds. Do A several times and time it. Even if they don't handle everything, as long as it's within the requirements, its fine.
Note: As this is about debug, you most likely can track the size of an executable as well. As soon as you start using streams, a lot of conversions to string ... your exe size will grow. (And you'll find it a bless as you will immediately find commits which add 10% to the image size)
And than the final question: You have a buggy compiler, it doesn't meet my requirements. Here the only answer is: Replace it. In the end, you can use any compiler to compiler your code to bare metal, as long as the linker scripts work. If you need a start, C++Now 2018: Michael Caisse “Modern C++ in Embedded Systems” gives you a very good idea of what you need to use a different compiler. (Like a recent Clang or GCC, on which you even can log bugs if the optimization ain't good enough)
Insert a reference to external data or function into the block that should be verified to be optimised away. Like this:
extern void nop();
constexpr bool isDebugging = false; // somehow a global flag
void foo(int arg) {
if( isDebugging ) {
nop();
std::cout << "Arg was " << arg << std::endl; // may not appear in production binary!
}
// normal code here...
}
In Debug-Builds, link with an implementation of nop() in a extra compilation unit nop.cpp:
void nop() {}
In Release-Builds, don't provide an implementation.
Release builds will only link if the optimisable code is eliminated.
`- kisch
Here's another nice solution using inline assembly.
This uses assembler directives only, so it might even be kind of portable (checked with clang).
constexpr bool isDebugging = false; // somehow a global flag
void foo(int arg) {
if( isDebugging ) {
asm(".globl _marker\n_marker:\n");
std::cout << "Arg was " << arg << std::endl; // may not appear in production binary!
}
// normal code here...
}
This would leave an exported linker symbol in the compiled executable, if the code isn't optimised away. You can check for this symbol using nm(1).
clang can even stop the compilation right away:
constexpr bool isDebugging = false; // somehow a global flag
void foo(int arg) {
if( isDebugging ) {
asm("_marker=1\n");
std::cout << "Arg was " << arg << std::endl; // may not appear in production binary!
}
asm volatile (
".ifdef _marker\n"
".err \"code not optimised away\"\n"
".endif\n"
);
// normal code here...
}
This is not an answer to "How to ensure some code is optimized away?" but to your summary line "Can a unit test be written that e.g. whole loops are optimized away?".
First, the answer depends on how far you see the scope of unit-testing - so if you put in performance tests, you might have a chance.
If in contrast you understand unit-testing as a way to test the functional behaviour of the code, you don't. For one thing, optimizations (if the compiler works correctly) shall not change the behaviour of standard-conforming code.
With incorrect code (code that has undefined behaviour) optimizers can do what they want. (Well, for code with undefined behaviour the compiler can do it also in the non-optimizing case, but sometimes only the deeper analyses peformed during optimization make it possible for the compiler to detect that some code has undefined behaviour.) Thus, if you write unit-tests for some piece of code with undefined behaviour, the test results may differ when you run the tests with and without optimization. But, strictly speaking, this only tells you that the compiler translated the code both times in a different way - it does not guarantee you that the code is optimized in the way you want it to be.
Here's another different way that also covers the first example.
You can verify (at runtime) that the code has been eliminated, by comparing two labels placed around it.
This relies on the GCC extension "Labels as Values" https://gcc.gnu.org/onlinedocs/gcc/Labels-as-Values.html
before:
for(size_t i = 0; i<elements; i++)
buffer[i].~T();
behind:
if (intptr_t(&&behind) != intptr_t(&&before)) abort();
It would be nice if you could check this in a static_assert(), but sadly the difference of &&label expressions is not accepted as compile-time constant.
GCC insists on inserting a runtime comparison, even though both labels are in fact at the same address.
Interestingly, if you compare the addresses (type void*) directly, without casting them to intptr_t, GCC falsely optimises away the if() as "always true", whereas clang correctly optimises away the complete if() as "always false", even at -O1.
#include <iostream>
struct Foo { static auto foo() -> int { return 123; } };
int main() {
std::cout << static_cast<Foo*>(nullptr)->foo() << std::endl;
return 0;
}
I DO KNOW that this is not allowed by standard. But, how about specific compiler?
I only care GCC(G++) and Clang.
Is there any guarantee that these two compiler allow this as compiler feature/specification/extension?
I can find this neither in the list of gcc's C++ extensions nor in the corresponding list for clang. So I would say that you have no guarantee that this works for either compiler.
There is nothing saying that this will work. It is up to you, if you decide to use it, to ascertain that it really does work for the actual architecture in which you are using it. And it's important to reflect on the fact that code generation, debug traps for null pointer usage (and such) and optimisation in all compilers is "target system dependent" - so it may well work on x86, but not on MIPS, as an example. Or it may stop working in one way or another in version X+0.0.1 of the compiler.
Having said that, I'd expect this particular example to work perfectly fine on any architecture, because this is not used anywhere.
Note that just because something is undefined, doesn't necessarily mean that it will crash, fail or even "not work exactly as you expect it to". It does, however, allow the compiler to generate whatever code the compiler would like, including something that blows up your computer, formats your hard-drive, etc, etc.
I read this line in a book:
It is provably impossible to build a compiler that can actually
determine whether or not a C++ function will change the value of a
particular variable.
The paragraph was talking about why the compiler is conservative when checking for const-ness.
Why is it impossible to build such a compiler?
The compiler can always check if a variable is reassigned, a non-const function is being invoked on it, or if it is being passed in as a non-const parameter...
Why is it impossible to build such a compiler?
For the same reason that you can't write a program that will determine whether any given program will terminate. This is known as the halting problem, and it's one of those things that's not computable.
To be clear, you can write a compiler that can determine that a function does change the variable in some cases, but you can't write one that reliably tells you that the function will or won't change the variable (or halt) for every possible function.
Here's an easy example:
void foo() {
if (bar() == 0) this->a = 1;
}
How can a compiler determine, just from looking at that code, whether foo will ever change a? Whether it does or doesn't depends on conditions external to the function, namely the implementation of bar. There's more than that to the proof that the halting problem isn't computable, but it's already nicely explained at the linked Wikipedia article (and in every computation theory textbook), so I'll not attempt to explain it correctly here.
Imagine such compiler exists. Let's also assume that for convenience it provides a library function that returns 1 if the passed function modifies a given variable and 0 when the function doesn't. Then what should this program print?
int variable = 0;
void f() {
if (modifies_variable(f, variable)) {
/* do nothing */
} else {
/* modify variable */
variable = 1;
}
}
int main(int argc, char **argv) {
if (modifies_variable(f, variable)) {
printf("Modifies variable\n");
} else {
printf("Does not modify variable\n");
}
return 0;
}
Don't confuse "will or will not modify a variable given these inputs" for "has an execution path which modifies a variable."
The former is called opaque predicate determination, and is trivially impossible to decide - aside from reduction from the halting problem, you could just point out the inputs might come from an unknown source (eg. the user). This is true of all languages, not just C++.
The latter statement, however, can be determined by looking at the parse tree, which is something that all optimizing compilers do. The reason they do is that pure functions (and referentially transparent functions, for some definition of referentially transparent) have all sorts of nice optimizations that can be applied, like being easily inlinable or having their values determined at compile-time; but to know if a function is pure, we need to know if it can ever modify a variable.
So, what appears to be a surprising statement about C++ is actually a trivial statement about all languages.
I think the key word in "whether or not a C++ function will change the value of a particular variable" is "will". It is certainly possible to build a compiler that checks whether or not a C++ function is allowed to change the value of a particular variable, you cannot say with certainty that the change is going to happen:
void maybe(int& val) {
cout << "Should I change value? [Y/N] >";
string reply;
cin >> reply;
if (reply == "Y") {
val = 42;
}
}
I don't think it's necessary to invoke the halting problem to explain that you can't algorithmically know at compile time whether a given function will modify a certain variable or not.
Instead, it's sufficient to point out that a function's behavior often depends on run-time conditions, which the compiler can't know about in advance. E.g.
int y;
int main(int argc, char *argv[]) {
if (argc > 2) y++;
}
How could the compiler predict with certainty whether y will be modified?
It can be done and compilers are doing it all the time for some functions, this is for instance a trivial optimisation for simple inline accessors or many pure functions.
What is impossible is to know it in the general case.
Whenever there is a system call or a function call coming from another module, or a call to a potentially overriden method, anything could happen, included hostile takeover from some hacker's use of a stack overflow to change an unrelated variable.
However you should use const, avoid globals, prefer references to pointers, avoid reusing variables for unrelated tasks, etc. that will makes the compiler's life easier when performing aggressive optimisations.
There are multiple avenues to explaining this, one of which is the Halting Problem:
In computability theory, the halting problem can be stated as follows: "Given a description of an arbitrary computer program, decide whether the program finishes running or continues to run forever". This is equivalent to the problem of deciding, given a program and an input, whether the program will eventually halt when run with that input, or will run forever.
Alan Turing proved in 1936 that a general algorithm to solve the halting problem for all possible program-input pairs cannot exist.
If I write a program that looks like this:
do tons of complex stuff
if (condition on result of complex stuff)
{
change value of x
}
else
{
do not change value of x
}
Does the value of x change? To determine this, you would first have to determine whether the do tons of complex stuff part causes the condition to fire - or even more basic, whether it halts. That's something the compiler can't do.
Really surprised that there isn't an answer that using the halting problem directly! There's a very straightforward reduction from this problem to the halting problem.
Imagine that the compiler could tell whether or not a function changed the value of a variable. Then it would certainly be able to tell whether the following function changes the value of y or not, assuming that the value of x can be tracked in all the calls throughout the rest of the program:
foo(int x){
if(x)
y=1;
}
Now, for any program we like, let's rewrite it as:
int y;
main(){
int x;
...
run the program normally
...
foo(x);
}
Notice that, if, and only if, our program changes the value of y, does it then terminate - foo() is the last thing it does before exiting. This means we've solved the halting problem!
What the above reduction shows us is that the problem of determining whether a variable's value changes is at least as hard as the halting problem. The halting problem is known to be incomputable, so this one must be also.
As soon as a function calls another function that the compiler doesn't "see" the source of, it either has to assume that the variable is changed, or things may well go wrong further below. For example, say we have this in "foo.cpp":
void foo(int& x)
{
ifstream f("f.dat", ifstream::binary);
f.read((char *)&x, sizeof(x));
}
and we have this in "bar.cpp":
void bar(int& x)
{
foo(x);
}
How can the compiler "know" that x is not changing (or IS changing, more appropriately) in bar?
I'm sure we can come up with something more complex, if this isn't complex enough.
It is impossible in general to for the compiler to determine if the variable will be changed, as have been pointed out.
When checking const-ness, the question of interest seems to be if the variable can be changed by a function. Even this is hard in languages that support pointers. You can't control what other code does with a pointer, it could even be read from an external source (though unlikely). In languages that restrict access to memory, these types of guarantees can be possible and allows for more aggressive optimization than C++ does.
To make the question more specific I suggest the following set of constraints may have been what the author of the book may have had in mind:
Assume the compiler is examining the behavior of a specific function with respect to const-ness of a variable. For correctness a compiler would have to assume (because of aliasing as explained below) if the function called another function the variable is changed, so assumption #1 only applies to code fragments that don't make function calls.
Assume the variable isn't modified by an asynchronous or concurrent activity.
Assume the compiler is only determining if the variable can be modified, not whether it will be modified. In other words the compiler is only performing static analysis.
Assume the compiler is only considering correctly functioning code (not considering array overruns/underruns, bad pointers, etc.)
In the context of compiler design, I think assumptions 1,3,4 make perfect sense in the view of a compiler writer in the context of code gen correctness and/or code optimization. Assumption 2 makes sense in the absence of the volatile keyword. And these assumptions also focus the question enough to make judging a proposed answer much more definitive :-)
Given those assumptions, a key reason why const-ness can't be assumed is due to variable aliasing. The compiler can't know whether another variable points to the const variable. Aliasing could be due to another function in the same compilation unit, in which case the compiler could look across functions and use a call tree to statically determine that aliasing could occur. But if the aliasing is due to a library or other foreign code, then the compiler has no way to know upon function entry whether variables are aliased.
You could argue that if a variable/argument is marked const then it shouldn't be subject to change via aliasing, but for a compiler writer that's pretty risky. It can even be risky for a human programmer to declare a variable const as part of, say a large project where he doesn't know the behavior of the whole system, or the OS, or a library, to really know a variable won't change.
Even if a variable is declared const, doesn't mean some badly written code can overwrite it.
// g++ -o foo foo.cc
#include <iostream>
void const_func(const int&a, int* b)
{
b[0] = 2;
b[1] = 2;
}
int main() {
int a = 1;
int b = 3;
std::cout << a << std::endl;
const_func(a,&b);
std::cout << a << std::endl;
}
output:
1
2
To expand on my comments, that book's text is unclear which obfuscates the issue.
As I commented, that book is trying to say, "let's get an infinite number of monkeys to write every conceivable C++ function which could ever be written. There will be cases where if we pick a variable that (some particular function the monkeys wrote) uses, we can't work out whether the function will change that variable."
Of course for some (even many) functions in any given application, this can be determined by the compiler, and very easily. But not for all (or necessarily most).
This function can be easily so analysed:
static int global;
void foo()
{
}
"foo" clearly does not modify "global". It doesn't modify anything at all, and a compiler can work this out very easily.
This function cannot be so analysed:
static int global;
int foo()
{
if ((rand() % 100) > 50)
{
global = 1;
}
return 1;
Since "foo"'s actions depends on a value which can change at runtime, it patently cannot be determined at compile time whether it will modify "global".
This whole concept is far simpler to understand than computer scientists make it out to be. If the function can do something different based on things can change at runtime, then you can't work out what it'll do until it runs, and each time it runs it may do something different. Whether it's provably impossible or not, it's obviously impossible.
Let's say I have a function:
template <bool stuff>
inline void doSomething() {
if(stuff) {
cout << "Hello" << endl;
}
else {
cout << "Goodbye" << endl;
}
}
And I call it like this:
doSomething<true>();
doSomething<false>();
It would pring out:
Hello
Goodbye
What I'm really wondering is does the compiler fully optimize this?
When I call the templated function with true, will it create a function that just outputs "Hello" and avoids the if statement and the code for "Goodbye"?
This would be really useful for this one giant function I just wrote that's supposed to be very optimized and avoid as many unnecessary if statement checks as possible. I have a very good feeling it would, at least in a release build with optimizations if not in a debug build with no optimizations.
Disclaimer: Noone can guarantee anything.
That said, this an obvious and easy optimization for any compiler. It's quite safe to say that it will be optimized away, unless the optimizer is, well, practically useless.
Since your "true" and "false" are constants, you are unambiguously creating an obvious dead branch in each class, and the compiler should optimize it away. Should is taken literally here - I would consider it a major, major problem if an "optimising" compiler did not do dead branch removal.
In other words, if your compiler cannot optimize this, it is the use of that compiler that should be evaluated, not the code.
So, I would say your gut feeling is correct: while yes, no "guarantees" as such can be made on each and every compiler, I would not use a compiler incapable of performing simplistic optimizations in any production environment, and of course not in any performance critical one. (In release builds of course).
So, use it. Any modern optimizing compiler will optimize it away because it is a trivial optimization. If in doubt, check disassembly, and if it is not optimized, change the compiler to something more modern.
In general, if you are writing any kind of performance-critical code, you must rely, at least to some extent, on compiler optimizations.
This is inherently up to the compiler, so you'd have to check the compiler's documentation or the generated code. But in simple cases like this, you can easily implement the optimization yourself:
template <bool stuff>
inline void doSomething();
template<>
inline void doSomething<true>() {
cout << "Hello" << endl;
}
template<>
inline void doSomething<false>() {
cout << "Goodbye" << endl;
}
But "optimization" isn't really the right word to use since this might actually degrade performance. It's only an optimization if it actually benefits your code performance.
Indeed, it really createa two functions, but
premature optimization is the root of all evil
especially if your changing your code structure because of a simple if statement. I doubt that this will affect performance. Also the boolean must be static, that means you cant take a runtime evaluated var and pass it to the function. How should the linker know which function to call? In this case youll have to manually evaluate it and call the appropiate function on your own.
Compilers are really good at constant folding. That is, in this case it would surprise me if the check would stay until after optimization. A non-optimized build might still have the check. The easiest way to verify is to create assembler output and check.
That said, it is worth noting that the compiler has to check both branches for correctness, even if it only ever uses one branch. This frequently shows up, e.g., when using slightly different algorithms for Random Access Iterators and other iterators. The condition would depend on a type-trait and one of the branches may fail to compile depending on operations tested for by the traits. The committee has discussed turning off this checking under the term static if although there is no consensus, yet, on how the features would look exactly (if it gets added).
If I understand you correctly you want (in essence) end up with 'two' functions that are optimised for either a true or a false input so that they don't need check that flag?
Aside from any trivial optimisation that may yield (I'm against premature otimisation - I believe in maintainability before measurement before optimisation), I would say why not refactor your function to actually be two functions? If they have common code then then that code could be refactored out too. However if the requirement is such that the refactoring is non optimal then I'd replace that with a #define refactoring.
If i declare a variable but not use it later in the program, the complier will give me a warning, and since "every warning should not be ignored", why the warning is there? and how can it cause a error? thanks!
First, a minor point: declaring a variable that's never used is a waste of memory, and thus is itself a bug.
Second, and more importantly: you took the trouble of writing out a declaration for a variable you then never used. Since you would not have bothered to declare a variable if you had no plan to use it, this suggests you've forgotten to use it! Is it possible you typed the wrong variable name in its place? Is it possible you forgot to perform a critical calculation whose result you'd store in that variable?
Of course, you might just have declared something you ended up not needing, which is why it's a warning and not an error, but it's easy to see situations where that warning can point you to an important piece of missing code, which would indeed be a bug.
It's there because maybe you meant to use the variable. You generally don't declare a variable and then use not it :)
It's a helpful warning, and it exists in most languages.
It may assist in detecting typos, where you accidentally used another variable instead of the one you meant to, this warning will remind you of the one that you haven't used :)
It doesn't cause an error. It causes a warning (normally only at higher levels on most compilers) because the variable isn't doing anything, which might indicate that you intended to use it but didn't. Sometimes that might indicate that your code is behaving incorrectly.
It's not that the code could fail as it is, just that the compiler's trying to warn you that you've done something a little odd.
It may indicate the presence of a bug, a variable that is declared but unused is obviously a programming error, since why else would it be there at all?
Are you asking about something like this?
int function() {
double x;
return 42;
}
There's a warning saying the variable x is unused because, well, it's unused. It is a warning because if a variable is unused, it does nothing*.
You may fall into this warning if you mistype or accidentally shadow a variable. For instance:
void printArgument10Times(int i) {
for(int i = 0; i < 10; ++i) {
std::cout << i << std::endl;
}
}
// ...
printArgument10Times(42);
Instead of printing 42 ten times, it prints 0 through 9. i was shadowed and was also unused. The compiler will hopefully tell you both these things.
*In C++ constructors and deconstructors are called, of course.