I trying to send array to function, but my program gets stuck
int main()
{
int n, i;
bool random;
cout << "number of elements in array:"; cin >> n;
cout << "use random values?"; cin >> random;
int* arr = new int[n]; //create int array with n size
init_array(random, n, arr); //fill array using function
for (i = 0; i <= n; i++) //display array
cout << " " << arr[i];
return 0;
}
This function should fill array with random number or input from keyboard
void init_array(bool random, int n, int* arr)
{
int i;
if (random)
{
srand(time(0)); //initialize random;
for (i = 0; i < n; i++)
arr[i] = rand() % 100;
}
else
for (i = 0; i<n; i++)
cout << "arr[" << i << "]: "; cin >> arr[i];
}
Is there any way send dynamic array to function?
When you do not use brackets after your for-loop, only the first statement is used as a loop:
else
for (i = 0; i<n; i++)
cout << "arr[" << i << "]: "; cin >> arr[i];
This loop will attempt to print "arr[#]" n times, and then ask for an input (which will attempt to be placed in the item 1 after the last element in your array (UB).
What you want is this:
else
{
for (i = 0; i<n; i++)
{
cout << "arr[" << i << "]: ";
cin >> arr[i];
}
}
You also have a problem with your output:
for (i = 0; i < n; i++) // <= would attempt to print 1 more item than exists in the array
And just for completeness, most of these issues go away when you use a container that does all of this for you:
int main()
{
int n = 0;
bool useRandom = false;
std::cout << "number of elements in array:";
std::cin >> n;
std::cout << "use random values?";
std::cin >> useRandom;
std::vector<int> arr(n);
init_array(useRandom, arr);
std::copy(arr.begin(), arr.end(), std::ostream_iterator<int>(std::cout, " "));
return 0;
}
void init_array(bool useRandom, std::vector<int>& vec)
{
::srand(time(0)); //initialize random;
int n = 0;
std::transform(vec.begin(), vec.end(), vec.begin(), [&](int i)
{
if (useRandom)
{
i = rand() % 100;
}
else
{
std::cout << "arr[" << n++ << "]: ";
std::cin >> i;
}
return i;
});
}
Your code is asking for a number at the end because of last cin>>n
fix else part in init_array as:
else
for (i = 0; i<n; i++)
{ //Notice braces
cout << "arr[" << i << "]: ";
cin >> arr[i];
}
Also fix:
for (i = 0; i < n; i++) //display array from index 0 to n-1
cout << " " << arr[i];
Related
I have a problem with this piece of code, I'm trying to print the EVEN and ODD numbers, but there is a problem when it comes to show them, the vectors don't save the numbers as I'm expecting.
#include <iostream>
using namespace std;
int main() {
int n;
cin >> n;
int vect[n], even[n], odd[n]; // CREATING VECTORS LIMIT AFTER "n"
for(int i = 1; i <= n; ++i) { // ENTERING The ELEMENS IN VECTOR
cin >> vect[i];
}
for(int i = 1; i <= n; ++i) {
if(vect[i] % 2 != 0) {
odd[i] = vect[i]; // I think that here's the problem, the vectors don't save the right numbers.
} /// VERIFYING IF THE NUMBER IS ODD OR EVEN.
else if (vect[i] % 2 == 0) {
even[i] == vect[i];
}
}
for(int i = 1; i <= n; ++i) {
cout << even[i] << " " << endl; /// PRINTING THE ODD AND EVEN numbers.
cout << odd[i] << " " << endl;
}
return 0;x
}
I have fixed the problem, thanks all for help.
Now it works perfectly.
#include <iostream>
using namespace std;
int main() {
int n;
cin >> n;
int vect[n], even[n], odd[n], z = 0, x = 0; // CREATING VECTORS LIMIT AFTER "n"
for(int i = 1; i <= n; ++i) { // ENTERING The ELEMENS IN VECTOR
cin >> vect[i];
}
for(int i = 1; i <= n; ++i) {
if(vect[i] % 2 != 0) {
odd[1+z] = vect[i];
z++;
// I think that here's the problem, the vectors don't save the right numbers.
} /// VERIFYING IF THE NUMBER IS ODD OR EVEN.
else if (vect[i] % 2 == 0) {
even[1+x] = vect[i];
x++;
}
}
for(int i = 1; i <= x; i++) {
cout << even[i] << " ";
}
cout << endl;
for(int i = 1; i <= z; i++) {
cout << odd[i] << " ";
}
return 0;
}
Considering the hints of the comments, your program shall be changed into this:
#include <iostream>
#include <vector>
using namespace std;
int main() {
int n, number;
cin >> n;
vector<int> vect, even, odd; // CREATING DYNAMIC VECTORS
for(int i = 0; i < n; ++i) { // ENTERING THE ELEMENTS IN VECTOR
cin >> number;
vect.push_back(number);
}
for(int i = 0; i < n; ++i) {
if(vect[i] % 2 != 0) { /// VERIFYING IF THE NUMBER IS ODD OR EVEN.
odd.push_back(vect[i]);
}
else {
even.push_back(vect[i]);
}
}
for (int i = 0; i < n; ++i)
cout << vect[i] << " ";
cout << endl;
/// PRINTING THE ODD AND EVEN NUMBERS.
for (auto& val : odd)
cout << val << " ";
cout << endl;
for (auto& val : even)
cout << val << " ";
cout << endl;
return 0;
}
It uses the vector container of STL for your arrays, start the indexing at 0 and prints out the resulting arrays separately, as the number of odd and of even entries might be different.
Hope it helps?
With standard, you might use std::partition (or stable version) to solve your problem:
void print_even_odd(std::vector<int> v)
{
auto limit = std::stable_partition(v.begin(), v.end(), [](int n){ return n % 2 == 0; });
std::cout << "Evens:";
// Pre-C++20 span:
// for (auto it = v.begin(); it != limit; ++it) { int n = *it;
for (int n : std::span(v.begin(), limit)) {
std::cout << " " << n;
}
std::cout << std::endl;
std::cout << "Odds:";
for (int n : std::span(limit, v.end())) {
std::cout << " " << n;
}
std::cout << std::endl;
}
Demo
I am new, not that good with functions, and I am trying to solve this question:
Suppose A, B, C are arrays of integers of size [M], [N], and [M][N], respectively. The user will enter the values for the array A and B. Write a user defined function in C++ to calculate the third array C by adding the elements of A and B. If the elements have the same index number, they will be multiplied. C is calculated as the following: -
Use A, B and C as arguments in the function.
Below is my attempt at the problem.
#include<iostream>
using namespace std;
void Mix(int(&A)[], int(&B)[], int(&C)[][100], int N, int M);
//dont understand why you used Q
int main()
{
//variable declaration
int A[100], B[100], C[100][100], n, m, l = 0;
//input of size of elements for first ararys
cout << "Enter number of elements you want to insert in first array: ";
cin >> n;
cout << "-----------------" << endl;
cout << "-----------------" << endl;
cout << "Enter your elements in ascending order" << endl;
//input the elements of the array
for (int i = 0; i < n; i++)
{
cout << "Enter element " << i + 1 << ":";
cin >> A[i];
}
cout << endl << endl;
//input of size of elements for first ararys
cout << "Enter number of elements you want to insert in second array: ";
cin >> m;
cout << "-----------------" << endl;
cout << "-----------------" << endl;
cout << "Enter your elements in descending order" << endl;
//input the elements of the array
for (int i = 0; i < m; i++)
{
cout << "Enter element " << i + 1 << ":";
cin >> B[i];
}
Mix(A, B, C, n, m);
cout << "\nThe Merged Array in Ascending Order" << endl;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++)
{
cout << C[i][j] << " ";
}
cout << "\n"; //endline never use endl its 10 times slower
}
system("pause");
return 0;
}
void Mix(int(&A)[], int(&B)[], int(&C)[][100], int N, int M)
{
// rows is the index for the B array, cols is index for A array
int rows = 0;
int cols = 0;
while (rows < M) {
while (cols < N) {
if (rows == cols) { // remember ==
C[rows][cols] = B[rows] * A[cols];
}
else {
C[rows][cols] = B[rows] + A[cols];
}
cols++; // increment here
}
rows++; // increment here
}
return;
}
Here is an example of the output:
enter image description here
In order to make the C array two-dimensional, it needs to be expressed as C[100][100], instead of C[200]. That is the first step. Next, in your Mix() function, you need to cycle through each element of both A and B (ex. two for loops). Your rows change as B changes, and your columns change as A changes. Include a check for identical indices that will determine whether to add or multiply the two values together.
void Mix(int A[], int B[], int C[][], int N, int M) {
// rows is the index for the B array, cols is index for A array
for (int rows = 0; rows < M; rows++) {
for (int cols = 0; cols < N; cols++) {
if (rows == cols) { // remember ==
C[rows][cols] = B[rows] * A[cols];
} else {
C[rows][cols] = B[rows] + A[cols];
}
}
}
}
Make sure your arrays are properly defined and print out the C array by row and column to match the specifications.
UPDATE: If you want to use while loops, I would default to deconstructing the for loops and apply the same logic:
void Mix(int A[], int B[], int C[][], int N, int M) {
// rows is the index for the B array, cols is index for A array
int rows = 0;
int cols = 0;
while (rows < M) {
while (cols < N) {
if (rows == cols) { // remember ==
C[rows][cols] = B[rows] * A[cols];
} else {
C[rows][cols] = B[rows] + A[cols];
}
cols++; // increment here
}
rows++; // increment here
}
}
I would definitely recommend the for loop approach, as it is more compact, yet does the exact same operations.
There are a lot of things wrong with your code. First off an 2D array must be declared with 2 squared brackets so C[200][200]. In the Mix function the logical operator is == not = in if (A[I] = B[J])
Anyway here's the function that you need:
#include<iostream>
using namespace std;
void Mix(int A[], int B[], int C[], int N, int M) {
//dont understand why you used Q
int i, j;
for(i=0; i<N; i++) {
for(j=0; j<M; j++) {
if(i==j){
C[i][j] = A[i] * B[j];
}
else {
C[i][j] = A[i] + B[j];
}
}
}
return C[i][j];
}
int main()
{
//variable declaration
int A[100], B[100], C[200], j, i, n, m, l = 0;
string Comma;
//input of size of elements for first ararys
cout << "Enter number of elements you want to insert in first array: ";
cin >> n;
cout << "-----------------" << endl;
cout << "-----------------" << endl;
cout << "Enter your elements in ascending order" << endl;
//input the elements of the array
for (i = 0; i < n; i++)
{
cout << "Enter element " << i + 1 << ":";
cin >> A[i];
}
cout << endl << endl;
//input of size of elements for first ararys
cout << "Enter number of elements you want to insert in second array: ";
cin >> m;
cout << "-----------------" << endl;
cout << "-----------------" << endl;
cout << "Enter your elements in descending order" << endl;
//input the elements of the array
for (j = 0; j < m; j++)
{
cout << "Enter element " << j + 1 << ":";
cin >> B[j];
}
C = Mix(A, B, C, n, m);
cout << "\nThe Merged Array in Ascending Order" << endl;
for(i=0; i<n; i++) {
for(j=0; j<m; j++) {
cout<<C[i][j]<<" ";
}
cout<<"\n" //endline never use endl its 10 times slower
}
system("pause");
return 0;
}
Because M and N are defined at run time, you'll really want to use vectors to represent them. Additionally consider returning a 2D container so as to leverage return value optimization.
I'm going to write an example using a vector of vectors for simplicity (see What are the Issues with a vector-of-vectors? for more on why that's really just good for a toy example):
vector<vector<int>> Mix(const vector<int>& A, const vector<int>& B) {
vector<vector<int>> result(size(B), vector<int>(size(A)));
for(size_t i = 0U; i < size(B); ++i) {
for(size_t j = 0U; j < size(A); ++j) {
result[i][j] = A[j] * B[i];
}
}
return result;
}
Live Example
EDIT:
If you must use arrays you'll miss out on return value optimization. I'd only choose this as a good option in the situations:
That you weren't returning anything, in which case your function would probably look something like:
void Mix(const int* A, const int* B, const size_t size_A, const size_t size_B)
{
for(size_t i = 0U; i < size_B; ++i) {
for(size_t j = 0U; j < size_A; ++j) {
cout << '[' << i << "][" << j << "]: " << A[j] * B[i] << '\t';
}
cout << endl;
}
}
That you weren't calling a function and you'd already been given int A[M] and int B[N] as inputs and int C[N][M] as an output, in which case the code you'd inline would probably look something like this:
for(size_t i = 0U; i < size(B); ++i) {
for(size_t j = 0U; j < size(A); ++j) {
C[i][j] = A[j] * B[i];
}
}
I am trying to write a program that will count the number of 5s in a number and display it. So if the user enters:
52315
Then the program should ouptut:
Yes, there are '5's and they're 2
Here is my code but there is something wrong with it.
{
int n,m;
int count = 0;
cout << "Enter an: ";
cin >> n;
int *arr;
arr = new int[count];
// Getting digits from number
while (n > 0)
{
count++;
m = n%10;
n = n/10;
//Trying to put every digit in an array
for (int i = 0; i<count; i++)
{
cin>>arr[i];
arr[i] = m;
cout << arr[i];
}
int even = 5;
//Trying to see if there's a digit 5 in inputed number and how many if so.
for (int j = 0; j<count; j++)
{
if (arr[j]==even)
{
cout << "Yes, there's a digit '5' " << endl;
s++;
}
}
cout << "Count of '5's : " << even;
delete[] arr;
}
return 0;
}
This
for (int i = 0; i<count; i++)
{
cin >> arr[i];
}
You're trying to populate the array with the another user input rather than the existing one.
You also can do it without the array:
int count = 0;
int n;
cin >> n;
do {
if (n%10 ==5) count++;
n /= 10;
} while (n);
cout << "Count of '5's : " << count;
This should do it for every number.
If you only want to know a special number like 5, just remove the for-loop and print count[theNumberYouWantToKnow].
#define BASE 10
void output() {
int givenNumber;
cout << "Please enter a number: ";
cin >> givenNumber;
int count[BASE] = { 0 };
while(givenNumber > 0) {
int digit = givenNumber % BASE;
givenNumber /= BASE;
count[digit]++;
}
for(int i = 0; i < BASE; i++) {
cout << "Found the number " << i << " " << count[i] << " times." << endl;
}
}
The user enteres a number which is put in an array and then the array needs to be orinted backwadrds
int main()
{
int numbers[5];
int x;
for (int i = 0; i<5; i++)
{
cout << "Enter a number: ";
cin >> x;
numbers[x];
}
for (int i = 5; i>0 ; i--)
{
cout << numbers[i];
}
return 0;
}
You're very close. Hope this helps.
#include <iostream>
using namespace std;
int main(int argc, char *argv[]) {
int numbers[5];
/* Get size of array */
int size = sizeof(numbers)/sizeof(int);
int val;
for(int i = 0; i < size; i++) {
cout << "Enter a number: ";
cin >> val;
numbers[i] = val;
}
/* Start index at spot 4 and decrement until k hits 0 */
for(int k = size-1; k >= 0; k--) {
cout << numbers[k] << " ";
}
cout << endl;
return 0;
}
You are very close to your result but you did little mistakes, the following code is the correct solution of the code you have written.
int main()
{
int numbers[5];
int x;
for (int i = 0; i<5; i++)
{
cout << "Enter a number: ";
cin >> numbers[i];
}
for (int i = 4; i>=0; i--)
{
cout << numbers[i];
}
return 0;
}
#include<iostream>
using namespace std;
int main()
{
//get size of the array
int arr[1000], n;
cin >> n;
//receive the elements of the array
for (int i = 0; i < n; i++)
{
cin >> arr[i];
}
//swap the elements of indexes
//the condition is just at "i*2" be cause if we exceed these value we will start to return the elements to its original places
for (int i = 0; i*2< n; i++)
{
//variable x as a holder for the value of the index
int x = arr[i];
//index arr[n-1-i]: "-1" as the first index start with 0,"-i" to adjust the suitable index which have the value to be swaped
arr[i] = arr[n - 1 - i];
arr[n - 1 - i] = x;
}
//loop for printing the new elements
for(int i=0;i<n;i++)
{
cout<<arr[i];
}
return 0;
}
#include <iostream>
using namespace std;
int main() {
//print numbers in an array in reverse order
int myarray[1000];
cout << "enter size: " << endl;
int size;
cin >> size;
cout << "Enter numbers: " << endl;
for (int i = 0; i<size; i++)
{
cin >> myarray[i];
}
for (int i = size - 1; i >=0; i--)
{
cout << myarray[i];
}
return 0;
}
of course you can just delete the cout statements and modify to your liking
this one is more simple
#include<iostream>
using namespace std;
int main ()
{
int a[10], x, i;
cout << "enter the size of array" << endl;
cin >> x;
cout << "enter the element of array" << endl;
for (i = 0; i < x; i++)
{
cin >> a[i];
}
cout << "reverse of array" << endl;
for (i = x - 1; i >= 0; i--)
cout << a[i] << endl;
}
answer in c++. using only one array.
#include<iostream>
using namespace std ;
int main()
{
int array[1000] , count ;
cin >> count ;
for(int i = 0 ; i<count ; i++)
{
cin >> array[i] ;
}
for(int j = count-1 ; j>=0 ; j--)
{
cout << array[j] << endl;
}
return 0 ;
}
#include <iostream>
using namespace std;
int main ()
{
int array[10000];
int N;
cout<< " Enter total numbers ";
cin>>N;
cout << "Enter numbers:"<<endl;
for (int i = 0; i <N; ++i)
{
cin>>array[i];
}
for ( i = N-1; i>=0;i--)
{
cout<<array[i]<<endl;
}
return 0;
}
I am Having Problem with Passing a 2D array to a c++ Function. The function is supposed to print the value of 2D array. But getting errors.
In function void showAttributeUsage(int)
Invalid types for int(int) for array subscript.
I know the problem is with the syntax in which I am passing the particular array to function but I don't know how to have this particular problem solved.
Code:
#include <iostream>
using namespace std;
void showAttributeUsage(int);
int main()
{
int qN, aN;
cout << "Enter Number of Queries : ";
cin >> qN;
cout << "\nEnter Number of Attributes : ";
cin >> aN;
int attVal[qN][aN];
cout << "\nEnter Attribute Usage Values" << endl;
for(int n = 0; n < qN; n++) { //for looping in queries
cout << "\n\n***************** COLUMN " << n + 1 << " *******************\n\n";
for(int i = 0; i < aN; i++) { //for looping in Attributes
LOOP1:
cout << "Use(Q" << n + 1 << " , " << "A" << i + 1 << ") = ";
cin >> attVal[n][i];
cout << endl;
if((attVal[n][i] > 1) || (attVal[n][i] < 0)) {
cout << "\n\nTHE VALUE MUST BE 1 or 0 . Please Re-Enter The Values\n\n";
goto LOOP1; //if wrong input value
}
}
}
showAttributeUsage(attVal[qN][aN]);
cout << "\n\nYOUR ATTRIBUTE USAGE MATRIX IS\n\n";
getch();
return 0;
}
void showAttributeUsage(int att)
{
int n = 0, i = 0;
while(n != '\0') {
while(i != '\0') {
cout << att[n][i] << " ";
i++;
}
cout << endl;
n++;
}
}
I really suggest to use std::vector : live example
void showAttributeUsage(const std::vector<std::vector<int>>& att)
{
for (std::size_t n = 0; n != att.size(); ++n) {
for (std::size_t i = 0; i != att.size(); ++i) {
cout << att[n][i] << " ";
}
cout << endl;
}
}
And call it that way:
showAttributeUsage(attVal);
Looking at your code, I see no reason why you can't use std::vector.
First, your code uses a non-standard C++ extension, namely Variable Length Arrays (VLA). If your goal is to write standard C++ code, what you wrote is not valid standard C++.
Second, your initial attempt of passing an int is wrong, but if you were to use vector, your attempt at passing an int will look almost identical if you used vector.
#include <iostream>
#include <vector>
#include <algorithm>
#include <iterator>
typedef std::vector<int> IntArray;
typedef std::vector<IntArray> IntArray2D;
using namespace std;
void showAttributeUsage(const IntArray2D&);
int main()
{
int qN, aN;
cout << "Enter Number of Queries : ";
cin >> qN;
cout << "\nEnter Number of Attributes : ";
cin >> aN;
IntArray2D attVal(qN, IntArray(aN));
//... Input left out ...
showAttributeUsage(attVal);
return 0;
}
void showAttributeUsage(const IntArray2D& att)
{
for_each(att.begin(), att.end(),
[](const IntArray& ia) {std::copy(ia.begin(), ia.end(), ostream_iterator<int>(cout, " ")); cout << endl;});
}
I left out the input part of the code. The vector uses [] just like a regular array, so no code has to be rewritten once you declare the vector. You can use the code given to you in the other answer by molbdnilo for inputing the data (without using the goto).
Second, just to throw it into the mix, the showAttributeUsage function uses the copy algorithm to output the information. The for_each goes throw each row of the vector, calling std::copy for the row of elements. If you are using a C++11 compliant compiler, the above should compile.
You should declare the function like this.
void array_function(int m, int n, float a[m][n])
{
for (int i = 0; i < m; i++)
for (int j = 0; j < n; j++)
a[i][j] = 0.0;
}
where you pass in the dimensions of array.
This question has already been answered here. You need to use pointers or templates. Other solutions exists too.
In short do something like this:
template <size_t rows, size_t cols>
void showAttributeUsage(int (&array)[rows][cols])
{
for (size_t i = 0; i < rows; ++i)
{
std::cout << i << ": ";
for (size_t j = 0; j < cols; ++j)
std::cout << array[i][j] << '\t';
std::cout << std::endl;
}
}
You're using a compiler extension that lets you declare arrays with a size determined at runtime.
There is no way to pass a 2D array with such dimensions to a function, since all but one dimension for an array as a function parameter must be known at compile time.
You can use fixed dimensions and use the values read as limits that you pass to the function:
const int max_queries = 100;
const int max_attributes = 100;
void showAttributeUsage(int array[max_queries][max_attributes], int queries, int attributes);
int main()
{
int attVal[max_queries][max_attributes];
int qN = 0;
int aN = 0;
cout << "Enter Number of Queries (<= 100) : ";
cin >> qN;
cout << "\nEnter Number of Attributes (<= 100) : ";
cin >> aN;
cout << "\nEnter Attribute Usage Values" << endl;
for (int n = 0; n < qN; n++)
{
cout << "\n\n***************** COLUMN " << n + 1 <<" *******************\n\n";
for (int i = 0; i < aN; i++)
{
bool bad_input = true;
while (bad_input)
{
bad_input = false; // Assume that input will be correct this time.
cout << "Use(Q" << n + 1 << " , " << "A" << i + 1 << ") = ";
cin >> attVal[n][i];
cout << endl;
if (attVal[n][i] > 1 || attVal[n][i] < 0)
{
cout << "\n\nTHE VALUE MUST BE 1 or 0 . Please Re-Enter The Values\n\n";
bad_input = true;
}
}
}
}
cout << "\n\nYOUR ATTRIBUTE USAGE MATRIX IS\n\n";
showAttributeUsage(attVal, qN, aN);
getch();
return 0;
}
void showAttributeUsage(int att[max_queries][max_attributes], int queries, int attributes)
{
for (int i = 0; i < queries; i++)
{
for (int j = 0; j < attributes; j++)
{
cout << att[i][j] << " ";
}
cout << endl;
}
}
For comparison, the same program using std::vector, which is almost identical but with no size limitations:
void showAttributeUsage(vector<vector<int> > att);
int main()
{
cout << "Enter Number of Queries (<= 100) : ";
cin >> qN;
cout << "\nEnter Number of Attributes (<= 100) : ";
cin >> aN;
vector<vector<int> > attVal(qN, vector<int>(aN));
cout << "\nEnter Attribute Usage Values"<<endl;
for (int n = 0; n < qN; n++)
{
cout<<"\n\n***************** COLUMN "<<n+1<<" *******************\n\n";
for (int i = 0; i < aN; i++)
{
bool bad = true;
while (bad)
{
bad = false;
cout << "Use(Q" << n + 1 << " , " << "A" << i + 1 << ") = ";
cin >> attVal[n][i];
cout << endl;
if (attVal[n][i] > 1 || attVal[n][i] < 0)
{
cout << "\n\nTHE VALUE MUST BE 1 or 0 . Please Re-Enter The Values\n\n";
bad = true;
}
}
}
}
cout << "\n\nYOUR ATTRIBUTE USAGE MATRIX IS\n\n";
showAttributeUsage(attVal);
getch();
return 0;
}
void showAttributeUsage(vector<vector<int> > att);
{
for (int i = 0; i < att.size(); i++)
{
for (int j = 0; j < att[i].size(); j++)
{
cout << att[i][j] << " ";
}
cout << endl;
}
}
The Particular Logic worked for me. At last found it. :-)
int** create2dArray(int rows, int cols) {
int** array = new int*[rows];
for (int row=0; row<rows; row++) {
array[row] = new int[cols];
}
return array;
}
void delete2dArray(int **ar, int rows, int cols) {
for (int row=0; row<rows; row++) {
delete [] ar[row];
}
delete [] ar;
}
void loadDefault(int **ar, int rows, int cols) {
int a = 0;
for (int row=0; row<rows; row++) {
for (int col=0; col<cols; col++) {
ar[row][col] = a++;
}
}
}
void print(int **ar, int rows, int cols) {
for (int row=0; row<rows; row++) {
for (int col=0; col<cols; col++) {
cout << " | " << ar[row][col];
}
cout << " | " << endl;
}
}
int main () {
int rows = 0;
int cols = 0;
cout<<"ENTER NUMBER OF ROWS:\t";cin>>rows;
cout<<"\nENTER NUMBER OF COLUMNS:\t";cin>>cols;
cout<<"\n\n";
int** a = create2dArray(rows, cols);
loadDefault(a, rows, cols);
print(a, rows, cols);
delete2dArray(a, rows, cols);
getch();
return 0;
}
if its c++ then you can use a templete that would work with any number of dimensions
template<typename T>
void func(T& v)
{
// code here
}
int main()
{
int arr[][7] = {
{1,2,3,4,5,6,7},
{1,2,3,4,5,6,7}
};
func(arr);
char triplestring[][2][5] = {
{
"str1",
"str2"
},
{
"str3",
"str4"
}
};
func(triplestring);
return 0;
}