changing printf to cout

Question

I am still a bit unfamiliar with C++ and need some help with using cout.

int main()
{
    char letterGrades[25] = { 'a', 'b', 'c', 'd', 'e', 'e', 'e', 'e', 'e', 'e', 'e', 'e', 'e', 'e', 'e', 'e', 'e', 'e', 'e', 'e', };

    for (int i = 0; i < 25; i++)
    {
        printf("[%d] --> %c", i, letterGrades[i]);

        if (i == 3)      // how can I print \n when i == 7 , 11 , 15 , 19....
        {
            printf("\n");
        }
    }
}

This is what I am trying to do, and it works perfectly fine. However, I don't know how to write this code using cout.

Also, I would print the result in a 4 in a row tabulate format. so result can look something like this

[0] --> A [1] --> A [2] --> A [3] --> A
[4] --> A [5] --> A [6] --> A [7] --> A
[8] --> A [9] --> A [10] --> A [11] --> A

Answer 1

The class of which cout is an instance has clever overloads to << for many types, including char, int, and const char[] (for string literals):

So you can write

std::cout << "[" << i << "] --> " << letterGrades[i];

in place of your first printf and

std::cout << "\n";

for the second one. You can use "\t" to inject a tabulation character into the stream.

All this comes at a slight performance hit, which ought to be negligible cf. the I/O on your platform.

Also, consider reading C++: "std::endl" vs "\n" for further study.

Finally, use

if (i % 4 == 3){
    // i is 3, 7, 11, 15, 19, etc
}

for your conditional check for the periodic newlines. % is the remainder operator.