function that modifies object pointed to by std::unique_ptr<T> - c++11

Somewhere in my code I have a local std::unique_ptr<T>. I need to do stuff with the object pointed at, and I use a function for that:
std::unique_ptr<T> some_function( std::unique_ptr<T> &t )
{
// do stuff
return t;
}
I call the function like this:
std::unique_ptr<T> some_T_ptr( new T(/*args*/) );
vector_of_finalized_Ts.push_back( std::move(some_function(std::move(some_T_ptr))));
Now I wonder, is there a better way to get the necessary functionality? It just seems two moves are pretty superfluous and potentially dangerous. I do have error handling code I'm not showing here, but that's beside the point.

It is all about ownership. Do you want some_function to take ownership of the pointer or not? If not, you can just pass a raw pointer to some_function. If you want some_function to take ownership (and return ownership), then it should take the unique_ptr by value. Otherwise the return statement (which should be std::move(t)) will be moving from a reference of unknown origins.
std::unique_ptr<T> some_function( std::unique_ptr<T> t )
{
// I own t here and will delete it if an exception happens
// do stuff
// I'm transferring ownership back to the caller
// (who may or may not accept ownership)
return std::move(t);
}
vector_of_finalized_Ts.push_back( some_function(std::move(some_T_ptr)));
or:
void some_function( T* t )
{
// I don't own t and won't delete it if an exception happens
// do stuff
}
some_function(some_T_ptr.get());
vector_of_finalized_Ts.push_back( std::move(some_T_ptr));
Either design is fine. It just depends on what code should own the pointer (especially if an exception is thrown at some point).

(Ignoring the unrelated syntax error in your code. See my comment above for that.)
As far as your snippet goes, your code is valid. The verbosity of the moves is the price you pay for using std::unique_ptr in this manner, and for passing the unique_ptr into the function rather than a reference to the object itself.
I suppose you have your good reasons for wanting some_function to take a std::unique_ptr and, if that's the case, then as far as I can tell you can't really do any better.
If you don't have your good reasons then, well, there's your answer. :)
Hope that helps.

Your problem is that you both take the unique_ptr by reference and return it. It's a unique pointer- you're treating it liked a shared pointer, and you're left with a wasted nullptr value on the stack. If you took it by value or by rvalue reference, you could just call some_function directly, and you don't have to move the result.
std::unique_ptr<T> some_function( std::unique_ptr<T> &&t )
{
// do stuff
return t;
}
vector_of_finalized_Ts.push_back( some_function(std::unique_ptr<T>(new T(...))));

Related

When does c++ right value destruct in this scenario?

Here is the code:
class SomeType {
public:
SomeType() {}
~SomeType() {}
std::string xxx;
}
bool funtion_ab() {
SomeType(); // This is a right val;
// The right val destructs here when I test the code. I want to make sure that it would always destructs here.
int a = 0, b = 10;
....// other code
return true;
}
Please tell me if you know the truth. Thank you!
What you have is called a temporary object. From §6.7.7,
Temporary objects are created
when a prvalue is converted to an xvalue
or, more specifically,
[Note 3: Temporary objects are materialized:
...
when a prvalue that has type other than cv void appears as a discarded-value expression ([expr.context]).
— end note]
and, on the lifetime, the same section has this to say
Temporary objects are destroyed as the last step in evaluating the full-expression ([intro.execution]) that (lexically) contains the point where they were created.
You can read more about the expression semantics, but in your case "full-expression" is fairly unambiguous.
SomeType();
The "full-expression" containing your constructor call is... the constructor call itself. So the destructor will be called immediately after evaluating the constructor. There are some exceptions to this rule (such as if the temporary object is thrown as an exception or is bound as a reference), but none of those apply here.
As noted in the comments, compilers are free to inline your constructor and destructor calls and then are free to notice that they do nothing and omit them entirely. Optimizers can do fun stuff with your code, provided it doesn't change the semantics. But a strict reading of the standard states that the destructor is called exactly where you suggested.

Accessing object T in vector<unique_ptr<T>> without taking ownership

I have the following member variable in a class:
std::vector<std::unique_ptr<Object>> objects_;
I explicitly want the vector to maintain ownership at all times. I've seen suggestions that in order for a member function to access a pointer in the vector and make changes to the object T wrapped in the std::unique_ptr, we must move the pointer to calling code, i.e:
void foo(int i) {
auto object = std::move( vector.at( i ) ); // move object to caller (caller owns)
object->dosomething();
vector.at(i) = std::move(object); // move back into vector (vector owns)
}
Another method was to work with raw pointers:
void foo(int i) {
Object* object = vector.at( i ).get();
object->doSomething();
}
However, I've been working with this:
void foo(int i) {
auto& object = vector.at( i );
object->doSomething();
}
Which is the correct and most robust method for my case? Does this function ever take ownership of the data in the std::unique_ptr? Is there a way to access Object without playing with the std::unique_ptr?
(excuse me if my methods are incorrect, I hope I got the point across)
The first approach will not retain ownership of the object if object->dosomething() throws an exception (i.e. it is not exception safe) since the second std::move() statement will not be executed.
Assuming C++11, both of the other approaches are effectively equivalent, subject to the assumption that the owned pointer is not null. Under the same assumption, the code can be simplified to
void foo(int i)
{
vector.at(i)->doSomething();
}
which will work with all C++ standards (not just C++11 or later).
It is possible to access the object without monkeying with the unique_ptr - simply store the pointer elsewhere and use that. However, that does compromise the purpose of using std::unique_ptr in the first place. And is error-prone - for example, the std::unique_ptr can destroy the object, and leave those other pointers dangling.
If you are really that worried about the potential of your vector losing ownership, consider using a shared_ptr instead.

Is there a way to make a moved object "invalid"?

I've some code that moves an object into another object. I won't need the original, moved object anymore in the upper level. Thus move is the right choice I think.
However, thinking about safety I wonder if there is a way to invalidate the moved object and thus preventing undefined behaviour if someone accesses it.
Here is a nice example:
// move example
#include <utility> // std::move
#include <vector> // std::vector
#include <string> // std::string
int main () {
std::string foo = "foo-string";
std::string bar = "bar-string";
std::vector<std::string> myvector;
myvector.push_back (foo); // copies
myvector.push_back (std::move(bar)); // moves
return 0;
}
The description says:
The first call to myvector.push_back copies the value of foo into the
vector (foo keeps the value it had before the call). The second call
moves the value of bar into the vector. This transfers its content
into the vector (while bar loses its value, and now is in a valid but
unspecified state).
Is there a way to invalidate bar, such that access to it will cause a compiler error? Something like:
myvector.push_back (std::move(bar)); // moves
invalidate(bar); //something like bar.end() will then result in a compiler error
Edit: And if there is no such thing, why?
Accessing the moved object is not undefined behavior. The moved object is still a valid object, and the program may very well want to continue using said object. For example,
template< typename T >
void swap_by_move(T &a, T &b)
{
using std::move;
T c = move(b);
b = move(a);
a = move(c);
}
The bigger picture answer is because moving or not moving is a decision made at runtime, and giving a compile-time error is a decision made at compile time.
foo(bar); // foo might move or not
bar.baz(); // compile time error or not?
It's not going to work.. you can approximate in compile time analysis, but then it's going to be really difficult for developers to either not get an error or making anything useful in order to keep a valid program or the developer has to make annoying and fragile annotations on functions called to promise not to move the argument.
To put it a different way, you are asking about having a compile time error if you use an integer variable that contains the value 42. Or if you use a pointer that contains a null pointer value. You might be succcessful in implementing an approximate build-time code convention checker using clang the analysis API, however, working on the CFG of the C++ AST and erroring out if you can't prove that std::move has not been called till a given use of a variable.
Move semantics works like that so you get an object in any it's correct state. Correct state means that all fields have correct value, and all internal invariants are still good. That was done because after move you don't actually care about contents of moved object, but stuff like resource management, assignments and destructors should work OK.
All STL classes (and all classed with default move constructor/assignment) just swap it's content with new one, so both states are correct, and it's very easy to implement, fast, and convinient enough.
You can define your class that has isValid field that's generally true and on move (i. e. in move constructor / move assignment) sets that to false. Then your object will have correct state I am invalid. Just don't forget to check it where needed (destructor, assignment etc).
That isValid field can be either one pointer having null value. The point is: you know, that object is in predictable state after move, not just random bytes in memory.
Edit: example of String:
class String {
public:
string data;
private:
bool m_isValid;
public:
String(string const& b): data(b.data), isValid(true) {}
String(String &&b): data(move(b.data)) {
b.m_isValid = false;
}
String const& operator =(String &&b) {
data = move(b.data);
b.m_isValid = false;
return &this;
}
bool isValid() {
return m_isValid;
}
}

How can I move a shared_ptr's data?

I have an easy question about shared pointers and move semantics. Imagine that I have a class with a private member variable like this:
class C
{
private:
std::shared_ptr<std::vector<uint8_t>> buffer;
}
I need to provide public getters and setters. The getter seems obvious:
std::shared_ptr<std::vector<uint8_t>> C::GetBuffer()
{
return buffer;
}
However, being new to C++ I'm having trouble writing the setter. I could do something like this:
void C::SetBuffer(std::shared_ptr<std::vector<uint8_t>> input)
{
buffer = input;
}
However that results in a copy of input to buffer, but I don't really want the caller to have shared ownership. Instead I want to move the data. I tried to solve this with:
void C::SetBuffer(std::shared_ptr<std::vector<uint8_t>>& input)
{
buffer(std::move(input));
}
But this is an error: "call of an object of a class type without appropriate operator() or conversion functions to pointer-to-function type."
Can somebody help me understand:
1. What is going on here?
2. How to best implement the setter?
You can fix the error you're getting by writing this:
void C::SetBuffer( std::shared_ptr<std::vector<uint8_t> > &input ) {
buffer = move(input);
}
This will call shared_ptr's move-assignment operator, which will pilfer input. However, this won't really stop the caller from having shared ownership. Once you accept (or dispense) a shared_ptr from/to an unknown client, you don't have much in the way of control about who shares ownership. Even if input is pilfered, there's no reason to expect that input was the only copy of the shared_ptr you just received. If, for example, the function that called SetBuffer() took whatever became input from its caller by value, that higher-level copy of the pointer will continue to share ownership.
Note that your getter has a similar issue. You're returning a shared_ptr to your own internal object (and what's more, it's a shared_ptr-to-non-const, so the client can modify the shared state) and wherever that shared_ptr gets passed around after you provide it, those copies will also share (mutable) ownership.
If you really want to ensure you have exclusive ownership, you can hold a unique_ptr instead of a shared_ptr and have your getter pass back a const-reference, and your setter take either a unique_ptr or a value.
If your goal is to allow a caller to pass sole ownership of a buffer to your object, you should accept it by unique_ptr instead of shared_ptr:
void C::SetBuffer(std::unique_ptr<std::vector<uint8_t>> input)
{
buffer = std::move(input);
}
Rvalue unique_ptr is convertible to shared_ptr for exactly this purpose.

How to convert between shared_ptr<FILE> to FILE* in C++?

I am trying to use a FILE pointer multiple times through out my application
for this I though I create a function and pass the pointer through that. Basically I have this bit of code
FILE* fp;
_wfopen_s (&fp, L"ftest.txt", L"r");
_setmode (_fileno(fp), _O_U8TEXT);
wifstream file(fp);
which is repeated and now instead I want to have something like this:
wifstream file(SetFilePointer(L"ftest.txt",L"r"));
....
wofstream output(SetFilePointer(L"flist.txt",L"w"));
and for the function :
FILE* SetFilePointer(const wchar_t* filePath, const wchar_t * openMode)
{
shared_ptr<FILE> fp = make_shared<FILE>();
_wfopen_s (fp.get(), L"ftest.txt", L"r");
_setmode (_fileno(fp.get()), _O_U8TEXT);
return fp.get();
}
this doesn't simply work. I tried using &*fp instead of fp.get() but still no luck.
You aren't supposed to create FILE instances with new and destroy them with delete, like make_shared does. Instead, FILEs are created with fopen (or in this case, _wfopen_s) and destroyed with fclose. These functions do the allocating and deallocating internally using some unspecified means.
Note that _wfopen_s does not take a pointer but a pointer to pointer - it changes the pointer you gave it to point to the new FILE object it allocates. You cannot get the address of the pointer contained in shared_ptr to form a pointer-to-pointer to it, and this is a very good thing - it would horribly break the ownership semantics of shared_ptr and lead to memory leaks or worse.
However, you can use shared_ptr to manage arbitrary "handle"-like types, as it can take a custom deleter object or function:
FILE* tmp;
shared_ptr<FILE> fp;
if(_wfopen_s(&tmp, L"ftest.txt", L"r") == 0) {
// Note that we use the shared_ptr constructor, not make_shared
fp = shared_ptr<FILE>(tmp, std::fclose);
} else {
// Remember to handle errors somehow!
}
Please do take a look at the link #KerrekSB gave, it covers this same idea with more detail.

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