How to initialize a 2d array with -1 globally? - c++

The below is the basic code. I want to make the array globally so that i don't have to call it for every function.Now how do I initialize the 2d array with -1. I tried to use memset(arr,-1,sizeof(arr)) just below line 3, but it didn't worked out, so, can anyone tell me what am I doing wrong??
#include <bits/stdc++.h>
using namespace std;
int arr[10][10];
int func(){
//this function will be using the global arr
// if(arr[x][y]!=-1)
//do something
}
int main(){
//the code
}

I do not know the good way to initialize a built-in array in place without code repetition. I do, however, know a way to initialize std::array:
#include <array>
#include <utility>
#include <cstddef>
template<size_t... Ix>
auto constexpr make1array(int v, std::index_sequence<Ix...>) {
auto populate = [](int v, size_t) { return v; };
std::array<int, 10> a = { populate(v, Ix)... };
return a;
}
template<size_t... Ix1, size_t... Ix2>
auto constexpr make2array(int v, std::index_sequence<Ix1...> seq, std::index_sequence<Ix2...>) {
auto populate = [](auto v, size_t) { return v; };
std::array<std::array<int, 10>, 10> a = { populate(make1array(v, seq), Ix2)... };
return a;
}
std::array<std::array<int, 10>, 10> arr = make2array(-1, std::make_index_sequence<10>{}, std::make_index_sequence<10>{});
This code produces an array pre-populated with -1 as the value at compile time.

The function memset won't work because memset uses bytes and many integers occupy more than one byte.
IMHO, your best source is to use std::fill.
Example:
std::fill(&arr[0][0], &arr[9][9] + 1, -1);
Otherwise, you can always fall back on the nested loop:
for (int r = 0; r < MAX_ROWS; ++r)
{
for (int c = 0; c < MAX_COLUMNS; ++c)
{
arr[r][c] = -1;
}
}
Your best bet is to let the compiler optimize the nested loops.
There may be some micro-optimizations that you could employ, but the compiler probably already has them in its tool chest.

There is no direct way to initialize raw array by values that aren't result of default initialization. One of the reasons is that array cannot be returned from function and cannot be assigned directly from anything that is not a {}-list.
Simplest way (since C++14) is to make it part of class-type with constexpr constructor. In C++111 constructor with non-empty body cannot be constexpr.
#include <iostream>
struct MinusOneArray {
static constexpr int NX = 10;
static constexpr int NY = 10;
int arr[NX][NY];
constexpr MinusOneArray() : arr() {
for(int i = 0; i < NX; ++i)
for(int j = 0; j < NY; ++j)
arr[i][j] = -1;
}
};
int main()
{
MinusOneArray a;
auto &arr = a.arr;
for(auto &line: arr) {
for(auto val: line)
std::cout << val << ",";
std::cout << std::endl;
}
}
Alternative is to use standard structure std::array and initialize it with constexpr function, how SergeyA offered.

Related

Create matrix (2d-array) of size specified by parameter input in C++

I am learning C++ with experiencein mostly Python, R and SQL.
The way arrays (and vectors which differes somehow from 1d-arrays? and matrices which are 2d-arrays?) work in C++ seems quite different as I cannot specify the size of dimension of the array with an argument from the function.
A toy-example of my goal is some thing like this:
Have a function my_2d_array which takes two arguments M and N and returns a matrix or 2d-array of dimension (MxN) with elements indicating the position of that element. E.g. calling my_2d_array(4,3) would return:
[[00, 01, 02],
[10, 11, 12],
[20, 21, 22],
[30, 31, 32]]
The main function should execute my_2d_array and be able to potentially perform calculations with the result or modify it.
This is my attempt (with errors):
int my_2d_array(int N, int M) {
int A[N][M];
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++) {
std::string element = std::to_string(i) + std::to_string(j);
A[i][j] = element;
}
}
return A;
}
void main() {
int N, M;
N = 4;
M = 3;
int A[N][M] = my_2d_array(N, M);
// Print the array A
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++) {
std::cout << A[i][j] << " ";
}
std::cout << "\n";
}
}
One (1) dimensional attempt of #JustLearning's suggestion:
int my_array(int N) {
std::array<int, N> A;
for (int i = 0; i < N; i++) {
A[i] = i;
}
return A;
}
int main() {
int N = 4;
int A[N] = my_array(N);
// Print the array A
for (int i = 0; i < N; i++) {
std::cout << A[i] << " ";
}
}
You can use a 2d vector like this
vector<vector int> A;
It works the same way as a 2d array
Welcome to C++! Your function my_2d_array has a couple of issues:
the return type is int, however you are attempting to return an array of ints.
the identifier of an array in C++ is actually a pointer to the first element of that array. Therefore, when you return A, you must be aware of how it should be passed to a new variable in the main part of the code. In particular, your code is passing a reference to a temporary variable A, which is not permitted or safe.
In addition, in C++, unless you know what you're doing, main should always return an int:
int main() { ... }
What is not clear from your question is whether you are attempting to implement your own "array" class, or simply want to use arrays already established in the standard. For the latter, std::array is a good place to start. The advantage is that you can return std::arrays from functions like you return ints or doubles.
std::arrays are good if you plan to work with arrays of fixed size, as the size becomes part of the type: std::array<int, 3> my_array;. Then you can fill it in manually or with member functions of the class (see dox linked above).
If for some reason you prefer to work with arrays of dynamical size (sizes that will change during running your program), std::vector is the way to go.
Finally, if you are actually learning C++ by attempting to implement a container MyArray, you should specify that in your question and be a bit more specific in what help you need.
Here's a working example in 1d:
#include <iostream>
#include <array>
template <int N>
std::array<int, N> my_array() {
std::array<int, N> A;
for (int i = 0; i < N; i++) {
A[i] = i;
}
return A;
}
int main() {
const int N = 4;
std::array<int, N> arr = my_array<N>();
// Print the array A
for (int i = 0; i < N; i++) {
std::cout << arr[i] << " ";
}
}
Since the size of a std::array is included it its type, you need to create a function template, which is basically a function that works for different types. (In C++, std::array<int, 3> and std::array<int, 4> are considered different types.)
In order to use this in main, the index is promoted to a const int, as plain ints can vary during run time, and therefore are not suitable for defining types. (In C++ jargon, look up constant expressions).
Finally, note that both the return type and the type of the variable that receives the value returned by the function must be std::array, not int as you tried in your 1d code.
Following your comment, I can see why you are confused in your attempts to use a matrix in code.
There are many types of containers in C++. Many of them you can find in the standard library (std::vector, std::list, std::set, ...), others you can create yourself or use other libraries. Plain arrays (like int a[5]) are a somewhat unique case because they come from C and are part of the language itself.
A plain array lives on the stack (not very important but you might want to read up on stack vs heap allocations), and refers to a contiguous region of memory.
If you declare some array a like int a[5], you get a region of 5 integers one after the other, and you can point to the first one by just writing a. You can access each of them using a[i] or, equivalently, *(a+i).
If you declare a like int a[5][3], you now get a region of 15 integers, but you can access them slightly differently, like a[i][j], which is equivalent to *(a+i*3+j).
The important thing to you here is that the sizes (5 and 3) must be compile-time constants, and you cannot change them at runtime.
The same is true for std::array: you could declare a like std::array<std::array<int, 3, 5> a and get a similar region of 15 integers, that you can access the same way, but with some convenience (for example you can return that type, whereas you cannot return a plain array type, only a pointer, losing the size information in the process).
My advice is not to think of these arrays as having dimensionality, but as simple containers that give you some memory to work with however you choose. You can very well declare a like std::array<int, 15> a and access elements in a 2D way by indexing like this: a[i*3+j]. Memory-wise, it's the same.
Now, if you want the ability to set the sizes at runtime, you can use std::vector in a similar way. Either you declare a like std::vector<std::vector<int>> a(5, std::vector<int>(3)) and deal with the nested vectors (that initialization creates 5 std::vector<int> of size 3 each), or you declare a as a single vector like std::vector<int> a(15) and index it like a[i*3+j]. You can even make your own class that wraps a vector and helps with the indexing.
Either way, it's rare in C++ to need a plain array, and you should generally use some kind of container, with std::vector being a good choice for a lot of things.
Here is an example of how your code would look like using vectors:
#include <vector>
#include <string>
#include <iostream>
std::vector<std::string> my_2d_array(int N, int M) {
std::vector<std::string> A(N*M);
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++) {
std::string element = std::to_string(i) + std::to_string(j);
A[i*M+j] = element;
}
}
return A;
}
int main() {
int N, M;
N = 4;
M = 3;
std::vector<std::string> A = my_2d_array(N, M);
// Print the array A
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++) {
std::cout << A[i*M+j] << " ";
}
std::cout << "\n";
}
}
And here is a very crude example of a Matrix class used to wrap the vectors:
#include <vector>
#include <string>
#include <iostream>
template<typename T>
class Matrix {
public:
Matrix(int rowCount, int columnCount) : v(rowCount*columnCount), columnCount(columnCount) {}
T& operator()(int row, int column) {
return v[row*columnCount + column];
}
private:
std::vector<T> v;
int columnCount;
};
Matrix<std::string> my_2d_array(int N, int M) {
Matrix<std::string> A(N, M);
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++) {
std::string element = std::to_string(i) + std::to_string(j);
A(i, j) = element;
}
}
return A;
}
int main() {
int N, M;
N = 4;
M = 3;
Matrix<std::string> A = my_2d_array(N, M);
// Print the array A
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++) {
std::cout << A(i, j) << " ";
}
std::cout << "\n";
}
}

How to generate an inline constexpr plain array of array in c++17

The question
I am writing a software in c++17 for which performances are absolutely critical. I would like available in a few key functions constants in arrays themselves in arrays. It matters that both these array are accessible by a integer value in such (or similar) manner :
int main()
{
for (int i = 0; i < size_of_A ; i++)
{
for (int j = 0; j < size_of_B_in_A(i); j++)
{
std::cout << A[i][j];
}
}
}
This would be the kind of array we would like to create assuming some function int f(a, b)
A
{
// B1
{
f(1, 1),
f(1, 2),
f(1, 3),
...
f(1, large number)
},
// B2
{
f(2, 1),
...
f(2, some other large number)
},
... etc
}
The Twist
Each inner array may be of different size which we have will stored elsewhere, we have to find the size at compile time. I would rather not use std::vector for they are assumed
slightly slower
.
Also an I suppose a std::vector would be stored on the heap which would be a performance issue in my specific case.
Furthermore,
std::vector cannot be used as "inline constexpr"
which would be necessary as I expect to have a large amount of value in those array never going to change. I am fine with recompiling all those values each time but not keeping them in an external file by policy as I am to follow a strict coding style.
What I Have Tried
brace initializer
// A.hh
#pragma once
#include <iostream>
void test1();
void test2();
inline constexpr int B1[1] = {1};
inline constexpr int B2[2] = {2, 3};
inline constexpr int B3[3] = {4, 5, 6};
inline constexpr const int *A[3] = {B1, B2, B3};
// main.cc
#include "A.hh"
int main()
{
std::cout << "values : ";
for (int i = 0; i < 3; i++)
{
for (int j = 0; j <= i; j++)
{
std::cout << A[i][j];
}
}
std::cout << "\n\naddress test : \n";
std::cout << &A << '\n';
test1();
test2();
}
// somewhere.cc
#include "A.hh"
void test1()
{
std::cout << &A << '\n';
}
// elsewhere.cc
#include "A.hh"
void test2()
{
std::cout << &A << '\n';
}
which prints :
./a.out
values : 123456
address test :
0x56180505cd70
0x56180505cd70
0x56180505cd70
Therefore A has not been copied in main.cc, somewhere.cc and elsewhere.cc which is good. I would like to go further and be able to create a huge amount of values.
struct with constexpr
using tips found
here
, I do this to be able to perform operations during array construction.
// B.hh
#pragma once
#include <iostream>
template <int N>
struct X
{
int arr[N];
constexpr X(): arr()
{
for (int i = 0; i < N; i++)
{
arr[i] = i % 3;
}
}
};
inline constexpr auto A = X<500>();
// main.cc
#include "B.hh"
int main()
{
for (int i = 0; i < 500; i++)
{
std::cout << A.arr[i];
}
}
Which unsuspectingly prints out
012012 (etc)...
Finally an array of array
And this where I am stuck
#pragma once
#include <iostream>
template <int N>
struct sub_array
{
int arr[N];
constexpr sub_array() : arr()
{
for (int i = 0; i < N; i++)
{
arr[i] = i;
}
}
};
struct array
{
sub_array</*what here ?*/> arr[100];
constexpr array() : arr()
{
for (int i = 0; i < 100; i++)
{
int size = i * 2; // a very large number
// the value of 'size' is not usable in a constant expression
//
// I see why it is, but I can't think of any other way
arr[i] = sub_array<size>;
}
}
};
inline constexpr array A = array();
How can I build such kind of array ?
Thank you for your time and consideration.
Just use std::array<std::span<int>, N>, which is a fixed size array of spans of different sizes. To generate this, use an std::index_sequence
Header:
constexpr std::size_t size_of_A = 500;
extern const std::array<const std::span<const int>, size_of_A>& A;
Implementation:
constexpr std::size_t size_of_B_in_A(std::size_t i) { return i%10+1;}
constexpr int f(std::size_t i, std::size_t j) {return static_cast<int>(i%(j+1));}
template <int I, int N>
struct B
{
std::array<int,N> arr;
explicit constexpr B()
{
for (int j = 0; j < N; j++)
arr[j] = f(I, j);
}
constexpr operator const std::span<const int>() const {return {arr};}
};
template<class index_sequence>
class BGen;
template<std::size_t... I>
struct BGen<std::integer_sequence<std::size_t,I...>> {
static constexpr std::tuple<B<I, size_of_B_in_A(I)>...> bs{};
static constexpr std::array<const std::span<const int>, sizeof...(I)> A {std::get<I>(bs)...};
};
const std::array<const std::span<const int>, size_of_A>& A
= BGen<decltype(std::make_index_sequence<size_of_A>{})>::A;
Usage:
int main()
{
for (unsigned i = 0; i < A.size() ; i++)
{
for (unsigned j = 0; j < A[i].size(); j++)
{
std::cout << A[i][j];
}
}
}
http://coliru.stacked-crooked.com/a/d68b0e9fd6142f86
However, stepping back: This solution is NOT the normal way to go about solving this problem. Since it's all constexpr, this is all data not code. Ergo, the most performant solution is two programs. One generates the data and saves it to a file that ships with (inside?) your program. Then your program simply maps the file into memory, and uses the data directly.
Here's a way of implementing a constexpr jagged array which can be initialized without intermediates. It does require listing the row sizes as template arguments, but there are ways to make that easier too, depending on how the row sizes can be known at compile time.
#include <tuple>
#include <array>
#include <utility>
template <std::size_t ...Sizes>
struct jagged_array
{
const std::tuple<std::array<int,Sizes>...> data;
static constexpr std::size_t num_rows = sizeof...(Sizes);
static constexpr std::size_t length[num_rows]{Sizes...};
int const* const row_ptr[num_rows];
template <std::size_t ...I>
constexpr jagged_array(std::index_sequence<I...>,
const std::array<int, Sizes>& ...arrs)
: data{arrs...}, row_ptr{&std::get<I>(data)[0]...} {}
constexpr jagged_array(const std::array<int, Sizes>& ...arrs)
: jagged_array(std::make_index_sequence<num_rows>{}, arrs...)
{}
constexpr int const* operator[](std::size_t idx) const
{ return row_ptr[idx]; }
};
inline constexpr jagged_array<2,4> jarr = {{2,3}, {4,5,6,7}};

Passing unknown size std::Array in a function with template. How can I correct this code?

I wrote this code using 2d Vector and Array. But I wanted to use std::array this time and my code did not work because this was the first time I use std::array and template.
It gave me for this line:
array<array<int, sizeY>, sizeX> arr;
this error:
Error C2971 std::array: template parameter _Size: sizeY,sizeX: a variable with non-static storage duration cannot be used as a non-type argument
#include <iostream>
#include <array>
using namespace std;
template <size_t Y, size_t X>
bool IsMagicSquare(array<array<int, Y>, X>& ar)
{
int x = ar.size();
int y = ar[0].size();
if (x == y)
{
int ver[x] = { };
int hor[y] = { };
int cross0 = 0;
int cross1 = 0;
for (int i = 0; i < x; i++)
{
for (int j = 0; j < y; j++)
{
hor[i] += ar[i][j];
ver[j] += ar[i][j];
if (i == j)
cross0 += ar[i][j];
if (i + j == x - 1)
cross1 += ar[i][j];
}
}
if (cross0 != cross1)
return false;
else
{
for (int i = 0; i < x; i++)
if ((cross0 != ver[i]) || (cross1 != hor[i]))
return false;
}
}
else
return false;
return true;
}
int main()
{
int sizeX, sizeY;
cout << "Size of Matrix:";
cin >> sizeX >> sizeY;
**array<array<int, sizeY>, sizeX> arr;**
cout << "Elements of the Matrix:";
for (int i = 0; i < sizeX; i++)
for (int j = 0; j < sizeY; j++)
cin >> arr[i][j];
if (IsMagicSquare(arr))
{
for (int i = 0; i < sizeX; i++)
{
cout << "\n";
for (int j = 0; j < sizeY; j++)
cout << arr[i][j];
}
}
else
cout << "Matrix is not magical square!";
return 0;
}
The size of an array (or template arguments in general) has to be known at compile-time, so there is no way to use the runtime values sizeX, sizeY as size (template argument) for an array.
You have to use a variable-length container like std::vector instead.
For reference, here's how you can get a std::array with a size which is decided at runtime:
#include <array>
#include <cstddef>
#include <iostream>
#include <memory>
template<typename T>
struct DynArray {
virtual std::size_t size() const = 0;
virtual T * data() = 0;
virtual ~DynArray() {}
};
template<typename T, std::size_t Size>
struct DynArrayImpl : public DynArray<T> {
std::array<T, Size> array;
std::size_t size() const override {
return array.size();
}
T * data() override {
return array.data();
}
};
template<typename T, std::size_t Size>
struct DynArrayFactory {
static DynArray<T> * allocate(std::size_t const size) {
if (size > Size) {
// ERROR
return nullptr;
}
if (size == Size) {
return new DynArrayImpl<T, Size>();
}
return DynArrayFactory<T, Size - 1>::allocate(size);
}
};
template<typename T>
struct DynArrayFactory<T, 0> {
static DynArray<T> * allocate(std::size_t const size) {
if (size > 0) {
return nullptr;
}
return new DynArrayImpl<T, 0>();
}
};
int main() {
std::size_t size;
std::cin >> size;
std::unique_ptr<DynArray<int>> array{DynArrayFactory<int, 100>::allocate(size)};
std::cout << array->size() << std::endl;
}
This requires a maximum size (100 in this case) to be specified at compile time and is a really convoluted way of doing things; thus not recommended.
Accessing the std::array is nearly impossible though, unless with similar templated code which then generates code for each possible size (see below). This will generate a lot of code. One can easily access the contents of the array however, as seen in the example above. But really: use std::vector.
"similar templated code":
template<std::size_t Size>
struct FillWithNumbers {
static void run(std::array<int, Size> & array) {
int n = 0;
std::generate(begin(array), end(array), [&n](){ return n++; });
}
};
template<typename T, std::size_t Size>
struct DynArrayApply {
template<template<std::size_t S> class Fn>
static void apply(DynArray<T> & array) {
if (array.size() > Size) {
// ERROR
}
if (array.size() == Size) {
DynArrayImpl<T, Size> & real = dynamic_cast<DynArrayImpl<T, Size> &>(array);
Fn<Size>::run(real.array);
}
else {
DynArrayApply<T, Size - 1>::template apply<Fn>(array);
}
}
};
template<typename T>
struct DynArrayApply<T,0> {
template<template<std::size_t S> class Fn>
static void apply(DynArray<T> & array) {
if (array.size() > 0) {
// ERROR
}
DynArrayImpl<T, 0> & real = dynamic_cast<DynArrayImpl<T, 0> &>(array);
Fn<0>::run(real.array);
}
};
int main() {
std::size_t size;
std::cin >> size;
std::unique_ptr<DynArray<int>> array{DynArrayFactory<int, 100>::allocate(size)};
DynArrayApply<int, 100>::apply<FillWithNumbers>(*array);
std::cout << array->size() << std::endl;
std::cout << array->data()[array->size() / 2] << std::endl;
}
I wrote this code using 2d Vector and Array.
That is appropriate, as you do not know the size of the matrix until run time.
But I wanted to use std::array this time [...]
Well, that's a problem because the size of a std::array must be known at compile time. Moving away from C-style arrays is a recommended move, but you have to know where to go. Use the correct tool for the job at hand.
Fixed-size arrays: For arrays whose size is known by the compiler, a std::array is a reasonable replacement. In fact, the std::array is probably nothing more than the C-style array with a different interface.
Variable-size arrays: For arrays whose size is not known until run time, a std::vector is a reasonable replacement. Even though the name does not say "array", it is an array. It is a bit more complex than std::array, but that is because it supports sizes not known at compile time.
This distinction tends to be better-known by those not using gcc, as that compiler has an extension that supports declaring variable-size C-style arrays using the same syntax as declaring fixed-size C-style arrays. It is standard C++ to declare an array along the lines of int col[10]. However, it is not standard C++ to declare an array along the lines of int col[sizeY], where sizeY has a value supplied at run time. The latter syntax is supported by gcc as an extension, and some people use it without realizing it is an extension (ported from gcc's C support). To some extent, std::vector makes this extension available in a more portable form.

How to conditionally add element to std::array - C++11

I have a simple program:
#include <array>
#include <iostream>
#include <functional>
#include <algorithm>
using namespace std;
int main(){
array<int, 5> myArr = {3, 10, 0, 5, 7};
int badNum = 0;
for(int item : myArr){
cout << item << endl;
}
cout << "\n" << endl;
cout << "\n" << endl;
sort(myArr.begin(), myArr.end(), greater<int>());
for(int item : myArr){
cout << item << endl;
}
array<int, 4> goodFour;
for (unsigned int i = 0; i < myArr.size(); i++){
if(myArr[i] != badNum){
// goodThree.append(myArr[i]); <-- This is where I am stuck
}
}
}
I am stuck on trying to assign an element to a std::array. I know in std::vector I can use push_back method, but on a std:array, how to assign to the next (not yet assigned) element? I am coming from Python 3.x where we have the append method for a list. I am not attempting to change the size of the array, instead, I am trying to fill in the allocated spaces with values.
I have looked at:
http://www.cplusplus.com/forum/beginner/67707/
http://www.cplusplus.com/forum/beginner/86394/
http://www.cplusplus.com/forum/beginner/152125/
But these are all for vectors or the primitive int[5] myArr types, not std::array.
If you only want the first three good numbers you would maintain a pointer to the next index to insert at:
for (int i = 0, k = 0; k < 3 && i < myArr.size(); i++) {
if (myArr[i] != badNum) {
goodThree[k++] = myArr[i];
}
}
If you want all the numbers that are good you would use a std::vector and call its push_back method:
std::vector<int> goodThree;
for (int i = 0; i < myArr.size(); i++) {
if (myArr[i] != badNum) {
goodThree.push_back(myArr[i]);
}
}
The size of std::array is fixed at compile time. If you need to append another value at run time, then you have to use std::vector (or something like it). The closest you can get to "appending" to a std::array is if you copy its contents into an array with one more element that contains the "appended" value. You can use some template metaprogramming techniques to make this easier:
template <typename T, std::size_t N, typename X, std::size_t... I>
std::array<T, N + 1> push_back(std::array<T, N> const& a, X&& x, std::index_sequence<I...>) {
return {std::get<I>(a)..., std::forward<X>(x)};
}
template <typename T, std::size_t N, typename X>
std::array<T, N + 1> push_back(std::array<T, N> const& a, X&& x) {
return detail::push_back(a, std::forward<X>(x), std::make_index_sequence<N>());
}
Example of use:
std::array<int, 2> a = {1, 2};
std::array<int, 3> b = push_back(a, 3);
for (int x : b) {
std::cout << x << "\n";
}
The size of a std::array is set at compile time. This means that you cannot "append" values to a std::array at run time. However, you could track how many values your array "contains" by your own program logic. For example:
std::array<int, 5> a = {1, 2, 3, 4, 5};
std::array<int, 4> b;
std::size_t n = 0;
for (int x : a) {
if (x != 2) {
if (n < std::size(a) - 1) {
b[n++] = x;
} else {
throw std::out_of_range("array is full");
}
}
}

Fill in the int array from zero to defined number

i need to fill in the int[] array in C++ from zero to number defined by variable, but ISO C++ forbids variable length array...
How to easily fill in the array? Do i need to allocate/free the memory?
int possibilities[SIZE];
unsigned int i = 0;
for (i = 0; i < SIZE; i++) {
possibilities[i] = i;
}
btw. if you would ask - Yes, i need exactly standard int[] arrays, no vectors, no maps etc.
In c++11 you can use std::iota and std::array. Example below fills array sized 10 with values from 1 to 10.
std::array<int, 10> a;
std::iota(a.begin(), a.end(), 1);
Edit
Naturally std::iota works with vectors as well.
As you've found, you cannot create a variable-length array on the stack. So your choices are either to allocate it on the heap (introduces memory-management issues), or to use a std::vector instead of a C-style array:
std::vector<int> possibilities(SIZE);
for (int i = 0; i < SIZE; i++)
{
possibilities[i] = i;
}
If you want to get even more flashy, you can use STL to generate this sequence for you:
// This is a "functor", a class object that acts like a function with state
class IncrementingSequence
{
public:
// Constructor, just set counter to 0
IncrementingSequence() : i_(0) {}
// Return an incrementing number
int operator() () { return i_++; }
private:
int i_;
}
std::vector<int> possibilities(SIZE);
// This calls IncrementingSequence::operator() for each element in the vector,
// and assigns the result to the element
std::generate(possibilities.begin(), possibilities.end(), IncrementingSequence);
If you have access to boost then you already have access to an incrementing iterator.
#include <vector>
#include <boost/iterator/counting_iterator.hpp>
std::vector<int> possibilities(
boost::counting_iterator<int>(0),
boost::counting_iterator<int>(SIZE));
The counting iterator essentially wraps incrementing a value. So you can automatically tell it the begin and end values and vector will populate itself properly.
As mentioned elsewhere, the resulting vector can be used directly with std::next_permutation.
std::next_permutation(possibilities.begin(),possibilities.end());
std::vector<int> possibilities;
unsigned int i = 0;
for (i = 0; i < SIZE; i++) {
possibilities.push_back(i);
}
Use std::vector(you need to include <vector>)
If you want pass vector to std::next_permutation you need to write:
std::next_permutation(possibilities.begin(),possibilities.end());
also you can use vector as C style arrays. &vec[0] returns pointer to C style array.
You can use the std::generate_n function:
std::generate_n( myarray, SIZE, increment() );
Where increment is an object that generates numbers:
struct increment {
int value;
int operator() () { return ++value; }
increment():value(0){}
};
If you make SIZE a constant (macro or const), you can use it to specify the size of your static array. If it is not possible to use a constant, for example you are reading the intended size from outside the program, then yes you will need to allocate the memory.
In short, if you don't know the size at compile time you probably need to allocate the memory at runtime.
std::generate() can be used with a mutable lambda function to be more concise. The following C++ snippet will place the values 0..9 inclusive in a vector A of size 10 and print these values on a single line of output:
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int main() {
size_t N = 10;
vector<int> A(N);
generate(A.begin(), A.end(), [i = 0]() mutable { return i++; });
copy(A.begin(), A.end(), ostream_iterator<int>(cout, " "));
cout << endl;
return 0;
}
Assuming SIZE is a constant expression and you're allowed to use std::array, you can create a function template that will allow you to do this at compile time. Note from C++17, std::array<T, N>::begin is constexpr so that you can do all this at compile time.
C++17 Version
template<std::size_t N> std::array<int, N> constexpr make_array()
{
std::array<int, N> tempArray{};
int count = 0;
for(int &elem:tempArray)
{
elem = ++count;
}
return tempArray;
}
int main()
{
const int SIZE = 8;
//-------------------------------V-------->number of elements
constexpr auto arr = make_array<SIZE>();
//lets confirm if all objects have the expected value
for(const auto &elem: arr)
{
std::cout << elem << std::endl; //prints 1 2 3 4 5 6 7 8 with newline in between
}
}
C++17 demo
C++11 Version
But prior to C++17, std::array<T, N>::begin was not constexpr, we will need to modify the above example slightly for C++11 as shown below:
template<std::size_t N> std::array<int, N> make_array()
{
std::array<int, N> tempArray{};
int count = 0;
for(int &elem:tempArray)
{
elem = ++count;
}
return tempArray;
}
int main()
{
const int SIZE = 8;
//---------------------VVVV---->number of elements
auto arr = make_array<SIZE>();
//lets confirm if all objects have the expected value
for(const auto &elem: arr)
{
std::cout << elem << std::endl; //prints 1 2 3 4 5 6 7 8 with newline in between
}
}
C++11 demo
Simply use a dynamic arrays?
type * pointer;
pointer = new type[number_of_elements];
void main()
{
int limit = 0; // Your lucky number
int * pointer = NULL;
cout << "Please, enter limit number: ";
cin >> n;
pointer = new int[limit+1]; // Just to be sure.
for (int i = 0; i < n; i++)
{
pointer[i] = i; // Another way is: *(pointer+i) = i (correct me if I'm wrong)
}
delete [] pointer; // Free some memory
pointer = NULL; // If you are "pedant"
}
I don't pretend this is the best solution. I hope it helps.
it should be help u man
int* a = NULL; // Pointer to int, initialize to nothing.
int n; // Size needed for array
cin >> n; // Read in the size
a = new int[n]; // Allocate n ints and save ptr in a.
for (int i=0; i<n; i++) {
a[i] = 0; // Initialize all elements to zero.
}
. . . // Use a as a normal array
delete [] a; // When done, free memory pointed to by a.
a = NULL; // Clear a to prevent using invalid memory reference