Why does the indexing in an array start with zero in C and not with 1? Why is the first element of an array a at a[0] and not a[1]?

I am assigning values in a C++ program out of the bounds like this:

Consider a 2D array (with a compile or runtime constant for the first dimension, and a compile time constant for the second). e.g. int arr[10][3]; or int arr[n][3];

Consider a 2D array (with a compile or runtime constant for the first dimension, and a compile time constant for the second). e.g. int arr[10][3]; or int arr[n][3];

We start with array of string, for example:

When I enter the user inputs of “12345” and “0”, I see in the first loop arrOne is being populated the way I’m expecting 1, 2, 3, 4, 5. However after the 2nd loop (which populates arrTwo) when I go to print out the values of arrOne (using the 3rd loop) I see that the values are now 0, 2, 3, 4, 5.

I am reading C++ Primer and on page 128, it says to avoid array to pointer conversion to use reference like so:

#include <iostream> #include <array> using namespace std; int main(){ array<int,5> a {1,2,3,4,5}; for(int i = 0; i < 5; i++){ cout << a[i] << endl; } return 0; } When I go to compile this, I get this error: error: expected ‘;’ at end of declaration array<int,5> a{1,2,3,4,5}; ^ ; 1 error generated. But I […]

how do i convert t[i-3] and t[i-1] into an addable value? For example: i = 3, 3 % 2 != 0, cout<<3<<” “<<t[3-1] % 7 (so -> 1 % 7 = 1)<<endl;

I’m working on a lottery application: