I was reading some notes on C++ and came across the following question
Q: Can a constructor throws an exception? How to handle the error when the constructor fails?
A:The constructor never throws an error.
Now this answer kind of confused me and just to make sure I went online and read that you could always throw exceptions from a constructor. I just wanted to make sure if this way a typo in the notes and that I may not be missing something important.
The notes are wrong, if they're talking about constructors in general. Ctors can indeed throw normally. Perhaps that was discussing a particular class which guarantees nonthrow construction?
On the other hand, it's strongly encouraged to code so that your destructors never throw. It's legal for them to do so, but throwing an exception during stack unwinding results in immediate program termination (a call to std::terminate).
Of course, it can. Even standard containers (e.g. std::vector) throw std::bad_alloc from constructors. This FAQ is telling the truth.
You can throw an exception from a constructor, but be careful: if object is not properly constructed, destructor will not be called.
class Foo
{
public:
Foo(int i)
{
throw i;
}
~Foo()
{
std::cout << "~Foo()" << std::endl;
}
};
int main()
{
try
{
Foo f(42);
}
catch(...)
{
std::cout << "Catched" << std::endl;
}
return 0;
}
Output:
Catched
To fix this you should encapsulate one constructor into another:
Foo()
{
}
Foo(int i): Foo()
{
throw i;
}
Output:
~Foo()
Catched
Related
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Can a local variable’s memory be accessed outside its scope?
Code:
#include <iostream>
using namespace std;
class B{
public:
int b;
B():b(1){}
~B(){cout << "Destructor ~B() " << endl;}
};
class A{
public:
B ob;
A()try{throw 4;}
catch(...){cout << "Catched in A() handler : ob.b= " << ob.b<<endl;}
};
int main()try{
A t;
}
catch(...){cout << "CATCHED in Main" << endl;}
Output:
Destructor ~B()
Catched in A() handler : ob.b= 1
CATCHED in Main
My Question is how it's possible to access a member variable b of an object ob that its destructor call finished.
Using a destructed object is undefined behaviour. This means that you may be getting this behaviour now, but nothing guarantees that you will get it some other time. Undefined behaviour is more dangerous than a regular bug because it can be more difficult to detect, as this example shows.
UPDATE: Following some comments, I'll explain why OP's code produces that output.
try function blocks are a somewhat obscure feature in C++. You can surround the whole body of a function inside a try block, with its corresponding catch. This is, instead of:
void foo()
{
try
{
//...
}
catch (/*whatever*/)
{
//...
}
}
You can write:
void foo()
try
{
//...
}
catch (/*whatever*/)
{
//...
}
This is not too useful, really, but can be marginally useful for constructors, because this is the only way to include the initialisation list inside the try block. Thus, OP's code for the A constructor is equivalent to:
A()
try
: b()
{
throw 4;
}
catch(...)
{
cout << "Catched in A() handler : ob.b= " << ob.b<<endl;
}
I said this is just marginally useful because you cannot use the object you are constructing inside the catch block; if you throw inside the try block, the exception will leave the constructor body, so the object will have never been constructed and any constructed data member will be immediately destructed before entering the catch block. But it might have some use for logging purposes.
Now, as we all know, a constructor (asuming we are not using the nothrow version) can only do two things:
Return a constructed object
Throw an exception
In this constructor we throw, but the exception is caught in the catch block. So what happens now? What will be returned to the code calling the constructor? We cannot return a constructed object because we have none, so there is only one alternative: the catch block must throw. And this is actually what the standard mandates in this case. If we do not throw explicitly, the compiler will silently add a throw; instruction at the end of the catch block. So, elaborating a bit more, the constructor is equivalent to the following:
A()
try
: b()
{
throw 4;
}
catch(...)
{
cout << "Catched in A() handler : ob.b= " << ob.b<<endl;
throw;
}
And this is the reason why the exception is caught twice: once in A constructor and once in main().
Because while the object has been destroyed, the actual memory that it occupied still exists. However, it's undefined behavior, and may or may not work.
That might be a bug in your compiler.
When I ran your code, the destructor is called in the proper order. The output is:
Catched in A() handler : ob.b= 1
Destructor ~B()
And I can't imagine any reason why catch in main is executed. The exception was already caught in A::A().
Update
I was confused by the edit. Originally, it was initializer list try/catch syntax which makes a difference. Now I can reproduce OP's output and it is UB indeed. I got a crash.
C++ standard says ([except.handle]/9):
If no matching handler is found, the function std::terminate() is called; whether or not the stack is unwound before this call to std::terminate() is implementation-defined
For example, behavior of code below (will it print S::~S() or not) is implementation defined:
struct S {
S() { std::cout << "S::S()" << std::endl; }
~S() { std::cout << "S::~S()" << std::endl; }
};
int main() {
S s;
throw std::runtime_error("exception");
}
I would like to know in-depth: why is it implementation defined? Why a context cannot be unwinded up to its entry before std::terminate() is called if an exception is uncaught (which is similar to try{ ... } catch(...) { throw; } in the top-level function)? At a glance, such behavior is much clearer and safer in consistence with RAII.
If an exception isn't caught, std::terminate is called. We failed so bad the host environment needs to step in and (maybe) clean up after us. Unwinding the stack in this case is like giving a helmet to a kamikaze pilot.
So for a hosted environment, it may make more sense to just do nothing and let the host clean up.
Now, if you are in a stand alone implementation, and are throwing exceptions, then there is no one to clean up after you. An implementation should preform stack unwinding in this case, because that's what is supposed to clean up the mess.
The standard leaves it to the implementation to facilitate these two very different execution environments.
Like #Matteo pointed out, std::terminate is called before any potential unwinding because you can setup a handler for it. And that handler can do something useful with the stack state, so long as the stack isn't unwound yet.
Not the strongest reason per se, but leaving this up to the implementation allows for more optimizations. E.g.:
class Foo { /*...*/ };
void f();
int main()
{
for ( int i = 0; i < 1000000; ++i )
{
Foo myFoo;
f();
}
}
Here, an implementation may choose not to destroy myFoo if f() throws, which may reduce code size and/or increase performance. The rationale would be if you don't write an exception handler, you don't expect f() to throw anyway, and shortcuts may be taken. This may sound a bit weak, but this is similar to noexcept (C++11) vs. throw() clause (C++98) – removing the requirement to unwind the stack allows for more aggressive optimization.
Why is the destructor not invoked in this code?
#include <boost/scoped_ptr.hpp>
#include <iostream>
class MyClass {
boost::scoped_ptr<int> ptr;
public:
MyClass() : ptr(new int) { *ptr = 0; throw; std::cout<<"MyClass Allocated\n"; }
~MyClass() { std::cout<<"MyClass De-allocated\n"; }
int increment() { return ++*ptr; }
};
int main()
{
boost::scoped_ptr<MyClass> myinst(new MyClass);
std::cout << myinst->increment() << '\n';
std::cout << myinst->increment() << '\n';
}
EDIT
From the answers, In understand that when an exception happens in the constructor, destructor will not be invoked. But if the exception happens in the main(), ie after the MyClass object is fully instantiated, will the MyClass destructor be invoked? If not, then why it is a smart pointer?
Adding the code
#include <boost/scoped_ptr.hpp>
#include <iostream>
class MyClass {
boost::scoped_ptr<int> ptr;
public:
MyClass() : ptr(new int) { *ptr = 0; std::cout<<"MyClass Allocated\n"; }
~MyClass() { std::cout<<"MyClass De-allocated\n"; }
int increment() { return ++*ptr; }
};
int main()
{
boost::scoped_ptr<MyClass> myinst(new MyClass);
throw 3;
std::cout << myinst->increment() << '\n';
std::cout << myinst->increment() << '\n';
}
Output:
MyClass Allocated
terminate called after throwing an instance of 'int'
Aborted
A C++ object's lifetime begins only after its constructor completes successfully.
Since the exception was thrown before constructor call was complete you don't have an complete object and hence no destructor.
Herb Sutter explains this nicely, to quote him:
Q: What does emitting an exception from a constructor mean?
A: It means that construction has failed, the object never existed, its lifetime never began. Indeed, the only way to report the failure of construction -- that is, the inability to correctly build a functioning object of the given type -- is to throw an exception. (Yes, there is a now-obsolete programming convention that said, "if you get into trouble just set a status flag to 'bad' and let the caller check it via an IsOK() function." I'll comment on that presently.)
In biological terms,
conception took place -- the constructor began -- but despite best efforts it was followed by a miscarriage -- the constructor never ran to term(ination).
Incidentally, this is why a destructor will never be called if the constructor didn't succeed -- there's nothing to destroy. "It cannot die, for it never lived." Note that this makes the phrase "an object whose constructor threw an exception" really an oxymoron. Such a thing is even less than an ex-object... it never lived, never was, never breathed its first. It is a non-object.
We might summarize the C++ constructor model as follows:
Either:
(a) The constructor returns normally by reaching its end or a return statement, and the object exists.
Or:
(b) The constructor exits by emitting an exception, and the object not only does not now exist, but never existed as an object.
EDIT 1:
But if the exception happens in the main(), ie after the MyClass object is fully instantiated, will the MyClass destructor be invoked?
Yes, it will be!
That is the purpose of using scoped_ptr, Once an exception is thrown in main, Stack Unwinding would cause all local objects to be deallocated, this means that myinst(which resides on stack) will also be deallocated, which in turn will call the destructor of MyClass.
Refer the Boost doccumentation when in doubt:
The scoped_ptr class template stores a pointer to a dynamically allocated object. (Dynamically allocated objects are allocated with the C++ new expression.) The object pointed to is guaranteed to be deleted, either on destruction of the scoped_ptr, or via an explicit reset
EDIT 2:
Why does your edited program crash?
Your program shows crashes because, You throw an exception but you never catch it. when such a scenario occurs an special function called terminate() is called whose default behavior is to call abort().It is implementation defined behavior whether stack is Unwound before terminate() is called in this particular scenarioRef 1.Seems your implementation doesn't & you should not rely on this behavior as well.
You can modify your program as follows to handle the exception and you should get the behavior you were expecting:
#include <boost/scoped_ptr.hpp>
#include <iostream>
class MyClass {
boost::scoped_ptr<int> ptr;
public:
MyClass() : ptr(new int) { *ptr = 0; std::cout<<"MyClass Allocated\n"; }
~MyClass() { std::cout<<"MyClass De-allocated\n"; }
int increment() { return ++*ptr; }
};
void doSomething()
{
boost::scoped_ptr<MyClass> myinst(new MyClass);
throw 3;
}
int main()
{
try
{
doSomething();
}
catch(int &obj)
{
std::cout<<"Exception Handled";
}
}
Ref1C++03 15.5.1 The terminate() function
In the following situations exception handling must be abandoned for less subtle error handling techniques:
....
— when the exception handling mechanism cannot find a handler for a thrown exception (15.3),
....
In such cases,
void terminate();
is called (18.6.3). In the situation where no matching handler is found, it is implementation-defined whether or not the stack is unwound before terminate() is called. In all other situations, the stack shall not be unwound before terminate() is called. An implementation is not permitted to finish stack unwinding prematurely based on a determination that the unwind process will eventually cause a call to terminate().
Because calling the destructor doesn't make sense in this case.
You only destruct things which are constructed, yet your object never fully constructs. Your class members have been constructed, though, and will have their destructors called.
If a constructor throws exception, then the destructor of the class will not be called, because the object is not fully constructed.
See this link how to manage resources in such situation:
http://www.parashift.com/c++-faq-lite/exceptions.html#faq-17.10
When the exception is thrown from the constructor (beginning or half way or at the end of the call), then it's assured that the object is not constructed.
So it's well defined not to invoke the destructor of an object which was never constructed.
Here is one related FAQ from Bjarne's website.
The destructor for MyClass was never invoked because no objects of type MyClass were ever constructed. Each attempt to construct one was aborted, due to the exception being thrown.
As an aside, if you want your debug messages to display -- especially if you're dealing with the program crashing -- you really ought to flush the streams: i.e. using std::endl instead of '\n' at the end of line. (or inserting std::flush)
While merely using '\n' often works, there are enough situations where it fails and it's really, really confusing to debug if you don't make a habit of doing things right.
Please have a look at demo code :
class myError
{
const char* str;
public:
myError():str(NULL) {}
myError(const char* temp)
{
str = temp;
}
const char* what()
{
return str;
}
};
class ab
{
int x;
public:
ab() try :x(0)
{
throw myError("error occured in the constructor of class ab");
}
catch(myError& temp)
{
std::cout<<"Handler no. 1 of ab constructor"<<std::endl;
}
};
int main () try
{
ab bb;
cout << "Resumed execution!" << endl;
return 0;
}
catch(myError& temp)
{
std::cout<<"Handler below the main function"<<std::endl;
std::cout<<"And the error is :" <<temp.what();
}
My Questions :
Why only function try block's handler of ctor and dtor only rethows the exception? ,
and when you simply throw exception inside ctor , its handler doesn't rethrows the object? i.e
Ctor::Ctor()
{
try{
throw Excep1();
}
catch(Excep1& temp) {
std::cout<<"Doesn't rethrows the exception object";
}
}
I wanna know that how to resume the control back to cout << "Resumed execution!" << endl; , after handling rethrown object?
why is it often said that we shouldn't place function try block on dtor of base class?
The usual rule is that a catch block doesn't rethrow unless you ask it
to. How would you stop the exception from propagating otherwise. In
the case of a constructor, however, if something in the initialization
list throws, then you haven't got a fully constructed object; there is
nothing you could do with the object, not even call the destructor on
it. And if the function catch block of the constructor doesn't
rethrow, what is it going to do, since it cannot simply return (and
leave the variable on the stack)?
In all other cases, it's up do the function containing the catch block
to know what to do. In the case of your main, for example, you could
write:
try {
ab bb;
} catch (...) {
}
std::cout << "Resumed execution!" << std::endl;
What you can't do is execute code where bb would be in scope and
accessible, but not have been correctly constructed.
As for why you shouldn't place a function try block on the destructor of a
base class, I've never heard that rule. In general, destructors
shouldn't throw, so there's no point in wrapping them in a try block,
period.
For the second question, destructors shouldn't be throwing period. Consider a case where your destructor is freeing a lot of memory through delete. What would happen if your destructor threw an error before finishing the clean up? You now have a memory leak. If your destructor is causing a runtime error, then you probably have problems elsewhere in your code that you need to fix.
As usually, Herb Sutter knows and explains everything:
If the handler body contained the statement "throw;" then the catch block would obviously rethrow whatever exception A::A() or B::B() had emitted. What's less obvious, but clearly stated in the standard, is that if the catch block does not throw (either rethrow the original exception, or throw something new), and control reaches the end of the catch block of a constructor or destructor, then the original exception is automatically rethrown.
More in his article
Is it possible to handle exceptions in these scenarios:
thrown from constructor before entering main()
thrown from destructor after leaving main()
You can wrap up your constructor withing a try-catch inside of it.
No, you should never allow exception throwing in a destructor.
The funny less-known feature of how to embed try-catch in a constructor:
object::object( int param )
try
: optional( initialization )
{
// ...
}
catch(...)
{
// ...
}
Yes, this is valid C++. The added benefit here is the fact that the try will catch exceptions thrown by the constructors of the data members of the class, even if they're not mentioned in the ctor initializer or there is no ctor initializer:
struct Throws {
int answer;
Throws() : answer(((throw std::runtime_error("whoosh!")), 42)) {}
};
struct Contains {
Throws baseball;
Contains() try {} catch (std::exception& e) { std::cerr << e.what() << '\n'; }
};
Yes: don't use dangerous global objects!
It might be possible to set an exception handler before construction / destruction of the objects in question, that one should be able to handle those exceptions.
For constructors, there's some weird new syntax that allows exceptions to be caught within the constructor. Not sure how that works, and it's not commonly implemented in many compilers.
For destructors, you have to wrap the content of the destructor in a try { code(); } catch(...) {} block. Which may not always be the desired behavior, depending on what you want to achieve in that destructor.
Short answer: no.
Any global object that throws an exception in its constructor will cause an unhandled exception (that is, terminate be called).
Any class that throws an exception in its destructor is a broken class.
Using the singleton pattern rather than globals will give you more options.