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.
Related
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
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...)
I am learning C++ and I have trouble with pointers to structures stored in vector. The problem is that I need to keep the structure Student sorted twice. Once by student's id and another time by student's name, so it is easy to search the values in it. Because of this I created two vectors of pointers:
vector<Student *> sortedByID;
vector<Student *> sortedByName;
The structure looks like this and I keep it in vector as well (even though it is probably not a good idea):
struct Student {
int id;
string name;
};
vector <Student> students;
I am creating the new struct with push_back and filling it with parameters from a function (yes, I have a constructor). To keep the vector of pointers sorted I am using lower_bound as shown below:
students.push_back(Student(id, name));
it = lower_bound(sortedByID.begin(), sortedByID.end(), id, cmp());
sortedByID.insert(it, &(students.back()));
//the same for name
The problem is that everytime I add the structure with push_back it reallocs new vector and destroys the address of previous objects, so pointers in vector sortedByID point to invalid value. I think it would be the same with an array of structs, because once the array is full, there is no other way (as far as I know) to resize it, than to create a new array and copy all the data from previous one (so the address will change again).
Is there any clever way to solve this problem? Please note that I am only allowed to use vector and not any other containers from STL.
There are three options to solve this issue using only vectors and no other containers:
1) Avoid the reallocation. This can only be achieved if you know the maximum M of expected number of elements to be inserted in your vector. In this case you can students.reserve(M);.
2) Forget the the pointers for sortedByID and sortedByName. Use integers (or better said size_t) store the index of a student in students instead of a pointer. This supposes of course that the order of items in students is never changed.
3) Do not store the students themselves in a vector, but make students a vector of pointers to students (unsorted) that are allocated from free store. If this alternative meets all your criteria, i'd suggest to go fore shared_ptr<Student> instead of raw pointers.
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.
If I want to declare a vector of unknown size, then assign values to index 5, index 10, index 1, index 100, in that order. Is it easily doable in a vector?
It seems there's no easy way. Cause if I initialize a vector without a size, then I can't access index 5 without first allocating memory for it by doing resize() or five push_back()'s. But resize clears previously stored values in a vector. I can construct the vector by giving it a size to begin with, but I don't know how big the vector should.
So how can I not have to declare a fixed size, and still access non-continuous indices in a vector?
(I doubt an array would be easier for this task).
Would an std::map between integer keys and values not be an easier solution here? Vectors will require a contiguous allocation of memory, so if you're only using the occasional index, you'll "waste" a lot of memory.
Resize doesn't clear the vector. You can easily do something like:
if (v.size() <= n)
v.resize(n+1);
v[n] = 42;
This will preserve all values in the vector and add just enough default initialized values so that index n becomes accessible.
That said, if you don't need all indexes or contigous memory, you might consider a different data structure.
resize() doesn't clear previously stored values in a vector.
see this documentation
I would also argue that if this is what you need to do then its possible that vector may not be the container for you. Did you consider using map maybe?
Data structures which do not contain a contiguous set of values are known as sparse or compressed data structures. It seems that this is what you are looking for.
If this is case, you want a sparse vector. There is one implemented in boost, see link text
Sparse structures are typically used to conserve memory. It is possible from your problem description that you don't actually care about memory use, but about addressing elements that don't yet exist (you want an auto-resizing container). In this case a simple solution with no external dependencies is as follows:
Create a template class that holds a vector and forwards all vector methods to it. Change your operator[] to resize the vector if the index is out of bounds.
// A vector that resizes on dereference if the index is out of bounds.
template<typename T>
struct resize_vector
{
typedef typename std::vector<T>::size_type size_type;
// ... Repeat for iterator/value_type typedefs etc
size_type size() const { return m_impl.size() }
// ... Repeat for all other vector methods you want
value_type& operator[](size_type i)
{
if (i >= size())
resize(i + 1); // Resize
return m_impl[i];
}
// You may want a const overload of operator[] that throws
// instead of resizing (or make m_impl mutable, but thats ugly).
private:
std::vector<T> m_impl;
};
As noted in other answers, elements aren't cleared when a vector is resized. Instead, when new elements are added by a resize, their default constructor is called. You therefore need to know when using this class that operator[] may return you a default constructed object reference. Your default constructor for <T> should therefore set the object to a sensible value for this purpose. You may use a sentinel value for example, if you need to know whether the element has previously been assigned a value.
The suggestion to use a std::map<size_t, T> also has merit as a solution, provided you don't mind the extra memory use, non-contiguous element storage and O(logN) lookup rather than O(1) for the vector. This all boils down to whether you want a sparse representation or automatic resizing; hopefully this answer covers both.