How do I clone a jagged array in a c++ class? - c++

I have a templated class I use to create and manipulate a jagged array. I can create the jagged array; add elements; remove them; modify them; print the jagged array, etc. It seems to work at least as far my testing so far, until I try to pass a previously created jagged array to it to create another. My class currently takes what is passed, and creates what appears to be a clone, but it ends up being an alias for the same memory space. So, when I go to make a change in what I think is the clone, I end up making the change in 'both' of them. I'm kind of lost and was wondering if anyone might be able to give me some pointers?

It's hard to tell exactly what your problem is without a code example, but it sounds like your array contains pointers. For example:
std::vector<int*> original;
original.push_back(new int[]{1}); // note: for demo only (`new` without `delete` is bad)
std::vector<int*> copy = original;
copy[0][0] = 2; // now original[0][0] also equals 2
Since copy contains the same pointers as original, original[0] and copy[0] are the same pointer, and so original[0][0] and copy[0][0] refer to the same object.
You can fix this by copying the values being pointed to, rather than the pointers themselves, and storing pointers to the copies in your new array. In my example, the easiest way would be to use std::vector<int> instead of int*:
std::vector<std::vector<int>> original;
original.push_back({1});
std::vector<std::vector<int>> copy = original;
copy[0][0] = 2; // original[0][0] still equals 1

Related

How to get an array of an Element from an array of structs

I have an array of a struct, lets say
struct cell{
int pos;
int id;
};
std::vector<cell> myArray;
I want an array of the id element. I can't just iterate over my array as it would take too long.
I have to provide std::vector<int> to a function.
My thought process was: Since arrays are usually just a pointer to the first element and then an offset I thought of creating an array where i can provide the offset, such as it would point to the id element of the next cell in std::vector<cell> myArray.
One solution I can think of is having an array of pointers to that element, for example:
The final solution might be something like:
struct cell{
int pos;
int id;
};
std::vector<cell> myArray;
std::vector<int*> pointersToIds;
// Creating an array of int from an array of int*
std::vector<int> idsArray = std::something(pointersToIds);
myFunc(idsArray);
Since the std library has tons of stuff I supposed there would be a way to do this.
Is there a way to convert the array of pointers to an actual array of elements in a very optimized way? The pointers approach was the only i could think but it's doesn't necessarily have to be it.
Thank you all in advance :)
I tried iterating over the the array of pointers and creating an array of elements, but it would take too much time.
TLDR Get array of an element from array of struct
I suppose this might be an instance of The XY Problem, since it's not clear what you are actually trying to solve, do you:
Want to find a fast way to pass the list of struct to a function
Want a way to extract all the members from a list of struct into a list of members
First off, shoo away from your mind the idea of manually creating an array of addresses and then fiddling around with the offsets yourself, this is certainly doable, but probably hard to do yourself in a safe and portable way due to Struct Alignment, something that differs from machine to machine.
besides accessing cell.id is already doing that in a portable way by itself!
Problem 1.
If you want to pass a vector (or any object really) to a function in a fast way, you can use a reference, it would look something like this:
void foo(std::vector<cell>& in_vec);
notice the & operator, declaring that in_vec must be passed as a reference, what this does internally is pass in_vec by address, avoiding copying values one by one, C++ does all this by himself and you can treat in_vec normally in the function without a care in the world, and it's blazing fast.
Problem 2.
if your point is that you want to extract all the IDs before passing them to a function, first off, I still suggest you pass the cell, that way it is clear that foo is supposed to operate on cell IDs and not random integers, once again, paying the cost of unpacking the structs outside (which warrants an iteration) or inside (where you might not even need to access all cells depending on foo's nature) is equal if not worse.
If you must carry through, it's as easy as a for loop:
std::vector<int> ids;
for(auto const& cell : myArray)
{
ids.push_back(cell.id);
}
Or, if you want a elegant and modern solution, using lambdas and algorithm:
#include <algorithm>
std::vector<int> ids;
std::transform(myArray.begin(), myArray.end(),
std::back_inserter(ids), [](cell const& c) {
return c.id;
});
Or something to this effect.

How to build an array by a vector?

I want to write a function to create an array with a viewed vector.
Such as
int foo(vector<int> &A){
int N=A.size();
int myarray[N]={0}; //here is an error
}
Here are my questions:
I know to build an array must use const value but vector::size() isn't return a const value?
using const modify the expression doesn't work N still a variable.
I don't see why you ever should need to create an array. If you pass arrays anywhere into a function, you pass a pointer to such an array. But you easily can get the vector's internal data as such a pointer:
std::vector<int> v({1, 2, 3, 4});
int* values = v.data();
Done, you have your "array", and you don't even have to copy the data...
If you still need a copy, just copy the vector itself.
Side-note: The vector's content is guaranteed by the standard to be contiguous, so you won't get into any trouble iterating over your "arrays".
However, one problem actually exists: If you want to store pointers to the contents of your array, they might get invalidated if you add further elements into your vector, as it might re-allocate its contents. Be aware of that! (Actually, the same just as if you need to create a new array for any reason...)

Finding information in a vector?

In C++ I understand that in order to create a dynamic array you need to use vectors. However I have a problem when I need to find information I put in the vector.
For example:
Lets say I have a simple vector that stores the name of a person and a small message the wrote. In the vector how do I find where Bill is located.
I was also trying to understand how to do this in PHP when I posted this question.
Indeed you seam confused. Let me try to help you.
One thing that is maybe confusing you: std::vector is not a geometric vector. It's only a sequence of data of the same type that is contiguous in memory. So it's like an array.
a) Determine the size of a vector based on a variable. For example if
I was using an array it would look something like array [x][y] ( I
know it's not possible to do this). How would I do this with a vector
std::vector is basically a automatically managed dynamic array.
It means that it IS an array inside, but it's managed by code that will make sure that array grows (gets bigger) when you try to add more data than it current capacity can hold.
Actually, std::vector is a class template. It means that it's not a real class, it's code that the compiler will use to generate itself a real class. If I say
std::vector<int> my_ints; // this is a vector of ints
This vector can only hold ints. And then:
std::vector<std::string> name_list;
this one hold std::string objects.
As I was saying, inside, it's only code to manage an array dynamically. You can think the previous examples as if it was like that:
class
{
unsigned long size; // count of elements contained in this container
unsigned long capacity; // count of elements that the memory allocated by the array can hold
int* array; // array containing the values, created using new, destroyed using delete
}
my_ints;
This is an oversimplified view of how it is inside, so don't assume it's exactly like that, but it might be useful.
Now, when you add values, the value is copied in the memory of the array, in an element that is not used yet (through push_back() for example) or writing over an element already existing (using insert() for example).
If you add a value and the capacity of the vector is not enough to hold all values, then the the vector will automatically grow: it will create a much bigger array, copy it's current values inside, copy the additional value too, then delete the array it had before.
It's important to understand this: if a vector grows, then you can't assume that it's data is always at the same adress in memory, so pointers to it's data can't be trusted.
b) second how would I using the push back command to store the value
of a variable inside a specific spot. Again if I was using an array
it'd be like array[x][y] += q. Where x and y are the spot in the array
and q is the value.
You don't use push_back() to add a value between two values, you use insert().
The syntaxe array[x][y] += q Will certainly not do what you describe. It will add q to the value at the position array[x][y].
Arrays are different to std::vector because they are of a fixed size. All elements of the array exist while the array exists. When you create a std::vector with its default constructor, it is empty. It contains no elements, so you cannot index any elements.
However, std::vector does have a constructor that takes the initial size. If you pass a single int argument to the std::vector constructor, it will default initialise that many elements. For example:
std::vector<int> v(10); // Will have 10 ints
If you want the equivalent of a 2D array, then you'll need a std::vector<std::vector<T>>. If you want to construct it with a specific size, you will need specify the size of the outer std::vector as above, and pass it the std::vector that each element should be initialised to. For example, if you want a 10x20 vector:
// This will have 10x20 ints
std::vector<std::vector<int>> v(10, std::vector<int>(20));
Once these elements exist, you can index them just as you would an array:
int value = v[x][y];
It's worth noting that C++11 introduces std::array which has a compile-time fixed size. You could use it like this:
std::array<std::array<int, 20>, 10> arr;
However, you cannot use this if you want your array size to be determined by a variable. The dimensions must be compile-time constants.

issues related to the copy of vector with pointer item

I want to ask whether there are some problems with the copy for the vector of pointer items. Do I need to strcpy or memcpy because there may be depth copy problem?
For instance:
Class B;
Class A
{
....
private:
std::vector<B*> bvec;
public:
void setB(std::vector<B*>& value)
{
this->bvec = value;
}
};
void main()
{
....
std::vector<const B*> value; // and already has values
A a;
a.setB(value);
}
This example only assign the value to the class variable bvec inside A class. Do I need to use memcpy since I found that std::vector bvec; has pointer items? I am confused with the depth copy in C++, could you make me clear about that? Thank you.
Think about this, if you remove and delete an item from the vector value after you call setB, then the vector in A will have a pointer that is no longer valid.
So either you need to do a "deep copy", have guarantees that the above scenario will never happen, or use shared smart pointers like std::shared_ptr instead of raw pointers. If you need pointers, I would recommend the last.
There is another alternative, and that is to store the vector in A as a reference to the real vector. However, this has other problems, like the real vector needs to be valid through the lifetime of the object. But here too you can use smart pointers, and allocate the vector dynamically.
It is unlikely you need strcpy or memcpy to solve your problem. However, I'm not sure what your problem is.
I will try to explain copying as it relates to std::vector.
When you assign bvev to value in setB you are making a deep copy. This means all of the elements in the vector are copied from value to bvec. If you have a vector of objects, each object is copied. If you have a vector of pointers, each pointer is copied.
Another option is to simply copy the pointer to the vector if you wish to reference the elements later on. Just be careful to manage the lifetimes properly!
I hope that helps!
You probably want to define your copy constructor for class A to ensure the problem your asking about is handled correctly (though not by using memcpy or strcpy). Always follow the rule of three here. I'm pretty sure with std::vector your good, but if not, then use a for loop instead of memcpy

Pass nested C++ vector as built-in style multi-dimensional array

If I have a vector in C++, I know I can safely pass it as an array (pointer to the contained type):
void some_function(size_t size, int array[])
{
// impl here...
}
// ...
std::vector<int> test;
some_function(test.size(), &test[0]);
Is it safe to do this with a nested vector?
void some_function(size_t x, size_t y, size_t z, int* multi_dimensional_array)
{
// impl here...
}
// ...
std::vector<std::vector<std::vector<int> > > test;
// initialize with non-jagged dimensions, ensure they're not empty, then...
some_function(test.size(), test[0].size(), test[0][0].size(), &test[0][0][0]);
Edit:
If it is not safe, what are some alternatives, both if I can change the signature of some_function, and if I can't?
Short answer is "no".
Elements here std::vector<std::vector<std::vector<int> > > test; are not replaced in contiguous memory area.
You can only expect multi_dimensional_array to point to a contiguos memory block of size test[0][0].size() * sizeof(int). But that is probably not what you want.
It is erroneous to take the address of any location in a vector and pass it. It might seem to work, but don't count on it.
The reason why is closely tied to why a vector is a vector, and not an array. We want a vector to grow dynamically, unlike an array. We want insertions into a vector be a constant cost and not depend on the size of the vector, like an array until you hit the allocated size of the array.
So how does the magic work? When there is no more internal space to add a next element to the vector, a new space is allocated twice the size of the old. The old space is copied to the new and the old space is no longer needed, or valid, which makes dangling any pointer to the old space. Twice the space is allocated so the average cost of insertion to the vector that is constant.
Is it safe to do this with a nested vector?
Yes, IF you want to access the inner-most vector only, and as long you know the number of elements it contains, and you don't try accessing more than that.
But seeing your function signature, it seems that you want to acess all three dimensions, in that case, no, that isn't valid.
The alternative is that you can call the function some_function(size_t size, int array[]) for each inner-most vector (if that solves your problem); and for that you can do this trick (or something similar):
void some_function(std::vector<int> & v1int)
{
//the final call to some_function(size_t size, int array[])
//which actually process the inner-most vectors
some_function(v1int.size(), &v1int[0]);
}
void some_function(std::vector<std::vector<int> > & v2int)
{
//call some_function(std::vector<int> & v1int) for each element!
std::for_each(v2int.begin(), v2int.end(), some_function);
}
//call some_function(std::vector<std::vector<int> > & v2int) for each element!
std::for_each(test.begin(), test.end(), some_function);
A very simple solution would be to simply copy the contents of the nested vector into one vector and pass it to that function. But this depends on how much overhead you are willing to take.
That being sad: Nested vectorS aren't good practice. A matrix class storing everything in contiguous memory and managing access is really more efficient and less ugly and would possibly allow something like T* matrix::get_raw() but the ordering of the contents would still be an implementation detail.
Simple answer - no, it is not. Did you try compiling this? And why not just pass the whole 3D vector as a reference? If you are trying to access old C code in this manner, then you cannot.
It would be much safer to pass the vector, or a reference to it:
void some_function(std::vector<std::vector<std::vector<int>>> & vector);
You can then get the size and items within the function, leaving less risk for mistakes. You can copy the vector or pass a pointer/reference, depending on expected size and use.
If you need to pass across modules, then it becomes slightly more complicated.
Trying to use &top_level_vector[0] and pass that to a C-style function that expects an int* isn't safe.
To support correct C-style access to a multi-dimensional array, all the bytes of all the hierarchy of arrays would have to be contiguous. In a c++ std::vector, this is true for the items contained by a vector, but not for the vector itself. If you try to take the address of the top-level vector, ala &top_level_vector[0], you're going to get an array of vectors, not an array of int.
The vector structure isn't simply an array of the contained type. It is implemented as a structure containing a pointer, as well as size and capacity book-keeping data. Therefore the question's std::vector<std::vector<std::vector<int> > > is more or less a hierarchical tree of structures, stitched together with pointers. Only the final leaf nodes in that tree are blocks of contiguous int values. And each of those blocks of memory are not necessarily contiguous to any other block.
In order to interface with C, you can only pass the contents of a single vector. So you'll have to create a single std::vector<int> of size x * y * z. Or you could decide to re-structure your C code to handle a single 1-dimensional stripe of data at a time. Then you could keep the hierarchy, and only pass in the contents of leaf vectors.