How to print 13 number array in C++ [closed] - c++

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I'm trying to create an array of 13 numbers and fill it in a loop with even numbers descending, starting from 48. I can't tell exactly what to do with my code to make it work.
#include <iostream>
using namespace std;
int main() {
const int size = 13;
int arr[size] = {0,0,0,0,0,0,0,0,0,0,0,0,0};
int a = 50;
while (a > 0){
a-=2;
cout << a << " " << endl;
for (int i = 0; i < size; i++)
{
arr[i] = a;
cout << i << " " << arr[i] << endl;
}
}
}
It's giving bunch of nonsense which i don't understand.

Answer 1

A no loop solution, using iterator and algorithms
using namespace std;
int main(int argc, char * argv[])
{
constexpr int N= 13;
array<int, N> t;
iota(t.begin(), t.end(),0);
transform(t.begin(), t.end(), t.begin(), [](int x) { return 48-2*x;});
copy(t.begin(), t.end(), ostream_iterator<int>(cout, " "));
return 0;
}

Answer 2

I do not know why you do have the Outer loop, but it seems you are New to Coding all in all, So I Won't give you anything that you wouldn't understand at this point. But I'll explain why you are getting "a bunch of nonsense which you don't understand."
while (a > 0){
a-=2;
cout << a << " " << endl;
this will run while a is greater than 0, but every time the Loops restart. it does this.
for (int i = 0; i < size; i++)
{
arr[i] = a;
cout << i << " " << arr[i] << endl;
}
Hence a lot of numbers will show up, you aren't really "inserting" a number into the array, you are "REPLACING" the whole array every loop. Have a look at this, I inserted a few outputs which would show what is happening in your code.
Without changing too much of your code, here is one way to do it in that is still in your depth. Instead of doing a while(a > 0), simply loop through the Array with a for loop, then decrement a by 2 each loop, since your goal is only to fill the array with a size of 13 and don't need to wait until a is exhausted of positive numbers.
int main() {
const int size = 13;
int arr[size] = {0,0,0,0,0,0,0,0,0,0,0,0,0};
int a = 50;
for (int i = 0; i < size; i++)
{
a-=2;
arr[i] = a;
cout << i << " " << arr[i] << endl;
}
}
Lastly, Please indent your code properly next time, it makes it much easier to understand your code, not only for us but for you as well.

Answer 3

It is enough to use two range-based for loops. For example
const int size = 13;
int arr[size];
int value = 48;
for ( auto &item : arr )
{
item = value;
value -= 2;
}
for ( const auto &item : arr )
{
std::cout << item << ' ';
}
std::cout << '\n';
If you want to fill the array with even numbers in the descending range [48, 0) then the array needs to have 24 elements instead of 13 elements.
Here is a demonstration program where instead of the range-based for loop there are used standard algorithms (of course you may use the range-based for loops shown above).
#include <iostream>
#include <iterator>
#include <algorithm>
int main()
{
const int N = 48;
int arr[N / 2];
int value = N + 2;
std::generate( std::begin( arr ), std::end( arr ),
[&value] { return value -= 2; } );
std::copy( std::begin( arr ), std::end( arr ),
std::ostream_iterator<int>( std::cout, " " ));
std::cout << '\n';
}
The program output is
48 46 44 42 40 38 36 34 32 30 28 26 24 22 20 18 16 14 12 10 8 6 4 2
Again if you need an array with only 13 elements then declare it explicitly like
int arr[13];
As for your code when within the outer while loop are are outputting anew the array in the inner for loop setting its all elements to the current value of a.
while (a > 0){
a-=2;
cout << a << " " << endl;
for (int i = 0; i < size; i++)
{
arr[i] = a;
cout << i << " " << arr[i] << endl;
}
}
that does not make sense.