Whats the cout << i << " \n"[ i == n ] term doing?

Question

In this statement

for (i = 1; i <= n; i++) {
    cout << i << " \n"[ i == n ];
}

what is the last term in cout statement [i==n] doing? This loop prints space separate numbers I guess.

Monofilament

Possible duplicate of How can I get the nth character of a string? – Nap

@underscore_d: Definitely not. Note the unusual expression in [ i==n ] . – Reticulation

and through standard boolean evaluation, i == n becomes 0 or 1. So, once someone learns what "foo"[bar] does, "foo"[i == n] seems trivial to me. – Nap

Answer 1

It is a silly way to index either the character ' ' or the character '\n'. This does the same idea and prints "Hello World":

#include <iostream>

int main() {
        for (int i = 0; i < 11; i++)
                std::cout << "Hello World"[i];
        return 0;
}

i == n is either going to be true or false. When cast to an integer for indexing using [i == n] you get either the first or second element

Manifesto

Realy right use of the silly qualifier (It should be standardized)! – Superabundant

Wonderful! Only, I would swap the arguments to [] for added quality. i["Hello World"]. ;-) – Barbrabarbuda

Answer 2

It is an obtuse way of writing:

(i == n ? '\n' : ' ')

That is, when i == n, a newline is printed, otherwise a space is printed.

The idea is to separate the numbers by spaces, and to put a newline after all the numbers have been printed.

Prototype

Never seen this before! – Kraft