I am using the the armadillo matrix library in c++, and I want to create a vec that uses "auxiliare memory". The standard way to do this is
vec qq(6); qq<<1<<2<<3<<4<<5<<6;
double *qqd = qq.memptr();
vec b1(qqd, 6, false);
So here, if I change the elements in b1, the elements in qq is changed, which is what I want. However, in my program, I declare the b1 vector globaly, so when I define it, I cant call the constructor that makes b1 use "auxiliare memory". Is there a function in armadillo that does what I want? Why do I get different results when I run the codes below?
vec b1= vec(qq.memptr(), 3, false); //changing b1 element changes one of qq's element
vec b1;
b1= vec(qq.memptr(), 3, false); //changing b1 element does not chagne qq's
So, how can I make a vector use the memory from a other vector, when it is declared globaly?
Using global variables is generally a bad idea.
However if you insist on using them, here is one possible way of using auxiliary memory with global Armadillo vectors:
#include <armadillo>
using namespace arma;
vec* x; // global declaration as a pointer; points to garbage by default
int main(int argc, char** argv) {
double data[] = { 1, 2, 3, 4 };
// create vector x and make it use the data array
x = new vec(data, 4, false);
vec& y = (*x); // use y as an "easier to use" alias of x
y.print("y:");
// as x is a pointer pointing to an instance of the vec class,
// we need to delete the memory used by the vec class before exiting
// (this doesn't touch the data array, as it's external to the vector)
delete x;
return 0;
}
Related
Is there a way to initialize first few elements of a vector after defining the size of the vector like -
vector<int> vec (10);
This doesn't work and produces a compiler error -
vector<int> vec(10) {1,2,3};
For example with arrays we can do the same thing like -
int arr[5] {1,2,3}; // This will initialize the first 3 elements of the array to 1,2,3 and the remaining two to 0.
In short, no. Your can fill out the entire list of things you want to be in the vector:
vector<int> vec{1, 2, 3, 0, 0, 0, 0, 0, 0, 0};
Which will give you a vector of 10 elements.
Or, you can create the vector, then call resize to make it larger (filling the remaining elements with 0):
vector<int> vec{1, 2, 3};
vec.resize(10);
You generally don't need to do this kind of thing to vector though, because unlike array, you can extend vector as needed, after creation:
vector<int> vec{1, 2, 3};
vec.push_back(4);
There isn't a way to do it all in one line like you can with an array. You can use
vector<int> vec{1,2,3};
vec.resize(10);
but that does make the code a little less easy to use. Another option is to wrap that in a function like
template <typename T>
auto make_sized_vector(std::intializer_list<T> il, std::size_t size = 0)
{
const auto vec_size = std::max(size, il.size());
vector<T> vec; // create vector
vec.reserve(vec_size); // allocate all the storage needed
vec.assign(il); // assign the elements
vec.resize(vec_size); // set the rest to zero
return vec;
}
and then you can use that like
auto vec = make_sized_vector<int>({1, 2, 3}, 10);
If you are concerned about passing the std::intializer_list by value see why is `std::initializer_list` often passed by value? for why that really isn't a concern.
In case you want to initialize a vector the way you describe, all at once, so that it can become (e.g.) a const member, this is always possible in C++, with just a bit of ugliness and twisting. Let’s say you have a class:
struct SomeClass {
SomeClass(const std::vector<int> &start, int rest, std::size_t size);
const std::vector<int> some_vector_; // This is const!
};
What the constructor could look like:
SomeClass::SomeClass(const std::vector<int> &start, int rest, std::size_t size)
: some_vector_{[&start, rest, size] {
std::vector<int> some_vector;
some_vector.reserve(size);
some_vector.insert(some_vector.end(), start.begin(), start.end());
some_vector.insert(some_vector.end(), size - start.size(), rest);
return some_vector;
}()} {}
Basically the problem boils down to: How do I do “something procedural” in an initializer list? To which the answer is: You invoke a function that returns the desired type.
To test the construct above:
#include <cstdint>
#include <iostream>
#include <vector>
namespace { /* SomeClass stuff from above goes here. */ }
int main() {
SomeClass sc{{1, 2, 3}, 0, 10};
for (int i : sc.some_vector_) std::cout << i << '\n';
}
There are (of course) plenty of ways to make it (slightly) more efficient if needed, such as
a templated variadic constructor to create the initial part of the vector,
a templated perfect-forwarding constructor to benefit from R-value containers, and
as a combined benefit of the above, arbitrary iterable containers as inputs and as the const member.
I am fairly new to C++ and have been avoiding pointers. From what I've read online I cannot return an array but I can return a pointer to it. I made a small code to test it and was wondering if this was the normal / correct way to do this:
#include <iostream>
using namespace std;
int* test (int in[5]) {
int* out = in;
return out;
}
int main() {
int arr[5] = {1, 2, 3, 4, 5};
int* pArr = test(arr);
for (int i = 0; i < 5; i++) cout<<pArr[i]<<endl;
cout<<endl;
return 0;
}
Edit: This seems to be no good. How should I rewrite it?
int* test (int a[5], int b[5]) {
int c[5];
for (int i = 0; i < 5; i++) c[i] = a[i]+b[i];
int* out = c;
return out;
}
Your code as it stands is correct but I am having a hard time figuring out how it could/would be used in a real world scenario. With that said, please be aware of a few caveats when returning pointers from functions:
When you create an array with syntax int arr[5];, it's allocated on the stack and is local to the function.
C++ allows you to return a pointer to this array, but it is undefined behavior to use the memory pointed to by this pointer outside of its local scope. Read this great answer using a real world analogy to get a much clear understanding than what I could ever explain.
You can still use the array outside the scope if you can guarantee that memory of the array has not be purged. In your case this is true when you pass arr to test().
If you want to pass around pointers to a dynamically allocated array without worrying about memory leaks, you should do some reading on std::unique_ptr/std::shared_ptr<>.
Edit - to answer the use-case of matrix multiplication
You have two options. The naive way is to use std::unique_ptr/std::shared_ptr<>. The Modern C++ way is to have a Matrix class where you overload operator * and you absolutely must use the new rvalue references if you want to avoid copying the result of the multiplication to get it out of the function. In addition to having your copy constructor, operator = and destructor, you also need to have move constructor and move assignment operator. Go through the questions and answers of this search to gain more insight on how to achieve this.
Edit 2 - answer to appended question
int* test (int a[5], int b[5]) {
int *c = new int[5];
for (int i = 0; i < 5; i++)
c[i] = a[i]+b[i];
return c;
}
If you are using this as int *res = test(a,b);, then sometime later in your code, you should call delete []res to free the memory allocated in the test() function. You see now the problem is it is extremely hard to manually keep track of when to make the call to delete. Hence the approaches on how to deal with it where outlined in the answer.
Your code is OK. Note though that if you return a pointer to an array, and that array goes out of scope, you should not use that pointer anymore. Example:
int* test (void)
{
int out[5];
return out;
}
The above will never work, because out does not exist anymore when test() returns. The returned pointer must not be used anymore. If you do use it, you will be reading/writing to memory you shouldn't.
In your original code, the arr array goes out of scope when main() returns. Obviously that's no problem, since returning from main() also means that your program is terminating.
If you want something that will stick around and cannot go out of scope, you should allocate it with new:
int* test (void)
{
int* out = new int[5];
return out;
}
The returned pointer will always be valid. Remember do delete it again when you're done with it though, using delete[]:
int* array = test();
// ...
// Done with the array.
delete[] array;
Deleting it is the only way to reclaim the memory it uses.
New answer to new question:
You cannot return pointer to automatic variable (int c[5]) from the function. Automatic variable ends its lifetime with return enclosing block (function in this case) - so you are returning pointer to not existing array.
Either make your variable dynamic:
int* test (int a[5], int b[5]) {
int* c = new int[5];
for (int i = 0; i < 5; i++) c[i] = a[i]+b[i];
return c;
}
Or change your implementation to use std::array:
std::array<int,5> test (const std::array<int,5>& a, const std::array<int,5>& b)
{
std::array<int,5> c;
for (int i = 0; i < 5; i++) c[i] = a[i]+b[i];
return c;
}
In case your compiler does not provide std::array you can replace it with simple struct containing an array:
struct array_int_5 {
int data[5];
int& operator [](int i) { return data[i]; }
int operator const [](int i) { return data[i]; }
};
Old answer to old question:
Your code is correct, and ... hmm, well, ... useless. Since arrays can be assigned to pointers without extra function (note that you are already using this in your function):
int arr[5] = {1, 2, 3, 4, 5};
//int* pArr = test(arr);
int* pArr = arr;
Morever signature of your function:
int* test (int in[5])
Is equivalent to:
int* test (int* in)
So you see it makes no sense.
However this signature takes an array, not pointer:
int* test (int (&in)[5])
A variable referencing an array is basically a pointer to its first element, so yes, you can legitimately return a pointer to an array, because thery're essentially the same thing. Check this out yourself:
#include <assert.h>
int main() {
int a[] = {1, 2, 3, 4, 5};
int* pArr = a;
int* pFirstElem = &(a[0]);
assert(a == pArr);
assert(a == pFirstElem);
return 0;
}
This also means that passing an array to a function should be done via pointer (and not via int in[5]), and possibly along with the length of the array:
int* test(int* in, int len) {
int* out = in;
return out;
}
That said, you're right that using pointers (without fully understanding them) is pretty dangerous. For example, referencing an array that was allocated on the stack and went out of scope yields undefined behavior:
#include <iostream>
using namespace std;
int main() {
int* pArr = 0;
{
int a[] = {1, 2, 3, 4, 5};
pArr = a; // or test(a) if you wish
}
// a[] went out of scope here, but pArr holds a pointer to it
// all bets are off, this can output "1", output 1st chapter
// of "Romeo and Juliet", crash the program or destroy the
// universe
cout << pArr[0] << endl; // WRONG!
return 0;
}
So if you don't feel competent enough, just use std::vector.
[answer to the updated question]
The correct way to write your test function is either this:
void test(int* a, int* b, int* c, int len) {
for (int i = 0; i < len; ++i) c[i] = a[i] + b[i];
}
...
int main() {
int a[5] = {...}, b[5] = {...}, c[5] = {};
test(a, b, c, 5);
// c now holds the result
}
Or this (using std::vector):
#include <vector>
vector<int> test(const vector<int>& a, const vector<int>& b) {
vector<int> result(a.size());
for (int i = 0; i < a.size(); ++i) {
result[i] = a[i] + b[i];
}
return result; // copy will be elided
}
In a real app, the way you returned the array is called using an out parameter. Of course you don't actually have to return a pointer to the array, because the caller already has it, you just need to fill in the array. It's also common to pass another argument specifying the size of the array so as to not overflow it.
Using an out parameter has the disadvantage that the caller may not know how large the array needs to be to store the result. In that case, you can return a std::vector or similar array class instance.
Your code (which looks ok) doesn't return a pointer to an array. It returns a pointer to the first element of an array.
In fact that's usually what you want to do. Most manipulation of arrays are done via pointers to individual elements, not via pointers to the array as a whole.
You can define a pointer to an array, for example this:
double (*p)[42];
defines p as a pointer to a 42-element array of doubles. A big problem with that is that you have to specify the number of elements in the array as part of the type -- and that number has to be a compile-time constant. Most programs that deal with arrays need to deal with arrays of varying sizes; a given array's size won't vary after it's been created, but its initial size isn't necessarily known at compile time, and different array objects can have different sizes.
A pointer to the first element of an array lets you use either pointer arithmetic or the indexing operator [] to traverse the elements of the array. But the pointer doesn't tell you how many elements the array has; you generally have to keep track of that yourself.
If a function needs to create an array and return a pointer to its first element, you have to manage the storage for that array yourself, in one of several ways. You can have the caller pass in a pointer to (the first element of) an array object, probably along with another argument specifying its size -- which means the caller has to know how big the array needs to be. Or the function can return a pointer to (the first element of) a static array defined inside the function -- which means the size of the array is fixed, and the same array will be clobbered by a second call to the function. Or the function can allocate the array on the heap -- which makes the caller responsible for deallocating it later.
Everything I've written so far is common to C and C++, and in fact it's much more in the style of C than C++. Section 6 of the comp.lang.c FAQ discusses the behavior of arrays and pointers in C.
But if you're writing in C++, you're probably better off using C++ idioms. For example, the C++ standard library provides a number of headers defining container classes such as <vector> and <array>, which will take care of most of this stuff for you. Unless you have a particular reason to use raw arrays and pointers, you're probably better off just using C++ containers instead.
EDIT : I think you edited your question as I was typing this answer. The new code at the end of your question is, as you observer, no good; it returns a pointer to an object that ceases to exist as soon as the function returns. I think I've covered the alternatives.
you can (sort of) return an array
instead of
int m1[5] = {1, 2, 3, 4, 5};
int m2[5] = {6, 7, 8, 9, 10};
int* m3 = test(m1, m2);
write
struct mystruct
{
int arr[5];
};
int m1[5] = {1, 2, 3, 4, 5};
int m2[5] = {6, 7, 8, 9, 10};
mystruct m3 = test(m1,m2);
where test looks like
struct mystruct test(int m1[5], int m2[5])
{
struct mystruct s;
for (int i = 0; i < 5; ++i ) s.arr[i]=m1[i]+m2[i];
return s;
}
not very efficient since one is copying it delivers a copy of the array
In C++, I can statically initialize an array, e.g.:
int a[] = { 1, 2, 3 };
Is there an easy way to initialize a dynamically-allocated array to a set of immediate values?
int *p = new int[3];
p = { 1, 2, 3 }; // syntax error
...or do I absolutely have to copy these values manually?
You can in C++0x:
int* p = new int[3] { 1, 2, 3 };
...
delete[] p;
But I like vectors better:
std::vector<int> v { 1, 2, 3 };
If you don't have a C++0x compiler, boost can help you:
#include <boost/assign/list_of.hpp>
using boost::assign::list_of;
vector<int> v = list_of(1)(2)(3);
You have to assign each element of the dynamic array explicitly (e.g. in a for or while loop)
However the syntax int *p = new int [3](); does initialize all elements to 0 (value initialization $8.5/5)
To avoid endless push_backs, I usually initialize a tr1::array and create a std::vector (or any other container std container) out of the result;
const std::tr1::array<T, 6> values = {T(1), T(2), T(3), T(4), T(5), T(6)};
std::vector <T> vec(values.begin(), values.end());
The only annoyance here is that you have to provide the number of values explicitly.
This can of course be done without using a tr1::array aswell;
const T values[] = {T(1), T(2), T(3), T(4), T(5), T(6)};
std::vector <T> vec(&values[0], &values[sizeof(values)/sizeof(values[0])]);
Althrough you dont have to provide the number of elements explicitly, I prefer the first version.
No, you cannot initialize a dynamically created array in the same way.
Most of the time you'll find yourself using dynamic allocation in situations where static initialization doesn't really make sense anyway. Such as when you have arrays containing thousands of items. So this isn't usually a big deal.
Using helper variable:
const int p_data[] = {1, 2, 3};
int* p = (int*)memcpy(new int[3], p_data, sizeof(p_data));
or, one line
int p_data[] = {1, 2, 3}, *p = (int*)memcpy(new int[3], p_data, sizeof(p_data));
Never heard of such thing possible, that would be nice to have.
Keep in mind that by initializing the array in the code that way
int a[] = { 1, 2, 3 };
..... only gains you easier code writing and NOT performance.
After all, the CPU will do the work of assigning values to the array, either way you do it.
Does the following create x new objects, or simply allocate space for x objects?:
Vector3D* binArray = new Vector3D[size];
I need to build an array with space for x Vector3D objects on the heap. However, a Vector3D object can only be created when an "add" function is called - this will take the parameters, construct the object on the heap and add its address to the array of Vector3D pointers.
This does create an array of Vector3D objects on the heap.
Each vector is created by calling the Vector3D constructor.
Put a little debugging print statement in the default constructor for Vector3D, and watch the constructor get called the same number of times as you have vectors in your array.
Example:
#include <iostream>
using namespace std;
class C {
public:
C() { cout << "Hello default constructor here\n"; }
};
int main() {
C* cs = new C[5];
}
Output is:
Hello default constructor here
Hello default constructor here
Hello default constructor here
Hello default constructor here
Hello default constructor here
If your class does not have a default constructor, you cannot allocate the array in one shot (thank you for the comment #Everyone), so in this case consider using a std::vector or a std::array and adding your Vector3D objects dynamically --- or even "statically"! Example:
#include <iostream>
#include <vector>
using namespace std;
class Vector3D {
double i, j, k;
public:
Vector3D(double i, double j, double k): i(i), j(j), k(k) {}
};
int main() {
vector<Vector3D> v = {
Vector3D(3, 4, 5),
Vector3D(6, 8, 10),
Vector3D(7, 24, 25)
};
v.push_back(Vector3D(1, 2, 3));
cout << v.size() << '\n';
}
This outputs 4.
You can also make your vector contain pointers to Vector3D objects.
Just to add based on the asker's comments on RayToal's excellent answer. If you don't know the size of the binArray prior to runtime then you must use std::vector. If you want to allocate each item alone, I would recommend using std::vector<Vector3D*>.
This way you can resize the std::vector at runtime and when you do, it will hold a bunch of nullptrs that are not allocated. Then you can allocate each one of them separately.
std::vector<Vector3D*> binArray;
binArray.resize(x); // now you have binArray of size x and no allocated elements
binArray[0] = new Vector3D(...);
Please keep in mind that you need to delete them after you're not using them in order to not have a memory leak:
for(size_t i=0;i<binArray.size(); i++)
if(binArray[i]!=nullptr) delete binArray[i];
Well I am questioning myself if there is a way to pass a vector directly in a parameter, with that I mean, like this:
int xPOS = 5, yPOS = 6, zPOS = 2;
//^this is actually a struct but
//I simplified the code to this
std::vector <std::vector<int>> NodePoints;
NodePoints.push_back(
std::vector<int> {xPOS,yPOS,zPOS}
);
This code ofcourse gives an error; typename not allowed, and expected a ')'
I would have used a struct, but I have to pass the data to a Abstract Virtual Machine where I need to access the node positions as Array[index][index] like:
public GPS_WhenRouteIsCalculated(...)
{
for(new i = 0; i < amount_of_nodes; ++i)
{
printf("Point(%d)=NodeID(%d), Position(X;Y;Z):{%f;%f;%f}",i,node_id_array[i],NodePosition[i][0],NodePosition[i][1],NodePosition[i][2]);
}
return 1;
}
Ofcourse I could do it like this:
std::vector <std::vector<int>> NodePoints;//global
std::vector<int> x;//local
x.push_back(xPOS);
x.push_back(yPOS);
x.push_back(zPOS);
NodePoints.push_back(x);
or this:
std::vector <std::vector<int>> NodePoints;//global
std::vector<int> x;//global
x.push_back(xPOS);
x.push_back(yPOS);
x.push_back(zPOS);
NodePoints.push_back(x);
x.clear()
but then I'm wondering which of the two would be faster/more efficient/better?
Or is there a way to get my initial code working (first snippet)?
Use C++11, or something from boost for this (also you can use simple v.push_back({1,2,3}), vector will be constructed from initializer_list).
http://liveworkspace.org/code/m4kRJ$0
You can use boost::assign as well, if you have no C++11.
#include <vector>
#include <boost/assign/list_of.hpp>
using namespace boost::assign;
int main()
{
std::vector<std::vector<int>> v;
v.push_back(list_of(1)(2)(3));
}
http://liveworkspace.org/code/m4kRJ$5
and of course you can use old variant
int ptr[1,2,3];
v.push_back(std::vector<int>(ptr, ptr + sizeof(ptr) / sizeof(*ptr));
If you don't have access to either Boost or C++11 then you could consider quite a simple solution based around a class. By wrapping a vector to store your three points within a class with some simple access controls, you can create the flexibility you need. First create the class:
class NodePoint
{
public:
NodePoint( int a, int b, int c )
{
dim_.push_back( a );
dim_.push_back( b );
dim_.push_back( c );
}
int& operator[]( size_t i ){ return dim_[i]; }
private:
vector<int> dim_;
};
The important thing here is to encapsulate the vector as an aggregate of the object. The NodePoint can only be initialised by providing the three points. I've also provided operator[] to allow indexed access to the object. It can be used as follows:
NodePoint a(5, 6, 2);
cout << a[0] << " " << a[1] << " " << a[2] << endl;
Which prints:
5 6 2
Note that this will of course throw if an attempt is made to access an out of bounds index point but that's still better than a fixed array which would most likely seg fault. I don't see this as a perfect solution but it should get you reasonably safely to where you want to be.
If your main goal is to avoid unnecessary copies of vector<> then here how you should deal with it.
C++03
Insert an empty vector into the nested vector (e.g. Nodepoints) and then use std::swap() or std::vector::swap() upon it.
NodePoints.push_back(std::vector<int>()); // add an empty vector
std::swap(x, NodePoints.back()); // swaps contents of `x` and last element of `NodePoints`
So after the swap(), the contents of x will be transferred to NodePoints.back() without any copying.
C++11
Use std::move() to avoid extra copies
NodePoints.push_back(std::move(x)); // #include<utility>
Here is the explanation of std::move and here is an example.
Both of the above solutions have somewhat similar effect.