C++ pointers prints out the memory address and not numbers - c++

Question

I really can't understand why this prints out the memory address and not the say {1, 2, 3, 4, 5} if you input 5. How can I fix this? Thanks.
#include <iostream>
#include <iomanip>
using namespace std;
int main() {
int columnInput;
//int rowInput;
int** column = NULL;
cin >> columnInput;
column = new(int*[columnInput]);
for (int i = 0; i < columnInput; i++) {
column[i] = new(int[i + 1]);
}
cout << " |";
for (int i = 0; i < columnInput; i++){
cout << setw(3) << column[i] << "|";
}
cout << endl;
return 0;
}

Answer 1

Whether you are looking for a 1D or a 2D dynamic array is not very clear from your code. The pointer-to-pointer variable and the commented out rowInput hint towards the latter, however using simple (i.e. not nested) for loops points to the former. In the case you want a 1D dynamic array, Learner's answer is probably what you're looking for. Otherwise, take a close look at this:
#include <iostream>
#include <iomanip>
int main()
{
// I don't really like this, because it can easily lead to name clashes.
// If you really want it, not putting it in global scope is a good idea.
using namespace std;
int ** matrix = 0;
int rows;
int cols;
cin >> rows;
cin >> cols;
// The convention is that the first dimension denotes the number of rows.
// As an exercise, you can try to do the same with columns first.
matrix = new int * [rows]; // reserve pointers for all rows
for (int i = 0; i < rows; i ++) {
matrix[i] = new int[cols]; // reserve space for each row
}
// In order to access a 2D array you typically need nested for loops.
for (int i = 0; i < rows; i ++) {
for (int j = 0; j < cols; j ++) {
matrix[i][j] = 0;
matrix[i][j] += (i % 10) * 10; // <- try commenting this out
matrix[i][j] += j % 10 + 1; // <- same with this here
}
}
cout << endl;
// Same here.
for (int i = 0; i < rows; i ++) {
for (int j = 0; j < cols; j ++) {
cout << setw(3) << matrix[i][j] << "|";
}
cout << endl;
}
// Don't forget to do cleanup!
for (int i = 0; i < rows; i ++) {
delete [] matrix[i];
}
delete [] matrix;
return 0;
}
Online Demo
As others mentioned, since this is C++, using std::vector is a better option here. This nice article over at cplusplus.com explores this and other options.

Answer 2

You're printing a pointer. How could operator<< possibly know how many elements the pointer points to? It can't If you want to print the contents of an array or what a pointer points to then you'll have to write that yourself.

Answer 3

Because column is a pointer to pointer to integer..int** column = NULL printing column directly would give you address instead of integer values.

Answer 4

If you want to allocate dynamic memory according to the user input and want to print it's content use this:
#include <iostream>
#include <iomanip>
using namespace std;
int main() {
int columnInput;
int* column = NULL;
cin >> columnInput;
column = new int[columnInput];
for (int i = 0; i < columnInput; i++) {
column[i] = i;
}
cout << " |";
for (int i = 0; i < columnInput; i++){
cout << setw(3) << column[i] << "|";
}
cout << endl;
return 0;
}

Answer 5

If you want to print a pointer's value, you have to put a * before the variable. This says to the compiler "Hey, here comes a pointer, give to me what is pointing at"
So, if ptr is a pointer, for example
*int ptr;
if you want to print ptr's value:
cout << *ptr;
This logic applies to every variable in C/C++.
If you have
int*** listOfListsOfListsOfInts;
And you want to print what's pointing at,
cout << ****listOfListsOfListsOfInts;
Whit a wildcard behind a variable, you get what a pointer is pointing at instead of the memory address that pointer is holding.