Hello all i am just beginning to learn c++ and just looked at templates. I was asked to make a template function print that would take an array and a int input and print out the elements in the array. Here is my code.
#include <iostream>
using namespace std;
template <typename T>
void print(T a[], int b)
{
for(int y=0; y < b; y++)
{
cout << a[y] <<" ";
}
}
int main()
{
int arr[5];
int size = sizeof(arr);
for(int h=0; h<=size; h++)
{
arr[h]=0;
}
print(arr,size);
}
The program compiles with no errors but when I run it there is no output. Can anyone help? or point me in the right direction?
Your program has Undefined Behavior at this point:
for (int h = 0; h <= size; h++)
{
arr[h] = 0;
}
You defined size to be the number of bytes held in the array arr, not the actual number of integers present. Therefore, an integer on your platform is most likely greater than 1 byte, so size will be equal to a number far greater than 5.
And when you do arr[h], you're eventually going to access an address beyond the bounds of the array. It is at that point that your program contracts a case of Undefined Behavior. At that moment, anything in your program can happen, including the output not appearing.
The fix to this problem is to divide the bytes present in arr by the size in bytes of a single int. That will tell you the number of elements:
int size = sizeof(arr)/sizeof(int);
An even better solution is to use std::array so that the size is known and can be accessed through its member function size:
std::array<int, 5> arr{{}};
for (int i = 0; i < arr.size(); ++i)
arr[i] = 0;
Note that using aggregate-initialization on std::array causes zero-initialization of each element. Thus, there is no use for the subsequent for loop.
You'll also need to adapt your function so that it can access an array object:
template<typename T, unsigned N>
void print(const std::array<T, N>& a)
{
for (const auto& x : a) // range-based for loop
{
std::cout << x;
}
}
The error is in the value of the variable size.
int size = sizeof(arr);
gives the size of the array in bytes, it is not 5.
int size = sizeof(arr)/sizeof(arr[0]);
will give the right answer.
It would also be a good idea to add
cout << endl;
at the end of main.
Related
I have passed an array of size 10 to a funtion to sort the array reversely, but it's going wrong after rightly sorting first five elements of the array.
I want to sort the array 'std' reversely here,
# include <iostream>
using namespace std;
int reverse(int a[]); //funtion prototype
int main()
{
int std[10] = {0,1,2,3,4,5,6,7,8,9};
reverse(std);
}
int reverse(int a[]) //funtion defination
{
int index = 0;
for (int i = 9; i >= 0; i--)
{
a[index] = a[i]; //swaping values of the array
cout << a[index] << " ";
index++;
}
}
There's basically three things wrong with your code.
You aren't swapping anything
You have to swap the first half of the array with the second half, not swap the whole array. If you do that then everything gets swapped twice, so that nothing changes
You should print the reversed array after you have finished the reverse, not while you are doing the reverse.
Here's some code that fixes all these problems
# include <iostream>
# include <utility>
void reverse(int a[]);
int main()
{
int std[10] = {0,1,2,3,4,5,6,7,8,9};
reverse(std);
// print the array after reversing it
for (int i = 0; i < 10; ++i)
std::cout << std[i] << ' ';
std::cout << '\n';
}
void reverse(int a[])
{
for (int i = 0; i < 5; ++i) // swap the first half of the array with the second half
{
std::swap(a[i], a[9 - i]); // real swap
}
}
Yes you can.
I usually don't use "C" style arrays anymore (they can still be useful, but the don't behave like objects). When passing "C" style arrays to functions you kind of always have to manuall pass the size of the array as well (or make assumptions). Those can lead to bugs. (not to mention pointer decay)
Here is an example :
#include <array>
#include <iostream>
// using namespace std; NO unlearn trhis
template<std::size_t N>
void reverse(std::array<int, N>& values)
{
int index = 0;
// you only should run until the middle of the array (size/2)
// or you start swapping back values.
for (int i = values.size() / 2; i >= 0; i--, index++)
{
// for swapping objects/values C++ has std::swap
// using functions like this shows WHAT you are doing by giving it a name
std::swap(values[index], values[i]);
}
}
int main()
{
std::array<int,10> values{ 0,1,2,3,4,5,6,7,8,9 };
reverse(values);
for (const int value : values)
{
std::cout << value << " ";
}
return 0;
}
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";
}
}
#include <iostream>
using namespace std;
void value(int array[],int size){
int minimum;
int maximum;
minimum = array[0];
for(int x = 0; x < size; x++){
if(minimum > array[x+1]){
minimum = array[x+1];
}
}
maximum = array[0];
for(int x = 0; x < size; x++){
if(maximum < array[x+1]){
maximum = array[x+1];
}
}
cout << "Minimum Value is: " << minimum << endl;
cout << "Maximum Value is: " << maximum;
}
int main(){
int size;
cout << "Number of values you want to input: ";
cin >> size;
cout << "Input " << size << " values" << endl;
int array[size];
for(int x = 0; x < size; x++){
cout << "Input #" << x+1 <<": ";
cin >> array[x];
}
value(array,size);
return 0;
How can I output the maximum value inside the array? whenever I print the value the maximum value always return a number that is not present inside the array but the minimum seems fine, its only the maximum value that I am encountering a problem, I tried every possible answer that I know but it doesn't work, I hope ya'll can help Thank you in advance
for(int x = 0; x < size; x++){
If you have an array with ten values, size will be 10. If you work out, with paper and pencil, what this for loop does, you will see that it iterates for values of x 0 through 9, that's what this says. x starts with 0. When it reaches 10, x < size will be false and the loop ends, so the loop runs with x ranging from 0 to 9.
if(minimum > array[x+1]){
Since x will range from 0-9, it logically follows that x+1 will range from 1 to 10, and so this if statement will check the values in array[1] through array[10].
In C++ array indexes start with 0, not 1. The values in your array are array[0] through array[9]. array[10] does not exist, so the above code is undefined behavior.
Furthermore:
int array[size];
This is not valid C++ either. Your C++ compiler may allow this as a non-standard C++ extension, but array sizes must be fixed, constant sizes in C++, determined at compile time. You can't use a non-constant variable to set the size of an array, C++ does not work this way. If you need to have an array of size that's determined at runtime then you need to use std::vector instead of a plain array, and change the rest of your code accordingly.
Mistake 1
Your example has undefined behavior because of the expression array[x+1]. That is, for the last iteration of the for loop, you're going out of bounds of the array and so have undefined behavior.
Undefined behavior means anything1 can happen including but not limited to the program giving your expected output. But never rely(or make conclusions based) on the output of a program that has undefined behavior.
So the output that you're seeing(maybe seeing) is a result of undefined behavior. And as i said don't rely on the output of a program that has UB. The program may just crash.
So the first step to make the program correct would be to remove UB. Then and only then you can start reasoning about the output of the program.
Mistake 2
In standard C++, the size of an array must be a compile time constant. So in your code:
int size;
cin >> size;
int array[size]; //NOT STANDARD C++
The statement int array[size]; is not standard C++ because size is not a constant expression.
Additionally you don't need 2 separate for loops when you can achieve the goal in 1 for loop as shown below.
Solution 1
You can use std::vector as shown below:
#include <iostream>
#include <vector>
#include <climits>
//this function take a vector as input
void value(const std::vector<int>& arr)
{
int max_num = INT_MIN;
int min_num = INT_MAX;
//iterate through the vector to find the max and min value
for(const int& element: arr)
{
if(element > max_num)
{
max_num = element;
}
if(element < min_num)
{
min_num = element;
}
}
std::cout<<"maximum is: "<<max_num<<std::endl;
std::cout<<"minimum is: "<<min_num; //return the difference of mx and min value
}
int main()
{
int n;
std::cout<<"elements: ";
std::cin >> n;
//create vector of int of size n
std::vector<int> arr(n);
//take elements from user
for(int i=0; i<n; i++)
{
std::cin >> arr[i];
}
value(arr);
return 0;
}
Demo
Solution 2
You can make the function a function template so that you don't need to pass a separate argument to the function as shown below:
#include <iostream>
#include <climits>
//N is a nontype template parameter
template<std::size_t N>
void value(const int (&array)[N]){
int max_num = INT_MIN;
int min_num = INT_MAX;
//iterate through the array to find the max and min value
for(const int& element: array)
{
if(element > max_num)
{
max_num = element;
}
if(element < min_num)
{
min_num = element;
}
}
std::cout<<"maximum is: "<<max_num<<std::endl;
std::cout<<"minimum is: "<<min_num;
}
int main(){
int array[3] = {};
for(int x = 0; x < sizeof (array) / (sizeof (array[0])); x++){
std::cout << "Input #" << x+1 <<": ";
std::cin >> array[x];
}
value(array); //no need to pass the second argument
return 0;
}
Demo
Also note that with C++17, you can use std::size instead of sizeof (array) / (sizeof (array[0])) to find the length of the array.
1For a more technically accurate definition of undefined behavior see this where it is mentioned that: there are no restrictions on the behavior of the program.
I'm trying to initialize an int matrix in C++, with user-inputted dimensions, with every element set to zero. I know there are some elegant ways to do that with a one-dimensional array so I was wondering if there are any similar ways to do it with a two-dimensional array without using for loops and iterating through every element.
I found a source that gave several different ways, including std::fill (I've modified the code so that the dimensions are read with cin):
#include <iostream>
using namespace std;
int main() {
int x;
cin >> x;
int matrix[x][x];
fill(*matrix, *matrix + x * 3, 0);
for (int i = 0; i < x; i++) {
for (int j = 0; j < 3; j++) {
cout << matrix[i][j] << " ";
}
cout << endl;
}
}
But why does this work, and why would the pointer to the matrix in the arguments for fill be necessary if it's not necessary for a one-dimensional array? That source said it was because matrixes in C++ are treated like one-dimensional arrays, which would make sense, but that is why I don't understand why the pointer is needed.
I don't know if this is relevant, but in case it can help, I've described my previous attempts below.
At first I thought I could initialize all elements to zero like in a one-dimensional array. For the matrix, this worked fine when the side lengths were not read with cin (i.e. when I declared the matrix as int matrix[3][3] = {{}}; as answered here) but when I tried getting the side lengths from cin I started getting errors.
This was my code:
#include <iostream>
using namespace std;
int main() {
int x;
cin >> x;
int matrix[x][x] = {{}};
for (int i = 0; i < 3; i++) {
for (int j = 0; j < 3; j++) {
cout << matrix[i][j] << " ";
}
cout << endl;
}
}
And when I tried to compile it, it threw this error:
matrix_test.cpp:7:14: error: variable-sized object may not be initialized
int matrix[x][x] = {{}};
^
1 error generated.
Why you're getting the error
c-style arrays (such as int matrix[3][3]) must have size specified at the point you declare it. They can't vary in size in C++.
What you could do instead.
If you use std::vector, there's a really elegant way to do it:
#include <vector>
#include <iostream>
int main() {
using namespace std;
int x;
cin >> x;
auto matrix = vector<vector<int>>(x, vector<int>(x, 0));
// This is how we can print it
for(auto& row : matrix) {
for(auto& elem : row) {
cout << elem << ' ';
}
cout << '\n';
}
}
In C++17, you can shorten this even further:
auto matrix = vector(x, vector(x, 0));
What vector(number, thing) means is "Create a vector of number, where each element is thing".
The second dimension of two-dimension array must be a compile time constant, but in your code x is not.
Actually if you write a function with a two-dimension parameter, the second dimension must also be a compile time constant. That's because the array is stored linearly in the memory and the compiler must know the second dimension to calculate the offset correctly.
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