I ran into a problem with a dirty shutdown of my program, because of a shared pointer. I found a solution, but I'm not sure, if I have the right answer.
This is the minimalistic example:
double var;
boost::shared_ptr<const double> ptr1 (&var); // creates an invalid pointer on program exit
boost::shared_ptr<const double> ptr2 (new double); // works fine
int main (int argc, char **argv)
{
return 0;
}
Here is my answer which I would like to varify:
In the case of ptr1 the pointed-to object will be deleted before the pointer, which then points to an invalid address. But in the case of ptr2 the "smartness" of the shared pointer handles the issue above.
True?
Small extra question:
Is there a way to make ptr1 work (I tried reset()) or is that bad programming practice (if so why)?
Thanks for the clarification!
in the first case, the pointer isn't dynamically allocated, so the shared_ptr should not attempt to destroy the underlying object. that can be done by using a custom no-op deleter functor:
http://www.boost.org/doc/libs/1_51_0/libs/smart_ptr/sp_techniques.html#static
Related
I was wondering if there is a way to access a data member within a struct that is being pointed to by a void*? What I'm trying to explain will hopefully be more apparent in my example code:
int main()
{
struct S
{
int val;
};
S s;
s.val = 5;
void* p;
p = malloc(sizeof(S));
*(struct S*) p = s;
std::cout<< *(struct S*)p.val << std::endl;
}
I have ran this exact code casting p as *(int*)p and it printed fine, however, using exact code above results in a compilation error. Haven't been able to find an example that quite accomplishes this task. Is it possible to access the data members of the struct after it is casted? why or why not? if so, how?
The . operator has higher precedence than a C-style cast. So *(struct S*)p.val is treated as *((struct S*)(p.val)), which doesn't make sense since p is a pointer and does not have members.
So you need parentheses to specify what you intended:
std::cout<< (*(struct S*)p).val << std::endl;
Or equivalently,
std::cout<< static_cast<S*>(p)->val << std::endl;
[But also: the statement *(struct S*) p = s; technically has undefined behavior, even though all most implementations will allow it. This is because C++ has rules about when an object is created, and there was no object of type S previously at that address, and assignment does not create an object except for some cases involving union members. A similar statement that does not have this problem would be new(p) S{s};.
Also also: use of malloc or void* is usually not a good idea in C++ in the first place. malloc should only be used when interfacing with a C library that requires it. Anything for which void* seems useful can probably be done more safely using templates. In a few cases a void* might be the only way to do something or "cleverly" avoid code duplication or something, but still use it sparingly and always with extreme caution.]
Ok, muddling though Stack on the particulars about void*, books like The C Programming Language (K&R) and The C++ Programming Language (Stroustrup). What have I learned? That void* is a generic pointer with no type inferred. It requires a cast to any defined type and printing void* just yields the address.
What else do I know? void* can't be dereferenced and thus far remains the one item in C/C++ from which I have discovered much written about but little understanding imparted.
I understand that it must be cast such as *(char*)void* but what makes no sense to me for a generic pointer is that I must somehow already know what type I need in order to grab a value. I'm a Java programmer; I understand generic types but this is something I struggle with.
So I wrote some code
typedef struct node
{
void* data;
node* link;
}Node;
typedef struct list
{
Node* head;
}List;
Node* add_new(void* data, Node* link);
void show(Node* head);
Node* add_new(void* data, Node* link)
{
Node* newNode = new Node();
newNode->data = data;
newNode->link = link;
return newNode;
}
void show(Node* head)
{
while (head != nullptr)
{
std::cout << head->data;
head = head->link;
}
}
int main()
{
List list;
list.head = nullptr;
list.head = add_new("My Name", list.head);
list.head = add_new("Your Name", list.head);
list.head = add_new("Our Name", list.head);
show(list.head);
fgetc(stdin);
return 0;
}
I'll handle the memory deallocation later. Assuming I have no understanding of the type stored in void*, how do I get the value out? This implies I already need to know the type, and this reveals nothing about the generic nature of void* while I follow what is here although still no understanding.
Why am I expecting void* to cooperate and the compiler to automatically cast out the type that is hidden internally in some register on the heap or stack?
I'll handle the memory deallocation later. Assuming I have no understanding of the type stored in void*, how do I get the value out?
You can't. You must know the valid types that the pointer can be cast to before you can dereference it.
Here are couple of options for using a generic type:
If you are able to use a C++17 compiler, you may use std::any.
If you are able to use the boost libraries, you may use boost::any.
Unlike Java, you are working with memory pointers in C/C++. There is no encapsulation whatsoever. The void * type means the variable is an address in memory. Anything can be stored there. With a type like int * you tell the compiler what you are referring to. Besides the compiler knows the size of the type (say 4 bytes for int) and the address will be a multiple of 4 in that case (granularity/memory alignment). On top, if you give the compiler the type it will perform consistency checks at compilation time. Not after. This is not happening with void *.
In a nutshell, you are working bare metal. The types are compiler directives and do not hold runtime information. Nor does it track the objects you are dynamically creating. It is merely a segment in memory that is allocated where you can eventually store anything.
The main reason to use void* is that different things may be pointed at. Thus, I may pass in an int* or Node* or anything else. But unless you know either the type or the length, you can't do anything with it.
But if you know the length, you can handle the memory pointed at without knowing the type. Casting it as a char* is used because it is a single byte, so if I have a void* and a number of bytes, I can copy the memory somewhere else, or zero it out.
Additionally, if it is a pointer to a class, but you don't know if it is a parent or inherited class, you may be able to assume one and find out a flag inside the data which tells you which one. But no matter what, when you want to do much beyond passing it to another function, you need to cast it as something. char* is just the easiest single byte value to use.
Your confusion derived from habit to deal with Java programs. Java code is set of instruction for a virtual machine, where function of RAM is given to a sort of database, which stores name, type, size and data of each object. Programming language you're learning now is meant to be compiled into instruction for CPU, with same organization of memory as underlying OS have. Existing model used by C and C++ languages is some abstract built on top of most of popular OSes in way that code would work effectively after being compiled for that platform and OS. Naturally that organization doesn't involve string data about type, except for famous RTTI in C++.
For your case RTTI cannot be used directly, unless you would create a wrapper around your naked pointer, which would store the data.
In fact C++ library contains a vast collection of container class templates that are useable and portable, if they are defined by ISO standard. 3/4 of standard is just description of library often referred as STL. Use of them is preferable over working with naked pointers, unless you mean to create own container for some reason. For particular task only C++17 standard offered std::any class, previously present in boost library. Naturally, it is possible to reimplement it, or, in some cases, to replace by std::variant.
Assuming I have no understanding of the type stored in void*, how do I get the value out
You don't.
What you can do is record the type stored in the void*.
In c, void* is used to pass around a binary chunk of data that points at something through one layer of abstraction, and recieve it at the other end, casting it back to the type that the code knows it will be passed.
void do_callback( void(*pfun)(void*), void* pdata ) {
pfun(pdata);
}
void print_int( void* pint ) {
printf( "%d", *(int*)pint );
}
int main() {
int x = 7;
do_callback( print_int, &x );
}
here, we forget thet ype of &x, pass it through do_callback.
It is later passed to code inside do_callback or elsewhere that knows that the void* is actually an int*. So it casts it back and uses it as an int.
The void* and the consumer void(*)(void*) are coupled. The above code is "provably correct", but the proof does not lie in the type system; instead, it depends on the fact we only use that void* in a context that knows it is an int*.
In C++ you can use void* similarly. But you can also get fancy.
Suppose you want a pointer to anything printable. Something is printable if it can be << to a std::ostream.
struct printable {
void const* ptr = 0;
void(*print_f)(std::ostream&, void const*) = 0;
printable() {}
printable(printable&&)=default;
printable(printable const&)=default;
printable& operator=(printable&&)=default;
printable& operator=(printable const&)=default;
template<class T,std::size_t N>
printable( T(&t)[N] ):
ptr( t ),
print_f( []( std::ostream& os, void const* pt) {
T* ptr = (T*)pt;
for (std::size_t i = 0; i < N; ++i)
os << ptr[i];
})
{}
template<std::size_t N>
printable( char(&t)[N] ):
ptr( t ),
print_f( []( std::ostream& os, void const* pt) {
os << (char const*)pt;
})
{}
template<class T,
std::enable_if_t<!std::is_same<std::decay_t<T>, printable>{}, int> =0
>
printable( T&& t ):
ptr( std::addressof(t) ),
print_f( []( std::ostream& os, void const* pt) {
os << *(std::remove_reference_t<T>*)pt;
})
{}
friend
std::ostream& operator<<( std::ostream& os, printable self ) {
self.print_f( os, self.ptr );
return os;
}
explicit operator bool()const{ return print_f; }
};
what I just did is a technique called "type erasure" in C++ (vaguely similar to Java type erasure).
void send_to_log( printable p ) {
std::cerr << p;
}
Live example.
Here we created an ad-hoc "virtual" interface to the concept of printing on a type.
The type need not support any actual interface (no binary layout requirements), it just has to support a certain syntax.
We create our own virtual dispatch table system for an arbitrary type.
This is used in the C++ standard library. In c++11 there is std::function<Signature>, and in c++17 there is std::any.
std::any is void* that knows how to destroy and copy its contents, and if you know the type you can cast it back to the original type. You can also query it and ask it if it a specific type.
Mixing std::any with the above type-erasure techinque lets you create regular types (that behave like values, not references) with arbitrary duck-typed interfaces.
I'm still learning C++, and I'm doing some API work, but I'm, having trouble parsing this pointer arrangement.
void* data;
res = npt.receive(0x1007, params, 1, response, (void**)&data, size);
uint32_t* op = (uint32_t*)data;
uint32_t num = *op;
op++;
Can anyone explain what is going on with that void pointer? I see it being defined, it does something in the res line(maybe initialized?), then it's copied to an uint32 pointer, and dereferenced in num. Can anyone help me parse the (void**)&data declaration?
Pay attention when you use the void pointer:
The void type of pointer is a special type of pointer. In C++, void represents the absence of type. Therefore, void pointers are pointers that point to a value that has no type (and thus also an undetermined length and undetermined dereferencing properties).
This gives void pointers a great flexibility, by being able to point to any data type, from an integer value or a float to a string of characters. In exchange, they have a great limitation: the data pointed to by them cannot be directly dereferenced (which is logical, since we have no type to dereference to), and for that reason, any address in a void pointer needs to be transformed into some other pointer type that points to a concrete data type before being dereferenced.
From C++ reference
Firstly: What is npt?
Secondly: Guessing what npt could be some explanation:
// Declare a pointer to void named data
void* data;
// npt.receive takes as 5th parameter a pointer to pointer to void,
// which is why you provide the address of the void* using &data.
// The void ** appears to be unnecessary unless the data type of the
// param is not void **
// What is "npt"?
res = npt.receive(0x1007, params, 1, response, (void**)&data, size);
// ~.receive initialized data with contents.
// Now make the uint32_t data usable by casting void * to uint32_t*
uint32_t* op = (uint32_t*)data;
// Use the data by dereferencing it.
uint32_t num = *op;
// Pointer arithmetic: Move the pointer by sizeof(uint32_t).
// Did receive fill in an array?
op++;
Update
Signature of receive is:
<whatever return type> receive(uint16_t code, uint32_t* params, uint8_t nparam, Container& response, void** data, uint32_t& size)
So the data parameter is of type void** already so the explicit type cast to void** using (void**) is not necessary.
Considering the usage, the received data appears to be an array of uint32_t values IN THIS CASE!
Void as a type means no type and no type information regarding size and alignment is available, but is mandatory for lexical and syntactical consistency.
In conjunction with the *, it can be used as a pointer to data of unknown type and must be explicitly cast to another type (adds type information) before any use.
You usually have a void* or void** in an API, if you dont know the specific data type or only received plain byte data.
To understand this please read up C type erasure using void*
Please read up as basics before:
Dynamically allocated C arrays.
Pointers and Pointer Arithmetics.
From the code, ntp.receive tells you whether it receives anything successfully in the return code but it also needs to give you what it receives. It has a pointer that it wants to pass back, so you have to tell it where that pointer is so that it can fill it, hence (void **), a pointer to a pointer, being the address of your pointer, &data.
When you have received it, you know as the developer that what it points to is actually a uint_32 value so you copy the void pointer into one that points to a uint_32. In fact, this step is unnecessary since you could have cast the uint_32 pointer to void** in the above call but we'll let that slide.
Now that you have told the compiler that the pointer points to a 32 bit number, you can take the number on the other end of that pointer (*op) and store it in a local variable. Again, unnecessary, as *op could be used anywhere num is subsequently used.
Hope this helps.
I wrote simple code to help me understand smart pointers:
string s = "str";
vector <unique_ptr<string>> pv ;
pv.push_back(unique_ptr<string>(&s));
cout<<*(pv[0])<<endl;
This code compiles fine, but gets me a runtime error:
str
* Error in `...': munmap_chunk(): invalid pointer: 0x00007ffd956e57e0 * Aborted (core dumped)
What happened and what have I done wrong?
In the std::unique_ptr's destructor it will call delete on the &s pointer which was not allocated via new.
Just use:
std::vector<std::string> vector;
vector.emplace_back("str");
std::cout << pv[0] << std::endl;
There's no need for std::unique_ptr<std::string> there.
Your string is being destructed twice - once when your pv goes out of scope and is deleted, freeing all its contained unique_ptr elements, and once when s goes out of scope.
To use a vector of unique pointers (or to use unique pointers in general), it is essential that they are not aliased. So you could write:
auto *s = new std::string("str");
pv.push_back(std::unique_ptr<std::string>(s));
// do not write "delete s" anywhere...
Or, simpler and safer:
pv.push_back(std::make_unique<std::string>("str")); // make_unique is C++14
Or even:
std::unique_ptr<std::string> p{new std::string("str")};
pv.push_back(std::move(p));
// Do not attempt to use p beyond this point.
I would like to use MS function to send data.
I didnt find examples where they send other type of data other than const char * .
I tried to send a int, or other, but I failed.
WSASend() and send() both function only take a Char* parameters.
How should i proceed ?
Thanks
Its just a pointer to a buffer, this buffer may contains anything you want.
This char pointer is actually an address to a bytes array, this function requires a length parameter too.
An integer is a 2/4 (short/long) bytes value,
Then if you want to send an integer variable (for example) you have to pass its address, and its length.
WSASend and send are simple functions that send a memory block.
I assume you are talking about C, you have to understand that C's char variables are bytes - 8 bits block, char variables contain any value between 0 and 255.
A pointer to a char var is an address to a byte (which maybe the first cell of a bytes array).
I think thats what confuses you.
I hope you understand.
The const char* parameter indicates that the function is taking a pointer to bytes. Witch really seems to be the result of the original socket api designers being pedantic - C has a generic type to handle any kind of pointer without explicit casts: void*.
You could make a convenience wrapper for send like this - which would allow you to send any (contiguous) thing you can make a pointer to:
int MySend(SOCKET s, const void* buf, int len,int flags)
{
return send(s,(const char*)buf,len,flags);
}
Using void* in place of char* actually makes the api safer, as it can now detect when you do something stupid:
int x=0x1234;
send(s,(const char*)x,sizeof(x),0); // looks right, but is wrong.
mysend(s,x,sizeof(x),0); // this version correctly fails
mysend(s,&x,sizeof(x),0); // correct - pass a pointer to the buffer to send.
WSASend is a bit more tricky to make a convenience wapper for as you have to pass it an array of structs that contain the char*'s - but again its a case of defining an equivalent struct with const void*'s in place of the const char*'s and then casting the data structures to the WSA types in the convenience wrapper. Get it right once, and the rest of the program becomes much easier to determine correct as you don't need casts everywhere hiding potential bugs.