Under what situations would it be valid to compare shared_ptr instances instead of the underly lying type the shared_ptr manages?
As an example, would there ever be a situation where the size of personset being 2 would be valid after the following code has run?
shared_ptr<person> p0 = make_shared<person>(....);
shared_ptr<person> p1 = p0;
set<shared_ptr<person>> personset;
personset.insert(p0);
personset.insert(p1);
There is no viable reason to compare the instances. Infact shared_ptrs by default will perform equality/inequality comparators based via the underlying pointer to the control block (via .get method).
http://en.cppreference.com/w/cpp/memory/shared_ptr/operator_cmp
Related
For C++11, is there still a performance difference between the following?
(for std::map<Foo, std::vector<Bar> > as an example)
map[key] = myVector and map.emplace(key, myVector)
The part I'm not figuring out is the exact internal of operator[]. My understanding so far has been (when key doesn't exist):
Create a new key and the associated empty default vector in place inside the map
Return the reference of the associated empty vector
Assign myVector to the reference???
The point 3 is the part I couldn't understand, how can you assign a new value to a reference in the first place?
Though I cannot sort through point 3 I think somehow there's just a copy/move required. Assuming C++11 will be smart enough to know it's gonna be a move operation, is this whole "[]" assignment then already cheaper than insert()? Is it almost equivalent to emplace()? ---- default construction and move content over, versus construct vector with content directly in place?
There are a lot of differences between the two.
If you use operator[], then the map will default construct the value. The return value from operator[] will be this default constructed object, which will then use operator= to assign to it.
If you use emplace, the map will directly construct the value with the parameters you provide.
So the operator[] method will always use two-stage construction. If the default constructor is slow, or if copy/move construction is faster than copy/move assignment, then it could be problematic.
However, emplace will not replace the value if the provided key already exists. Whereas operator[] followed by operator= will always replace the value, whether there was one there or not.
There are other differences too. If copying/moving throws, emplace guarantees that the map will not be changed. By contrast, operator[] will always insert a default constructed element. So if the later copy/move assignment fails, then the map has already been changed. That key will exist with a default constructed value_type.
Really, performance is not the first thing you should be thinking about when deciding which one to use. You need to focus first on whether it has the desired behavior.
C++17 will provide insert_or_assign, which has the effect of map[] = v;, but with the exception safety of insert/emplace.
how can you assign a new value to a reference in the first place?
It's fundamentally no different from assigning to any non-const reference:
int i = 5;
int &j = i;
j = 30;
i == 30; //This is true.
Given the code:
#include <stdlib.h>
#include <stdint.h>
typedef struct { int32_t x, y; } INTPAIR;
typedef struct { int32_t w; INTPAIR xy; } INTANDPAIR;
void foo(INTPAIR * s1, INTPAIR * s2)
{
s2->y++;
s1->x^=1;
s2->y--;
s1->x^=1;
}
int hey(int x)
{
static INTPAIR dummy;
void *p = calloc(sizeof (INTANDPAIR),1);
INTANDPAIR *p1 = p;
INTPAIR *p2a = p;
INTPAIR *p2b = &p1->xy;
p2b->x = x;
foo(p2b,p2a);
int result= p2b->x;
free(p);
return result;
}
#include <stdio.h>
int main(void)
{
for (int i=0; i<10; i++)
printf("%d.",hey(i));
}
Behavior depends upon gcc optimization level, which implies that gcc thinks
this code invokes Undefined Behavior (the definition of "foo" collapses to nothing, but interestingly the definition of "hey" increments the value passed in). I'm not quite sure what if anything it does that runs afoul of the Standard's rules, though.
The code very deliberately and evilly constructs two pointers such that
s2a->y and s2b->x will alias, but the pointers are deliberately constructed in such a way that both identify legitimate potential objects of type INTPAIR. Because code used calloc to get the memory, all field members have legitimate initial defined values of zero. All accesses to the allocated memory are done via an int32_t member of an INTPAIR*.
I can understand why it would make sense for the Standard to forbid aliasing structure fields in this fashion, but I couldn't find anything in the Standard which actually does so. Is gcc operating in Standard-compliant fashion here, or is it violating some clause in the Standard which isn't referenced by Annex J.2 and doesn't use any of the terms I searched for?
UPDATE:
I felt this answer was OK, but not still a little imprecise, and not cut and dry as to what the UB was. After a lot of very interesting discussion and comments I have tried again with a new answer
The right part of the C99 standard is quoted in this answer. I'm copying it here for convenience. The question and several of the answers are quite thorough.
(C99; ISO/IEC 9899:1999 6.5/7:
An object shall have its stored value accessed only by an lvalue
expression that has one of the following types 73) or 88):
a type compatible with the effective type of the object,
a qualified version of a type compatible with the effective type of
the object,
a type that is the signed or unsigned type corresponding to the
effective type of the object,
a type that is the signed or unsigned type corresponding to a
qualified version of the effective type of the object,
an aggregate or union type that includes one of the aforementioned
types among its members (including, recursively, a member of a
subaggregate or contained union), or
a character type.
73) or 88) The intent of this list is to specify those circumstances in which an object may or may not be aliased.
What is an effective type then? (C99; ISO/IEC 9899:1999 6.5/6:
The effective type of an object for an access to its stored value is the declared type of the object, if any. 87) If a value is stored into an object having no declared type through an lvalue having a type that is not a character type, then the type of the lvalue becomes the effective type of the object for that access and for subsequent accesses that do not modify the stored value. If a value is copied into an object having no declared type using memcpy or memmove, or is copied as an array of character type, then the effective type of the modified object for that access and for subsequent accesses that do not modify the value is the effective type of the object from which the value is copied, if it has one. For all other accesses to an object having no declared type, the effective type of the object is simply the type of the lvalue used for the access.
87) Allocated objects have no declared type.
So at the line p2b->x = x the object at p+4 becomes of effective type INTPAIR. Is it aligned correctly? If it isn't then Undefined Behavior (UB). But to keep it interesting, assume it is as it must be in this case because of the layout of INTANDPAIR.
By the same analysis there are two 8 byte objects, p2a (s2) at #(p+4) and p2b #p. As your example is demonstrating the 2nd element of p2a and the first of p2b end up being aliased.
In the foo(), the object p2b #p+4 is accessed by the normal method via s1->x. But then the "stored value" of object p2b is also accessed by a side effect of modifying a different object p2a #p. Since this falls under none of the bullets of 6.5/7, it is UB. Note that 6.5/7 says only, so objects shall not be accessed in any other ways.
I think the main distinction is that the "object" in question is the whole structure p2a/s2 and p2b/s1, not the integer members. If you change the argument of the function to take the integers and alias them it works "fine" because the function can't know s1 and s2 alias. For example:
void foo2(int *s1, int *s2)
{
(*s2)++;
(*s1)^=1;
(*s2)--;
(*s1)^=1;
}
...
/*foo(p2b,p2a);*/
foo2((int*)p, (int*)p); /* or p+4 or whatever you want */
This more or less confirms that this is the way GCC chose to interpret things: modifying a member is modifying the whole struct object and that since side effects of modifying one object are not on the listed legal ways to indirectly modify a different object, whee! we can do whatever silly thing we feel like doing.
So whether GCC interprets the ambiguities in standard to decide that by deriving s1 and s2 pointers through different typed pointers and then accessing them constitutes indirectly accessing the memory via different original types via p1 and p or whether it interprets the standard in the way I'm suggesting that "object" s2->y modifies is not just the integer but the s2 object, it is UB either way. Or is GCC just being especially snarky and pointing out that if the standard doesn't very clearly specify the semantics of dynamically allocated yet overlapping objects, it is free to do whatever it wants because by definition it is "undefined".
I don't think at this microscopic level anyone other than the standards body can definitively answer whether this should be UB or not because at this level it requires some "interpretation". The GCC's implementers opinion's seem to favor very aggressive interpretations.
I like Linus's reaction to this whole thing. And it is true, why not just be conservative and let the programmer tell the compiler when it is safe? Very Excellent Linus Rant
My previous answer was lacking, maybe not completely wrong, but the sample program is deliberately designed to sidestep each of the more obvious explicit Undefined Behaviors (UB) dictated by the C99 standard, like 6.5/7. But with both GCC (and Clang) this example demonstrates strict aliasing failure like symptoms under optimization. They appear to be assuming s1->y and s2-x can't alias. So, is the compiler wrong? Is this a loophole in the strict aliasing legalese?
Short answer: No. I wouldn't be surprised if there was a loophole of some kind in the standard, given its complexity. But in this example, creating overlapping objects on the heap is explicitly undefined behavior, and there are several other things happening that the standard does not define.
I think the point of the example is not that it fails - it is obvious that "playing fast and loose" with pointers is a bad idea and relying on corner cases and legalese to prove the compile "wrong" is of little help if the code doesn't work. The key questions are: is GCC wrong? and what in the standard says so.
First, lets look at the obvious strict aliasing rules and how this example is trying to avoid them.
C99 6.5/7:
An object shall have its stored value accessed only by an lvalue expression that has one of the following types: 76)
a type compatible with the effective type of the object,
a qualified version of a type compatible with the effective type of the object,
a type that is the signed or unsigned type corresponding to the effective type of the object,
a type that is the signed or unsigned type corresponding to a qualified version of the effective type of the object,
an aggregate or union type that includes one of the aforementioned types among its members (including, recursively, a member of a subaggregate or contained union), or
a character type.
This is the main strict aliasing section. It means that accessing the same memory via two different type pointers is UB. This example sidesteps it by accessing both using INTPAIR pointers in foo().
The key problem with this is that it is talking about accessing the stored value via two different effective types (e.g. pointers). It doesn't talk about accessing via two different objects.
What is being accessed? is it the integer member or the entire object s1 / s2? Is accessing s2->x via s1->y access via "a type compatible with the effective type of the object". I believe an argument can be made that a) the access as a side effect of modifying a different object does not fall under the permissible methods in 6.5/7 and that b) modifying one member of the aggregate transitively modifies the aggregate (*s1 or *s2) also.
Since this is not specified, it is UB, but it is a bit hand-wavy.
How did we get pointers to two overlapping objects? Are the pointer casts leading to them OK? Section 6.3.2.3 contains the rules for casting pointers and the example carefully does not violate any of them. In particular, because p2b is a pointer to INTANDPAIR member xy the alignment is guaranteed to be right, otherwise it would definitely run afoul of 6.3.2.3/7.
Furthermore, &p1->xy is not a problem - it can't be - it is a perfectly legitimate pointer to an INTPAIR. Simply casting pointers and/or taking addresses is safely outside the definition of "access" (3.1/1).
It is clear that the problem comes about by accessing two integer members that overlay each other as different parts of overlapping objects. Any attempt to do this via pointers of different types would clearly run afoul of 6.5/7. If accessed by the same type pointer at the same address, there would be no problem whatsoever. So the only way left that they could alias this way is that if two objects at different addresses overlapped in some fashion.
Obviously this could occur as part of a union, but that is not the case for this example. Type punning through unions may not be UB in C99, but it would be a different question whether a variant of this example could be made misbehave via unions.
The example uses dynamic allocation and casts the resultant void pointer to two different types. Going from from a pointer to an object to void * and back again is valid (6.3.2.3/1). Several other ways of obtaining pointers to objects that would overlap are explicitly UB by the pointer conversion rules of 6.3.2.3, the aliasing rules of 6.5/7, and/or the compatible type rules 6.2.7.
So what else is wrong?
6.2.4 Storage durations of objects
1 An object has a storage duration that determines its lifetime. There are three storage durations: static, automatic, and allocated. Allocated storage is described in 7.20.3
The storage for each of the objects is allocated by calloc() so the duration we want is "allocated". So we check 7.20.3: (emphasis added)
7.20.3 Memory management functions
1 The order and contiguity of storage allocated by successive calls to the calloc, malloc, and realloc functions is unspecified. The pointer returned if the allocation succeeds is suitably aligned so that it may be assigned to a pointer to any type of object and then used to access such an object or an array of such objects in the space allocated (until the space is explicitly deallocated). The lifetime of an allocated object extends from the allocation until the deallocation. Each such allocation shall yield a pointer to an object disjoint from any other object.
...
2 The lifetime of an object is the portion of program execution during which storage is guaranteed to be reserved for it. An object exists, has a constant address, 25) and retains its last-stored value throughout its lifetime. 26) If an object is referred to outside of its lifetime, the behavior is undefined.
To avoid UB, the accesses to the two different objects must be to a valid object within its lifetime. You can get a single valid object (or an array) with malloc()/calloc(), but these guarantee that you will receive a pointer disjoint from all other objects. So is the object returned from calloc() p or is it p1? It can't be both.
The UB is triggered by attempting to reuse the same dynamically allocated object to hold two objects that are not disjoint. While calloc() guarantees it will return a pointer to a disjoint object, there is nothing that says it will still work if you then start using parts of the buffer for a 2nd overlapping one. In fact, it even explicitly says it is UB if you access an object outside its lifetime and there is only a single allocation ergo a single lifetime.
Also note:
4. Conformance
In this International Standard, ‘‘shall’’ is to be interpreted as a requirement on an implementation or on a program; conversely, ‘‘shall not’’ is to be interpreted as a prohibition.
If a ‘‘shall’’ or ‘‘shall not’’ requirement that appears outside of a constraint is violated, the behavior is undefined. Undefined behavior is otherwise indicated in this International Standard by the words ‘‘undefined behavior’’ or by the omission of any explicit definition
of behavior. There is no difference in emphasis among these three; they all describe ‘‘behavior that is undefined’’.
For this to be a compiler error it must fail on a program that only uses constructs explicitly defined. Anything else is outside the safe-harbor and is still undefined, even if it the standard doesn't explicitly state that it is Undefined Behavior.
How can a container be both contiguous and support move semantics at the same time?
An example with std::vector:
When push_back() is called using std::move on an lvalue:
std::vector<MyClass> v;
MyClass obj;
MyClass obj2;
vt.push_back(std::move(obj));
vt.push_back(std::move(obj2));
obj and obj2 are not necessarily allocated next to each other in memory. Now as vector must have its elements in contiguous memory, how would move semantics work in this case? It seems to me that it must copy the obj2 to vector v's contiguous memory region (next to obj), otherwise the contiguosity requirement would not be satisfied. But that requires a copy, not a move. What is the difference between the above and this then?:
MyClass obj;
MyClass obj2;
vt.push_back(std::move(obj));
vt.push_back(obj2);
You just need to read a little bit more about move semantics :-)
Moving an object does not change the address of the object itself. It merely calls the move constructor (or assignment operator, etc. depending on context) of another instance of the object, which is passed the one that is to be moved.
In this example, the vector does create two MyClass objects within its internal storage, one per push_back. But for the first one, instead of calling the copy constructor and passing obj (via MyClass const&), it calls the move constructor and passes the rvalue reference (MyClass&&) to obj. It is then up to that constructor to move the contents of the obj object into the one inside the vector.
In other words, the objects themselves are being created in the vector, and its only their contents that are moved (what it means to 'move' an object is potentially different for each type, hence it's the move constructor's job to do the actual moving).
Note that even with std::move, the move constructor may not be called -- there may not be one, for example, or there may be one that isn't noexcept (in which case std::vector can't use it in all circumstances without violating its exception safety guarantees).
Your std::move is a cast, it's an unconditional cast and it can easily decay in a copy if you are not using it with the proper semantics.
In other words writing std::move doesn't guarantee you anything other than a T&& type which by itself doesn't guarantee you move semantics .
I am using std::shared_ptr in C++11 and I would like to understand if it's better to assign structures of type T in this way:
T a_data;
std::shared_ptr<T> my_pointer(new T);
*my_pointer = a_data;
or like:
memcpy(&my_pointer, data, sizeof(T));
or like:
my_pointer.reset(a_data);
Regards
Mike
They each do a different thing.
1.
T a_data;
std::shared_ptr<T> my_pointer(new T);
*my_pointer = a_data;
Here, a new object (call it n) of type T will be allocated, managed by my_pointer. Then, object a_data will be copy-assigned into n.
2.
memcpy(&my_pointer, a_data, sizeof(T)); // I assume you meant a_data here, not data
That's total nonsense - tha's overwriting the shared_ptr itself with the contents of a_data. Undefined behaviour at its finest (expect a crash or memory corruption).
Perhaps you actually meant my_pointer.get() instead of &my_pointer (that is, you wanted to copy into the object being pointed to)? If that's the case, it can work, as long as T is trivially copyable - which means that it doesn't have non-trivial copy or move ctors, doesn't have non-trivial copy or move assignment operators, and has a trivial destructor. But why rely on that, when normal assignment (*my_pointer = a_data;) does exactly the same for that case, and also works for non-trivially-copyable classes?
3.
my_pointer.reset(a_data);
This normally won't compile as-is, it would need to be my_pointer.reset(&a_data);. That's disaster waiting to happen - you point my_pointer to the automatic (= local) variable a_data and give it ownership of that. Which means that when my_pointer goes out of scope (actually, when the last pointer sharing ownership wiht it does), it will call the deleter, which normally calls delete. On a_data, which was not allocated with new. Welcome to UB land again!
If you just need to manage a dynamically-allocated copy of a_data with a shared_ptr, do this:
T a_data;
std::shared_ptr<T> my_pointer(new T(a_data));
Or even better:
T a_data;
auto my_pointer = std::make_shared<T>(a_data);
I have a public method which uses a variable (only in the scope of the public method) I pass as a parameter we will call A, this method calls a private method multiple times which also requires the parameter.
At present I am passing the parameter every time but it looks weird, is it bad practice to make this member variable of the class or would the uncertainty about whether it is initialized out way the advantages of not having to pass it?
Simplified pseudo code:
public_method(parameter a)
do something with a
private_method(string_a, a)
private_method(string_b, a)
private_method(string_c, a)
private_method(String, parameter a)
do something with String and a
Additional information: parameter a is a read only map with over 100 entries and in reality I will be calling private_method about 50 times
I had this same problem myself.
I implemented it differently in 3 different contexts to see hands-on what are result using 3 different strategies, see below.
Note that I am type of programmer that makes many changes to the code always trying to improve it. Thus I settle only for the code that is amenable to changes, readbale, would you call this "flexible" code. I settle only for very clear code.
After experimentation, I came to these results:
Passing a as parameter is perfectly OK if you have one or two - short number - of such values. Passing in parmeters has very good visibility, clarity, clear passing lines, well visible lifetime (initialization points, destruction points), amenable to changes, easy to track.
If number of such values begin to grow to >= 5-6 values, I swithc to approach #3 below.
Passing values through class members -- did not do good to clarity of my code, eventually I got rid of it. It makes for less clear code. Code becomes muddled. I did not like it. It had no advantages.
As alternative to (1) and (2), I adopted Inner class approach, in cases when amount of such values is > 5 (which makes for too long argument list).
I pack those values into small Inner class and pass such object by reference as argument to all internal members.
Public function of a class usually creates an object of Inner class (I call is Impl or Ctx or Args) and passes it down to private functions.
This combines clarity of arg passing with brevity. It's perfect.
Good luck
Edit
Consider preparing array of strings and using a loop rather than writing 50 almost-identical calls. Something like char *strings[] = {...} (C/C++).
This really depends on your use case. Does 'a' represent a state that your application/object care about? Then you might want to make it a member of your object. Evaluate the big picture, think about maintenance, extensibility when designing structures.
If your parameter a is a of a class of your own, you might consider making the private_method a public method for the variable a.
Otherwise, I do not think this looks weird. If you only need a in just 1 function, making it a private variable of your class would be silly (at least to me). However, if you'd need it like 20 times I would do so :P Or even better, just make 'a' an object of your own that has that certain function you need.
A method should ideally not pass more than 7 parameters. Using the number of parameters more than 6-7 usually indicates a problem with the design (do the 7 parameters represent an object of a nested class?).
As for your question, if you want to make the parameter private only for the sake of passing between private methods without the parameter having anything to do with the current state of the object (or some information about the object), then it is not recommended that you do so.
From a performance point of view (memory consumption), reference parameters can be passed around as method parameters without any significant impact on the memory consumption as they are passed by reference rather than by value (i.e. a copy of the data is not created). For small number of parameters that can be grouped together you can use a struct. For example, if the parameters represent x and y coordinates of a point, then pass them in a single Point structure.
Bottomline
Ask yourself this question, does the parameter that you are making as a members represent any information (data) about the object? (data can be state or unique identification information). If the answer to his question is a clear no, then do not include the parameter as a member of the class.
More information
Limit number of parameters per method?
Parameter passing in C#