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I've seen code that uses a const std::vector, however can't see why it wouldn't make more sense to simply use an std::array instead.
The values for the vector seem to be initialized at compile-time.
What is the benefit of a const std::vector?
Edit: The vector was not big, only a couple of strings, however I see that this may be one advantage.
const std::vector lets you have a "fixed sized array" whose size you only know at run time but still allows you to have all the benefits of a standard container. If you used a raw/smart pointer in your code you will need to manually pass the size of the array that it points to into the function(s) that need to know the size of the array.
The std::vector could be used with a large list of data, because it could use Dynamic Allocated Array to store values and before C++20 it's doesn't have a constexpr constructor. But std::array uses a raw C-Array and it could be used at compile time with the stack size restricted list of data. So:
std::array const is good for:
Where your data size is less than the stack size.
Where data locality is important.
You want your list at the compile time (before C++20).
std::vector const is good for:
Where your data size is large than the stack size.
It's quite common to see C++ code that expects references to vectors, sometimes const, something like this:
auto do_sum(std::vector<int> const& numbers) -> int {
return std::accumulate(numbers.begin(), numbers.end(), 0);
}
Even though the good intentions are not about not copying memory, this code require the caller to allocate a std::vector, even though the amount of value and the actual values might be known at compile time.
This might be fixed by using a span instead:
auto do_sum(std::span<int const> numbers) -> int {
return std::accumulate(numbers.begin(), numbers.end(), 0);
}
Now this code don't require the users to allocate a vector for the array, and might use std::array, or plain C arrays.
Also, sometimes you don't know what are the numbers of element the array can have, and might be determined at runtime. Consider this:
auto create_vector() -> std::vector<int> {
auto vec = std::vector<int>{};
if (/* runtime condition */) {
vec.push_back(1);
vec.push_back(2);
vec.push_back(3);
} else {
vec.push_back(4);
vec.push_back(5);
}
return vec;
}
int main() {
std::vector<int> const vec = create_vector();
}
As you can see here, as the vector is moved from scope to scope, the const-ness changes and is expressing the intent of the developer to only make it mutable in some initializing scopes.
Because the std::array needs you to specify the size as part of the type. By using std::vector the compiler can work that out for your dynamically.
This matters in maintenance situations as it prevents errors. You only add/remove a string from the initializer of the object and the vector will have the correct size automatically. If you use an array you need to add/remove the string and change the type of the array.
const std::vector<std::string> dataV = { "A", "B", "C" };
const std::array<std::string, 3> dataA = { "A", "B", "C" };
If I now modify these to only have two values.
// The vector will auto resize
const std::vector<std::string> dataV = { "A", "B"};
// This will still be of size three.
// This is not usually what you want.
const std::array<std::string, 3> dataA = { "A", "B"};
// The person modifying the code has to manually spot that and
// change the type to explicitly have two member array. Note
// It may not be as obvious as you think as the type may
// be hidden with a type alias of some description
using DataStore = std::array<std::string, 3>;
/// Lots of code:
DataStore dataA = { "A", "B"}; // Would you have spotted.
I am following this example to make an adjacency list. However it seems like the vector size cannot be dynamic.
Visual studio throws an error
expression did not evaluate to a constant
on this line
vector<int> adj[V];
The strange thing is that the same exact code works correctly on codeblocks IDE.
I've tried replacing the above line with vector<int> adj; but then I cannot send the vector as a parameter to addEdge(adj, 0, 1); as it throws another error about pointers which I also don't know how to correct.
What could I do to dynamically create my vector?
C++ - How to create a dynamic vector
You don't need to do that for this example. But if you did need it, you could use std::make_unique.
The linked example program is ill-formed. I recommend to not try to learn from that. The issue that you encountered is that they use a non-const size for an array. But the size of an array must be compile time constant in C++. Simple fix is to declare the variable type as const:
const int V = 5;
I've tried replacing the above line with vector<int> adj;
You can't just replace an array of vectors with a single vector and expect the program to work without making other changes.
I need the size to be dynamic as it will only be known at compile time.
Assuming you meant to say that the size will only be known at runtime, the solution is to use a vector of vectors.
As written by eerorika, the example code isn't a good one, and you should avoid using raw arrays like that. An array in C/C++ is of static size, each vector in this array is dynamic, but the entire array is not!
There are two approaches for such a question. Either use adjacency lists (which is more common):
#include <vector>
#include <stdint.h>
class Vertix
{
public:
Vertix(uint64_t id_) : id(id_) {}
uint64_t get_id() const { return id; }
void add_adj_vertix(uint64_t id) { adj_vertices.push_back(id); }
const std::vector<uint64_t>& get_adj_vertices() const { return adj_vertices; }
private:
uint64_t id;
std::vector<uint64_t> adj_vertices;
};
class Graph
{
public:
void add_vertix(uint64_t id)
{
vertices[id] = Vertix(id);
}
void add_edge(uint64_t v_id, uint64_t u_id)
{
edges.emplace_back(u_id, v_id);
vertices[u_id].add_adj_vertix(v_id);
}
private:
std::vector<Vertix> vertices;
std::vector<std::pair<uint64_t, uint64_t>> edges;
};
or use double vector to represent the edges matrix:
std::vector<std::vector<uint64_t>> edges;
But it isn't a real matrix, and you cannot check if (u, v) is in the graph in O(1), which misses the point of having adjacency matrix. Assuming you know the size of Graph on compile time, you should write something like:
#include <array>
#include <stdint.h>
template <size_t V>
using AdjacencyMatrix = std::array<std::array<bool, V>, V>;
template <size_t V>
void add_edge(AdjacencyMatrix<V>& adj_matrix, uint64_t u, uint64_t v)
{
if (u < V && v < V)
{
adj_matrix[u][v] = true;
}
else
{
// error handling
}
}
Then you can use AdjacencyMatrix<5> instead of what they were using on that example, in O(1) time, and although it has static size, it does work as intended.
There’s no need to use C-style arrays in modern C++. Their equivalent is std::array, taking the size as a template parameter. Obviously that size can’t be a runtime variable: template parameters can be types or constant expressions. The compiler error reflects this: std::array is a zero cost wrapper over an internal, raw “C” array.
If the array is always small, you may wish to use a fixed-maximum-size array, such as provided by boost. You get all performance benefits of fixed size arrays and can still store down to zero items in it.
There are other solutions:
If all vectors have the same size, make a wrapper that takes two indices, and uses N*i1+i2 as the index to an underlying std::vector.
If the vectors have different sizes, use a vector of vectors: std::vector>. If there are lots of vectors and you often add and remove them, you may look into using a std::list of vectors.
Is there any problem with my code ?
std::vector<int[2]> weights;
int weight[2] = {1,2};
weights.push_back(weight);
It can't be compiled, please help to explain why:
no matching function for call to ‘std::vector<int [2], std::allocator<int [2]> >::push_back(int*&)’
The reason arrays cannot be used in STL containers is because it requires the type to be copy constructible and assignable (also move constructible in c++11). For example, you cannot do the following with arrays:
int a[10];
int b[10];
a = b; // Will not work!
Because arrays do not satisfy the requirements, they cannot be used. However, if you really need to use an array (which probably is not the case), you can add it as a member of a class like so:
struct A { int weight[2];};
std::vector<A> v;
However, it probably would be better if you used an std::vector or std::array.
You cant do that simply.
It's better you use either of these:
vector <vector<int>> (it's basically a two dimensional vector.It should work in your case)
vector< string > (string is an array of characters ,so you require a type cast later.It can be easily.).
you can declare an structure (say S) having array of int type within it i.e.
struct S{int a[num]} ,then declare vector of
vector< S>
So indirectly, you are pushing array into a vector.
Array can be added to container like this too.
int arr[] = {16,2,77,29};
std::vector<int> myvec (arr, arr + sizeof(arr) / sizeof(int) );
Hope this helps someone.
Arrays aren't copy constructable so you can't store them in containers (vector in this case). You can store a nested vector or in C++11 a std::array.
You should use std::array instead of simple array:
#include <vector>
#include <array>
std::vector<std::array<int, 2>> weights;
std::array<int, 2> weight = {1, 2};
weights.push_back(weight);
or with a constructor:
std::vector<std::array<int, 2>> weights;
weights.push_back(std::array<int, 2> ({1, 2});
One possible solution is:
std::vector<int*> weights;
int* weight = new int[2];
weight[0] =1; weight[1] =2;
weights.push_back(weight);
Just use
vector<int*> .That will definitely work.
A relevant discussion on the same topic : Pushing an array into a vector
Situation like:
int arr[3] = { 1, 2, 3 };
std::vector<int[]> v;
v.push_back(arr);
doesn't work with error "cannot initialize array in vector with .."
This, could be worked well
int * arr = new int[3] { 1, 2, 3 };
std::vector<int*> v;
v.push_back(arr);
To instantiate the vector, you need to supply a type, but int[2] is not a type, it's a declaration.
all
I've a legacy code which in draft does something like this:
// sadly I have to use this structure
struct LegacyStruct {
int* values;
}
LegacyStruct* LgStr;
....
std::vector<int> vec;
// fill vector in some way here
size_t sz = vec.size();
LgStr->values = new int[sz];
std::copy(vec.begin(), vec.end(), &LgStr->values[0]);
vec can be huge and I need to avoid copying it to int*.
Is there a way to do it?
I tried following:
// type of new operator explained in More Effective C++
LgStr->values = new (&vec[0])int[vec.size()];
Ok, values points to the beginning of vec inner array, but it destroyed when vec is out of scope. But I have to keep it..
&vec[0] = nullptr; // does not compile of course
So question is: is it possible to apply move semantics in this case?
Or maybe some other trick?
The short answer is that no, there isn't any way to transfer ownership of a vector's buffer outside the vector.
I think your best option is to make sure that the vector just doesn't die by using a wrapper:
class LegacyStructWrapper : private boost::noncopyable // Or declare private copy constructor/copy assignment or use `= delete` in C++11.
{
private:
std::vector<int> vec_;
LegacyStruct wrapped_;
}
Then anytime you need to use values, just assign it to &vec_[0]. This will stay constant if/until you add more items to the vector (so you will have to use care to make sure that vector resizes don't cause problems).
Yup, you can do so - with a small trick:
struct LegacyStruct {
std::vector<int> backingStore;
int* values;
LegacyStruct(std::vector<int>& aSource) {
// Steal memory
aSource.swap(backingStore);
// Set pointer
values = &backingStore[0];
};
}
The vector.swap operation doesn't copy the ints.
This question already has answers here:
Initialization of all elements of an array to one default value in C++?
(12 answers)
Closed 4 months ago.
I'm trying to initialize an int array with everything set at -1.
I tried the following, but it doesn't work. It only sets the first value at -1.
int directory[100] = {-1};
Why doesn't it work right?
I'm surprised at all the answers suggesting vector. They aren't even the same thing!
Use std::fill, from <algorithm>:
int directory[100];
std::fill(directory, directory + 100, -1);
Not concerned with the question directly, but you might want a nice helper function when it comes to arrays:
template <typename T, size_t N>
T* end(T (&pX)[N])
{
return pX + N;
}
Giving:
int directory[100];
std::fill(directory, end(directory), -1);
So you don't need to list the size twice.
I would suggest using std::array. For three reasons:
1. array provides runtime safety against index-out-of-bound in subscripting (i.e. operator[]) operations,
2. array automatically carries the size without requiring to pass it separately
3. And most importantly, array provides the fill() method that is required for
this problem
#include <array>
#include <assert.h>
typedef std::array< int, 100 > DirectoryArray;
void test_fill( DirectoryArray const & x, int expected_value ) {
for( size_t i = 0; i < x.size(); ++i ) {
assert( x[ i ] == expected_value );
}
}
int main() {
DirectoryArray directory;
directory.fill( -1 );
test_fill( directory, -1 );
return 0;
}
Using array requires use of "-std=c++0x" for compiling (applies to the above code).
If that is not available or if that is not an option, then the other options like std::fill() (as suggested by GMan) or hand coding the a fill() method may be opted.
If you had a smaller number of elements you could specify them one after the other. Array initialization works by specifying each element, not by specifying a single value that applies for each element.
int x[3] = {-1, -1, -1 };
You could also use a vector and use the constructor to initialize all of the values. You can later access the raw array buffer by specifying &v.front()
std::vector directory(100, -1);
There is a C way to do it also using memset or various other similar functions. memset works for each char in your specified buffer though so it will work fine for values like 0 but may not work depending on how negative numbers are stored for -1.
You can also use STL to initialize your array by using fill_n. For a general purpose action to each element you could use for_each.
fill_n(directory, 100, -1);
Or if you really want you can go the lame way, you can do a for loop with 100 iterations and doing directory[i] = -1;
If you really need arrays, you can use boosts array class. It's assign member does the job:
boost::array<int,N> array; // boost arrays are of fixed size!
array.assign(-1);
It does work right. Your expectation of the initialiser is incorrect. If you really wish to take this approach, you'll need 100 comma-separated -1s in the initialiser. But then what happens when you increase the size of the array?
use vector of int instead a array.
vector<int> directory(100,-1); // 100 ints with value 1
It is working right. That's how list initializers work.
I believe 6.7.8.10 of the C99 standard covers this:
If an object that has automatic
storage duration is not initialized
explicitly, its value is
indeterminate. If an object that has
static storage duration is not
initialized explicitly, then:
if it has pointer type, it is initialized to a null pointer;
if it has arithmetic type, it is initialized to (positive or unsigned)
zero;
if it is an aggregate, every member is initialized (recursively) according
to these rules;
if it is a union, the first named member is initialized (recursively)
according to these rules.
If you need to make all the elements in an array the same non-zero value, you'll have to use a loop or memset.
Also note that, unless you really know what you're doing, vectors are preferred over arrays in C++:
Here's what you need to realize about containers vs. arrays:
Container classes make programmers more productive. So if you insist on using arrays while those around are willing to use container classes, you'll probably be less productive than they are (even if you're smarter and more experienced than they are!).
Container classes let programmers write more robust code. So if you insist on using arrays while those around are willing to use container classes, your code will probably have more bugs than their code (even if you're smarter and more experienced).
And if you're so smart and so experienced that you can use arrays as fast and as safe as they can use container classes, someone else will probably end up maintaining your code and they'll probably introduce bugs. Or worse, you'll be the only one who can maintain your code so management will yank you from development and move you into a full-time maintenance role — just what you always wanted!
There's a lot more to the linked question; give it a read.
u simply use for loop as done below:-
for (int i=0; i<100; i++)
{
a[i]= -1;
}
as a result as u want u can get
A[100]={-1,-1,-1..........(100 times)}
I had the same question and I found how to do, the documentation give the following example :
std::array<int, 3> a1{ {1, 2, 3} }; // double-braces required in C++11 (not in C++14)
So I just tried :
std::array<int, 3> a1{ {1} }; // double-braces required in C++11 (not in C++14)
And it works all elements have 1 as value. It does not work with the = operator. It is maybe a C++11 issue.
Can't do what you're trying to do with a raw array (unless you explicitly list out all 100 -1s in the initializer list), you can do it with a vector:
vector<int> directory(100, -1);
Additionally, you can create the array and set the values to -1 using one of the other methods mentioned.
Just use this loop.
for(int i =0 ; i < 100 ; i++) directory[i] =0;
the almighty memset() will do the job for array and std containers in C/C++/C++11/C++14
The reason that int directory[100] = {-1} doesn't work is because of what happens with array initialization.
All array elements that are not initialized explicitly are initialized implicitly the same way as objects that have static storage duration.
ints which are implicitly initialized are:
initialized to unsigned zero
All array elements that are not initialized explicitly are initialized implicitly the same way as objects that have static storage duration.
C++11 introduced begin and end which are specialized for arrays!
This means that given an array (not just a pointer), like your directory you can use fill as has been suggested in several answers:
fill(begin(directory), end(directory), -1)
Let's say that you write code like this, but then decide to reuse the functionality after having forgotten how you implemented it, but you decided to change the size of directory to 60. If you'd written code using begin and end then you're done.
If on the other hand you'd done this: fill(directory, directory + 100, -1) then you'd better remember to change that 100 to a 60 as well or you'll get undefined behavior.
If you are allowed to use std::array, you can do the following:
#include <iostream>
#include <algorithm>
#include <array>
using namespace std;
template <class Elem, Elem pattern, size_t S, size_t L>
struct S_internal {
template <Elem... values>
static array<Elem, S> init_array() {
return S_internal<Elem, pattern, S, L - 1>::init_array<values..., pattern>();
}
};
template <class Elem, Elem pattern, size_t S>
struct S_internal<Elem, pattern, S, 0> {
template <Elem... values>
static array<Elem, S> init_array() {
static_assert(S == sizeof...(values), "");
return array<Elem, S> {{values...}};
}
};
template <class Elem, Elem pattern, size_t S>
struct init_array
{
static array<Elem, S> get() {
return S_internal<Elem, pattern, S, S>::init_array<>();
}
};
void main()
{
array<int, 5> ss = init_array<int, 77, 5>::get();
copy(cbegin(ss), cend(ss), ostream_iterator<int>(cout, " "));
}
The output is:
77 77 77 77 77
Just use the fill_n() method.
Example
int n;
cin>>n;
int arr[n];
int value = 9;
fill_n(arr, n, value); // 9 9 9 9 9...
Learn More about fill_n()
or
you can use the fill() method.
Example
int n;
cin>>n;
int arr[n];
int value = 9;
fill(arr, arr+n, value); // 9 9 9 9 9...
Learn More about fill() method.
Note: Both these methods are available in algorithm library (#include<algorithm>). Don't forget to include it.
Starting with C++11 you could also use a range based loop:
int directory[10];
for (auto& value: directory) value = -1;