Writing a function implementing a Jacobi Matrix for Newton's method I have noticed a really nasty error.
Calling the function
auto DF = [T](VectorXd y){
return PhiAndW(y(0),y(1),T).second - MatrixXd::Identity(2,2);
};
only returns the value of PhiAndW(y(0),y(1),T).second and omits the subtraction of MatrixXd::Identity(2,2). But if I change the code to
auto DF = [T](VectorXd y){
MatrixXd mat = PhiAndW(y(0),y(1),T).second - MatrixXd::Identity(2,2);
return mat;
};
everything works seamlessly.
I have tried to reproduce it and this is not the exact same behavior but it also doesn't behave as wanted:
Consider the following code:
MatrixXd FF(MatrixXd y){
return y;
}
int other(){
auto DF = [](MatrixXd y){
MatrixXd test = FF(y) - MatrixXd::Identity(2,2);
return test;
};
std::cout << DF(MatrixXd::Ones(2,2)) <<std::endl;
std::cout << std::endl;
std::cout << (MatrixXd::Ones(2,2) - MatrixXd::Identity(2,2))<< std::endl;
}
It will print
> 1 0
> 0 1
>
> 1 0
> 0 1
to the console.
However if I change to function DF to
auto DF = [](MatrixXd y){
return FF(y) - MatrixXd::Identity(2,2);
};
The console will print
> 2.22045e-15 1.63042e-322
> 2.37152e-322 -0.999998
for the second matrix. (Which is just some uninitialized junk from memory).
Could someone explain what is happening with my code and my example problem. I have truly no idea what is going on here. I am especially interested why saving the result of the calculation in a temporary variable fixes the problem.
Since the comments have pretty much solved my issue (thank you very much) I thought I'd go ahead and answer my question so that other people see that this problem has been solved.
Why does the problem occur?
The problem is that, for example the result type of an Eigen multiplication of two matrices is not an Eigen matrix, but some internal object which represents the multiplication and references the two matrices we are trying to multiply.
Hence, if we use the auto keyword the compiler will very likely not give the variable we are setting the type MatrixXd but the type of some internal object.
For more information refer to the Eigen documentation which explicitly states:
In short: do not use the auto keywords with Eigen's expressions, unless you are 100% sure about what you are doing. In particular, do not use the auto keyword as a replacement for a Matrix<> type
How do I prevent it from happening?
do not use the auto keyword, use explicit types.
for a lambda function always specify the return type: Write auto DF = []() -> MatrixXd {...} instead of auto DF = []() {...}
Related
I'm using the ranges library to help filer data in my classes, like this:
class MyClass
{
public:
MyClass(std::vector<int> v) : vec(v) {}
std::vector<int> getEvens() const
{
auto evens = vec | ranges::views::filter([](int i) { return ! (i % 2); });
return std::vector<int>(evens.begin(), evens.end());
}
private:
std::vector<int> vec;
};
In this case, a new vector is constructed in the getEvents() function. To save on this overhead, I'm wondering if it is possible / advisable to return the range directly from the function?
class MyClass
{
public:
using RangeReturnType = ???;
MyClass(std::vector<int> v) : vec(v) {}
RangeReturnType getEvens() const
{
auto evens = vec | ranges::views::filter([](int i) { return ! (i % 2); });
// ...
return evens;
}
private:
std::vector<int> vec;
};
If it is possible, are there any lifetime considerations that I need to take into account?
I am also interested to know if it is possible / advisable to pass a range in as an argument, or to store it as a member variable. Or is the ranges library more intended for use within the scope of a single function?
This was asked in op's comment section, but I think I will respond it in the answer section:
The Ranges library seems promising, but I'm a little apprehensive about this returning auto.
Remember that even with the addition of auto, C++ is a strongly typed language. In your case, since you are returning evens, then the return type will be the same type of evens. (technically it will be the value type of evens, but evens was a value type anyways)
In fact, you probably really don't want to type out the return type manually: std::ranges::filter_view<std::ranges::ref_view<const std::vector<int>>, MyClass::getEvens() const::<decltype([](int i) {return ! (i % 2);})>> (141 characters)
As mentioned by #Caleth in the comment, in fact, this wouldn't work either as evens was a lambda defined inside the function, and the type of two different lambdas will be different even if they were basically the same, so there's literally no way of getting the full return type here.
While there might be debates on whether to use auto or not in different cases, but I believe most people would just use auto here. Plus your evens was declared with auto too, typing the type out would just make it less readable here.
So what are my options if I want to access a subset (for instance even numbers)? Are there any other approaches I should be considering, with or without the Ranges library?
Depends on how you would access the returned data and the type of the data, you might consider returning std::vector<T*>.
views are really supposed to be viewed from start to end. While you could use views::drop and views::take to limit to a single element, it doesn't provide a subscript operator (yet).
There will also be computational differences. vector need to be computed beforehand, where views are computed while iterating. So when you do:
for(auto i : myObject.getEven())
{
std::cout << i;
}
Under the hood, it is basically doing:
for(auto i : myObject.vec)
{
if(!(i % 2)) std::cout << i;
}
Depends on the amount of data, and the complexity of computations, views might be a lot faster, or about the same as the vector method. Plus you can easily apply multiple filters on the same range without iterating through the data multiple times.
In the end, you can always store the view in a vector:
std::vector<int> vec2(evens.begin(), evens.end());
So my suggestions is, if you have the ranges library, then you should use it.
If not, then vector<T>, vector<T*>, vector<index> depending on the size and copiability of T.
There's no restrictions on the usage of components of the STL in the standard. Of course, there are best practices (eg, string_view instead of string const &).
In this case, I can foresee no problems with handling the view return type directly. That said, the best practices are yet to be decided on since the standard is so new and no compiler has a complete implementation yet.
You're fine to go with the following, in my opinion:
class MyClass
{
public:
MyClass(std::vector<int> v) : vec(std::move(v)) {}
auto getEvens() const
{
return vec | ranges::views::filter([](int i) { return ! (i % 2); });
}
private:
std::vector<int> vec;
};
As you can see here, a range is just something on which you can call begin and end. Nothing more than that.
For instance, you can use the result of begin(range), which is an iterator, to traverse the range, using the ++ operator to advance it.
In general, looking back at the concept I linked above, you can use a range whenever the conext code only requires to be able to call begin and end on it.
Whether this is advisable or enough depends on what you need to do with it. Clearly, if your intention is to pass evens to a function which expects a std::vector (for instance it's a function you cannot change, and it calls .push_back on the entity we are talking about), you clearly have to make a std::vector out of filter's output, which I'd do via
auto evens = vec | ranges::views::filter(whatever) | ranges::to_vector;
but if all the function which you pass evens to does is to loop on it, then
return vec | ranges::views::filter(whatever);
is just fine.
As regards life time considerations, a view is to a range of values what a pointer is to the pointed-to entity: if the latter is destroied, the former will be dangling, and making improper use of it will be undefined behavior. This is an erroneous program:
#include <iostream>
#include <range/v3/view/filter.hpp>
#include <string>
using namespace ranges;
using namespace ranges::views;
auto f() {
// a local vector here
std::vector<std::string> vec{"zero","one","two","three","four","five"};
// return a view on the local vecotor
return vec | filter([](auto){ return true; });
} // vec is gone ---> the view returned is dangling
int main()
{
// the following throws std::bad_alloc for me
for (auto i : f()) {
std::cout << i << std::endl;
}
}
You can use ranges::any_view as a type erasure mechanism for any range or combination of ranges.
ranges::any_view<int> getEvens() const
{
return vec | ranges::views::filter([](int i) { return ! (i % 2); });
}
I cannot see any equivalent of this in the STL ranges library; please edit the answer if you can.
EDIT: The problem with ranges::any_view is that it is very slow and inefficient. See https://github.com/ericniebler/range-v3/issues/714.
It is desirable to declare a function returning a range in a header and define it in a cpp file
for compilation firewalls (compilation speed)
stop the language server from going crazy
for better factoring of the code
However, there are complications that make it not advisable:
How to get type of a view?
If defining it in a header is fine, use auto
If performance is not a issue, I would recommend ranges::any_view
Otherwise I'd say it is not advisable.
In trying to track down a bug I've been able to come up with the following simple program which illustrates the issue:
#include <iostream>
#include <Eigen/Dense>
using Mat = Eigen::Matrix<double, 1, 1, Eigen::AutoAlign>;
template <typename Derived>
auto Modify(const Eigen::DenseBase<Derived>& value)
-> decltype(2*value.derived().array())
{
return 2*value.derived().array();
}
Mat Make()
{
Mat mat;
mat << 2;
return mat;
}
int main()
{
auto mat = Make();
std::cout
<< "first: " << Modify(mat) << std::endl
<< "second: " << Modify(Modify(mat));
return 0;
}
The output of this program is:
first: 4
second: 3.95253e-323
running the resulting code under valgrind produces a Uninitialised value was created by a stack allocation error.
Is there anything wrong with the above code? As the original mat matrix is in scope at the point where the matrix is evaluated, why does it matter that I'm creating temporaries of Eigen::DenseBase expressions?
I have tried compiling with Eigen 3.2.1 as well as Eigen 3.2.2 and got the same result.
T.C.'s answers was true but that behavior was unintended as lightweight expressions are supposed to be nested by values, not by reference. There was a little shortcoming in ArrayWrapper that made it nested by reference in some cases. This is now fixed in both the 3.2 and default branches. So your example works fine now.
value.derived().array() returns a temporary ArrayWrapper object. 2 * value.derived().array() constructs an expression object, which stores a reference to that ArrayWrapper, but that temporary ArrayWrapper is destroyed right afterwards. Thus the expression object you returned from Modify contains a dangling reference.
I would like to build a space-efficient modular arithmetic class. The idea is that the modulus M is an immutable attribute that gets fixed during instantiation, so if we have a large array (std::vector or another container) of values with the same M, M only needs to be stored once.
If M can be fixed at compile time, this can be done using templates:
template <typename num, num M> class Mod_template
{
private:
num V;
public:
Mod_template(num v=0)
{
if (M == 0)
V = v;
else
{
V = v % M;
if (V < 0)
V += M;
}
}
// ...
};
Mod_template<int, 5> m1(2); // 2 mod 5
However, in my application, we should be able to express M runtime. What I have looks like this:
template <typename num> class Mod
{
private:
const num M;
num V;
public:
Mod(num m, num v=0): M(abs(m))
{
if (M == 0)
V = v;
else
{
V = v % M;
if (V < 0)
V += M;
}
}
// ...
};
Mod<int> m2(5, 2); // 2 mod 5
Mod<int> m3(3); // 0 mod 3
This works, but a large vector of mod M values uses 2x the space it needs to.
I think the underlying conceptual problem is that Mod's of different moduli are syntactically of the same type even though they "should" be different types. For example, a statement like
m2 = m3;
should raise a runtime error "naturally" (in my version, it does so "manually": check is built into the copy constructor, as well as every binary operator I implement).
So, is there a way to implement some kind of dynamic typing so that the Mod object's type remembers the modulus? I'd really appreciate any idea how to solve this.
This is a recurring problem for me with various mathematical structures (e.g. storing many permutations on the same set, elements of the same group, etc.)
EDIT: as far as I understand,
templates are types parametrized by a class or literal.
what I want: a type parametrized by a const object (const num in this case, const Group& or const Group *const for groups, etc.).
Is this possible?
It will be difficult to do it in zero storage space if the class needs to know what M should be without any outside help. Likely the best you can do is store a pointer to a shared M, which may be a little better depending on how large num is. But it's not as good as free.
It will be easier to design if M is a passed-in value to all the functions that need it. Then you can do things like make a pool of objects that all share the same M (there are plenty of easy ways to design this; e.g. map<num, vector<num> >) and only store M once for the pool. The caller will need to know which pool the Mod object came from, but that's probably something it knows anyway.
It's hard to answer this question perfectly in isolation... knowing more about the calling code would definitely help you get better answers.
For lambda expressions, I don't quite get the usefulness of closures in C++11.
auto f = [] (int n, int m) { return n + m };
std::cout << f(2,2);
versus.
int n = 2;
auto f = [n] (int m) { return n + m };
std::cout << f(2);
This is a very basic and primitive example. I'm guessing that closures play an important part in other kinds of statements, but my C++ book doesn't clarify this (so far).
Why not include the closure as a parameter?
OK, a simple example, remove all the x's from a string
char x = 'x';
std::string s = "Text to remove all 'x's from";
s.erase(std::remove_if(s.begin(), s.end(), [x](char c) {return x == c;}), s.end());
Borrowed and modifed from http://en.cppreference.com/w/cpp/algorithm/remove
In this example, remove_if() only takes a single parameter, but I need two values for the comparison.
Closures are not always called immediately. They are objects which can be stored and called later when the data necessary to successfully execute the lambda function may no longer be in scope or easily accessible from the call site.
It's possible to to store any necessary data along with the closure but it's so much simpler for the closure to grab anything it needs when it's created and use it when it's eventually called. It provides a form of encapsulation.
This also decreases code coupling because if you were to store the data along with the code then the caller could only work with the specific objects you decided to store. Since a closure carries its own data along with it, it can work with any data it needs.
Here's an greatly oversimplified real-life example. I built a database server which needed to support fields with multiple values. The problem was that when results were displayed, it was important to highlight which values actually caused a record to match the search criteria. So, the query parser would spit out a predicate in the form of a closure which would indicate whether or not it was a matching value.
It looked something like this:
std::function< bool(int value) > parser::match_int(int search_val) {
return [=](int value) { value == search_val; };
}
That closure got stored in a collection. When it was time to render the record, I could easily determine which values needed to be highlighted. Keep in mind that the parser and any associated data is now gone:
void render_values(std::function< bool(int value) > pred, std::vector<int> values) {
for (int value : values) {
if (pred(value))
render_highlight(value);
else
render_normal(value);
}
}
I have a function, f(a,b), that accepts two inputs. I do not know ahead of time which values of a and b will be used. I'm okay with being a little wasteful on memory (I care about speed). I want to be able to check if the output of f(a,b) has already been delivered, and if so, deliver that output again without re-running through the f(a,b) process.
Trivially easy to do in Python with decorators, but C++ is way over my head here.
I would use a std::map (or maybe an std::unordered_map) whose key is a std::pair, or perhaps use a map of maps.
C++11 improvements are probably helpful in that case. Or maybe some Boost thing.
The poster asks:
I want to be able to check if the output of f(a,b) has already been delivered, and if so, deliver that output again without re-running through the f(a,b) process.
It's pretty easy in C++ using a std::map. The fact that the function has exactly two parameters means that we can use std::pair to describe them.
#include <map>
#include <iostream>
uint64_t real_f(int a, int b) {
std::cout << "*";
// Do something tough:
return (uint64_t)a*b;
}
uint64_t memo_f(int a, int b) {
typedef std::pair<int, int> key;
typedef std::map<key, uint64_t> map;
static map m;
key k(a,b);
map::iterator it = m.find(k);
if(it == m.end()) {
return m[k] = real_f(a, b);
}
return it->second;
}
int main () {
std::cout << memo_f(1, 2) << "\n";
std::cout << memo_f(3, 4) << "\n";
std::cout << memo_f(1, 2) << "\n";
std::cout << memo_f(3, 4) << "\n";
std::cout << memo_f(5, 6) << "\n";
}
The output of the above program is:
*2
*12
2
12
*30
The lines without asterisks represent cached results.
With C++11, you could use tasks and futures. Let f be your function:
int f(int a, int b)
{
// Do hard work.
}
Then you would schedule the function execution, which returns you a handle to the return value. This handle is called a future:
template <typename F>
std::future<typename std::result_of<F()>::type>
schedule(F f)
{
typedef typename std::result_of<F()>::type result_type;
std::packaged_task<result_type> task(f);
auto future = task.get_future();
tasks_.push_back(std::move(task)); // Queue the task, execute later.
return std::move(future);
}
Then, you could use this mechanism as follows:
auto future = schedule(std::bind(&f, 42, 43)); // Via std::bind.
auto future = schedule([&] { f(42, 43); }); // Lambda alternative.
if (future.has_value())
{
auto x = future.get(); // Blocks if the result of f(a,b) is not yet availble.
g(x);
}
Disclaimer: my compiler does not support tasks/futures, so the code may have some rough edges.
The main point about this question are the relative expenses in CPU and RAM between calculating f(a,b) and keeping some sort of lookup table to cache results.
Since an exhaustive table of 128 bits index length is not (yet) feasable, we need to reduce the lookup space into a manageable size - this can't be done without some considerations inside your app:
How big is the really used space of function inputs? Is there a pattern in it?
What about the temporal component? Do you expect repeated calculations to be close to one another or ditributed along the timeline?
What about the distribution? Do you assume a tiny part of the index space to consume the majority of function calls?
I would simply start with a fixed-size array of (a,b, f(a,b)) tuples and a linear search. Depending on your pattern as asked above, you might want to
window-slide it (drop oldest on a cache miss): This is good for localized reocurrences
have (a,b,f(a,b),count) tuples with the tuple with the smallest count being expelled - this is good for non-localized occurrences
have some key-function determine a position in the cache (this is good for tiny index space usage)
whatever else Knuth or Google might have thought of
You might also want to benchmark repeated calculation against the lookup mechanism, if the latter becomes more and more complex: std::map and freinds don't come for free, even if they are high-quality implementations.
The only easy way is to use std::map. std::unordered_map does not work. We cannot use std::pair as the key in unordered map. You can do the following,
std::map<pair<int, int>, int> mp;
int func(int a, int b)
{
if (mp.find({a, b}) != mp.end()) return mp[{a, b}];
// compute f(a, b)...
mp[{a, b}] = // computed value;
return mp[{a, b}];
}