Lambdas can't be compared - c++11

I was wondering if this is Standard, or a bug in my code. I'm trying to compare a pair of my homegrown function objects. I rejected the comparison if the type of function object is not the same, so I know that the two lambdas are the same type. So why can't they be compared?

Every C++0x lambda object has a distinct type, even if the signature is the same.
auto l1=[](){}; // one do-nothing lambda
auto l2=[](){}; // and another
l1=l2; // ERROR: l1 and l2 have distinct types
If two C++0x lambdas have the same type, they must therefore have come from the same line of code. Of course, if they capture variables then they won't necessarily be identical, as they may have come from different invocations.
However, a C++0x lambda does not have any comparison operators, so you cannot compare instances to see if they are indeed the same, or just the same type. This makes sense when you think about it: if the captured variables do not have comparison operators then you cannot compare lambdas of that type, since each copy may have different values for the captured variables.

Is the equality operator overloaded for your lambda object? If not I'm assuming you'll need to implement it.

Related

How are the C++11 variadic templates used in practice?

I recently came across the topic of C++11 variadic templates, and I'm wondering in what situations are they actually used? Are they primarily intended for recursion, where the number of parameters can vary (or is unknown)? Are there any other use cases apart from recursion?
A good case for variadic templates is emplace (and emplace_back, emplace_front et.al.), i.e. constructing an element of a container in the storage the container allocated, with whatever constructor arguments the type takes.
Without that you would have to copy or move an existing object into that storage.
E.g.
std::vector<std::string> strings;
strings.emplace_back(10, 'a');
More generally, whenever you have something that wants to pass along arguments it was given to an expression that can take varying amounts of parameters, you'll want to use a variadic template.

Is there a difference between fun(n::Integer) and fun(n::T) where T<:Integer in performance/code generation?

In Julia, I most often see code written like fun(n::T) where T<:Integer, when the function works for all subtypes of Integer. But sometimes, I also see fun(n::Integer), which some guides claim is equivalent to the above, whereas others say it's less efficient because Julia doesn't specialize on the specific subtype unless the subtype T is explicitly referred to.
The latter form is obviously more convenient, and I'd like to be able to use that if possible, but are the two forms equivalent? If not, what are the practicaly differences between them?
Yes Bogumił Kamiński is correct in his comment: f(n::T) where T<:Integer and f(n::Integer) will behave exactly the same, with the exception the the former method will have the name T already defined in its body. Of course, in the latter case you can just explicitly assign T = typeof(n) and it'll be computed at compile time.
There are a few other cases where using a TypeVar like this is crucially important, though, and it's probably worth calling them out:
f(::Array{T}) where T<:Integer is indeed very different from f(::Array{Integer}). This is the common parametric invariance gotcha (docs and another SO question about it).
f(::Type) will generate just one specialization for all DataTypes. Because types are so important to Julia, the Type type itself is special and allows parameterization like Type{Integer} to allow you to specify just the Integer type. You can use f(::Type{T}) where T<:Integer to require Julia to specialize on the exact type of Type it gets as an argument, allowing Integer or any subtypes thereof.
Both definitions are equivalent. Normally you will use fun(n::Integer) form and apply fun(n::T) where T<:Integer only if you need to use specific type T directly in your code. For example consider the following definitions from Base (all following definitions are also from Base) where it has a natural use:
zero(::Type{T}) where {T<:Number} = convert(T,0)
or
(+)(x::T, y::T) where {T<:BitInteger} = add_int(x, y)
And even if you need type information in many cases it is enough to use typeof function. Again an example definition is:
oftype(x, y) = convert(typeof(x), y)
Even if you are using a parametric type you can often avoid using where clause (which is a bit verbose) like in:
median(r::AbstractRange{<:Real}) = mean(r)
because you do not care about the actual value of the parameter in the body of the function.
Now - if you are Julia user like me - the question is how to convince yourself that this works as expected. There are the following methods:
you can check that one definition overwrites the other in methods table (i.e. after evaluating both definitions only one method is present for this function);
you can check code generated by both functions using #code_typed, #code_warntype, #code_llvm or #code_native etc. and find out that it is the same
finally you can benchmark the code for performance using BenchmarkTools
A nice plot explaining what Julia does with your code is here http://slides.com/valentinchuravy/julia-parallelism#/1/1 (I also recommend the whole presentation to any Julia user - it is excellent). And you can see on it that Julia after lowering AST applies type inference step to specialize function call before LLVM codegen step.
You can hint Julia compiler to avoid specialization. This is done using #nospecialize macro on Julia 0.7 (it is only a hint though).

C++ Eigen: dereferencing an InnerIterator

Matrix<T, Dynamic, Dynamic>::InnerIterator doesn't seem to use the standard way (operator *) to dereference it.
It has a member function value().
I am rather surprised, as this would not interface well with third party algorithms, like STL algorithms.
Why it doesn't declare operator *? Even worse is that value() does not return by reference, so you can't change the underlying value of the pointed element. What is the proper way to use it?
This is because most of the times, the value alone in useless without the respective inner/row/column indices, as returned by index()/row()/col(). In other cases, you only care about the indices. Anyway, you can easily write a little wrapper to make it compatible with STL if needed.
Finally, if you want to modify the value, there is a valueRef() method returning by non const reference.

Why should I not use identity based operations on Optional in Java8?

The java.util.Optional javadoc states that:
This is a value-based class; use of identity-sensitive operations (including reference equality (==), identity hash code, or synchronization) on instances of Optional may have unpredictable results and should be avoided.
However, this junit snippet is green. Why? It seems to contradict the javadoc.
Optional<String> holder = Optional.ofNullable(null);
assertEquals("==", true, holder == Optional.<String>empty());
assertEquals("equals", true, holder.equals(Optional.<String>empty()));
You shouldn’t draw any conclusions from the observed behavior of one simple test ran under a particular implementation. The specification says that you can’t rely on this behavior, because the API designers reserve themselves the option to change the behavior at any time without notice.
The term Value-based Class already provides a hint about the intended options. Future versions or alternative implementations may return the same instance when repeated calls with the same value are made, or the JVM might implement true value types for which identity based operations have no meaning.
This is similar to instances returned by the boxing valueOf methods of the wrapper types. Besides the guaranty made for certain (small) values, it is unspecified whether new instances are created or the same instance is returned for the same value.
“A program may produce unpredictable results if it attempts to distinguish two references to equal values of a value-based class…” could also imply that the result of a reference comparison may change during the lifetime of two value-based class instances. Think of a de-duplication feature. Since the JVM already has such a feature for the internal char[] array of Strings, the idea of expanding this feature to all instances of “value-based classes” isn’t so far-fetched.
Optional.of() makes a new Object for every non-null value. So it is very likely that comparing 2 Optionals will be two references even if the Optional refers to the same value.
However, your example shows that Optional.empty() re-uses the same singleton instance. That is probably the only time the same Optional instance is ever returned twice.

Move Semantics in Golang

This from Bjarne Stroustrup's The C++ Programming Language, Fourth Edition 3.3.2.
We didn’t really want a copy; we just wanted to get the result out of
a function: we wanted to move a Vector rather than to copy it.
Fortunately, we can state that intent:
class Vector {
// ...
Vector(const Vector& a); // copy constructor
Vector& operator=(const Vector& a); // copy assignment
Vector(Vector&& a); // move constructor
Vector& operator=(Vector&& a); // move assignment
};
Given that definition, the compiler will choose the move constructor
to implement the transfer of the return value out of the function.
This means that r=x+y+z will involve no copying of Vectors. Instead,
Vectors are just moved.As is typical, Vector’s move constructor is
trivial to define...
I know Golang supports traditional passing by value and passing by reference using Go style pointers.
Does Go support "move semantics" the way C++11 does, as described by Stroustrup above, to avoid the useless copying back and forth? If so, is this automatic, or does it require us to do something in our code to make it happen.
Note: A few answers have been posted - I have to digest them a bit, so I haven't accepted one yet - thanks.
The breakdown is like here:
Everything in Go is passed by value.
But there are five built-in "reference types" which are passed by value as well but internally they hold references to separately maintained data structure: maps, slices, channels, strings and function values (there is no way to mutate the data the latter two reference).
Your own answer, #Vector, is incorrect is that nothing in Go is passed by reference. Rather, there are types with reference semantics. Values of them are still passed by value (sic!).
Your confusion suppsedly stems from the fact your mind is supposedly currently burdened by C++, Java etc while these things in Go are done mostly "as in C".
Take arrays and slices for instance. An array is passed by value in Go, but a slice is a packed struct containing a pointer (to an underlying array) and two platform-sized integers (the length and the capacity of the slice), and it's the value of this structure which is copied — a pointer and two integers — when it's assigned or returned etc. Should you copy a "bare" array, it would be copied literally — with all its elements.
The same applies to channels and maps. You can think of types defining channels and maps as declared something like this:
type Map struct {
impl *mapImplementation
}
type Slice struct {
impl *sliceImplementation
}
(By the way, if you know C++, you should be aware that some C++ code uses this trick to lower exposure of the details into header files.)
So when you later have
m := make(map[int]string)
you could think of it as m having the type Map and so when you later do
x := m
the value of m gets copied, but it contains just a single pointer, and so both x and m now reference the same underlying data structure. Was m copied by reference ("move semantics")? Surely not! Do values of type map and slice and channel have reference semantincs? Yes!
Note that these three types of this kind are not at all special: implementing your custom type by embedding in it a pointer to some complicated data structure is a rather common pattern.
In other words, Go allows the programmer to decide what semantics they want for their types. And Go happens to have five built-in types which have reference semantics already (while all the other built-in types have value semantics). Picking one semantics over the other does not affect the rule of copying everything by value in any way. For instance, it's fine to have pointers to values of any kind of type in Go, and assign them (so long they have compatible types) — these pointers will be copied by value.
Another angle to look at this is that many Go packages (standard and 3rd-party) prefer to work with pointers to (complex) values. One example is os.Open() (which opens a file on a filesystem) returning a value of the type *os.File. That is, it returns a pointer and expects the calling code to pass this pointer around. Surely, the Go authors might have declared os.File to be a struct containing a single pointer, essentially making this value have reference semantics but they did not do that. I think the reason for this is that there's no special syntax to work with the values of this type so there's no reason to make them work as maps, channels and slices. KISS, in other words.
Recommended reading:
"Go Data Structures"
"Go Slices: Usage and Internals"
Arrays, slices (and strings): The mechanics of 'append'"
A thead on golang-nuts — pay close attention to the reply by Rob Pike.
The Go Programming Language Specification
Calls
In a function call, the function value and arguments are evaluated in
the usual order. After they are evaluated, the parameters of the call
are passed by value to the function and the called function begins
execution. The return parameters of the function are passed by value
back to the calling function when the function returns.
In Go, everything is passed by value.
Rob Pike
In Go, everything is passed by value. Everything.
There are some types (pointers, channels, maps, slices) that have
reference-like properties, but in those cases the relevant data
structure (pointer, channel pointer, map header, slice header) holds a
pointer to an underlying, shared object (pointed-to thing, channel
descriptor, hash table, array); the data structure itself is passed by
value. Always.
Always.
-rob
It is my understanding that Go, as well as Java and C# never had the excessive copying costs of C++, but do not solve ownership transference to containers. Therefore there is still copying involved. As C++ becomes more of a value-semantics language, with references/pointers being relegated to i) smart-pointer managed objects inside classes and ii) dependence references, move semantics solves the problem of excessive copying. Note that this has nothing to do with "pass by value", nowadays everyone passes objects by Reference (&) or Const Reference (const &) in C++.
Let's look at this (1) :
BigObject BO(big,stuff,inside);
vector<BigObject> vo;
vo.reserve(1000000);
vo.push_back(BO);
Or (2)
vector<BigObject> vo;
vo.reserve(1000000);
vo.push_back(BigObject(big,stuff,inside));
Although you're passing by reference to the vector vo, in C++03 there was a copy inside the vector code.
In the second case, there is a temporary object that has to be constructed and then is copied inside the vector. Since it can only be accessed by the vector, that is a wasteful copy.
However, in the first case, our intent could be just to give control of BO to the vector itself. C++17 allows this:
(1, C++17)
vector<BigObject> vo;
vo.reserve(1000000);
vo.emplace_back(big,stuff,inside);
Or (2, C++17)
vector<BigObject> vo;
vo.reserve(1000000);
vo.push_back(BigObject(big,stuff,inside));
From what I've read, it is not clear that Java, C# or Go are exempt from the same copy duplication that C++03 suffered from in the case of containers.
The old-fashioned COW (copy-on-write) technique, also had the same problems, since the resources will be copied as soon as the object inside the vector is duplicated.
Stroustrup is talking about C++, which allows you to pass containers, etc by value - so the excessive copying becomes an issue.
In Go, (like in Delphi, Java, etc) when you pass a container type, etc they are always references, so it's a non-issue. Regardless, you don't have to deal with it or worry about in GoLang - the compiler just does what it needs to do, and from what I've seen thus far, it's doing it right.
Tnx to #KerrekSB for putting me on the right track.
#KerrekSB - I hope this is the right answer. If it's wrong, you bear no responsibility.:)

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