extract every digit for arithmetic calculation

[amoureux]

I'm trying to display a c++ output with the image link. this was a sample question given before anything was taught. so I was hoping for someone to give me a sample of how to get this using the for loop and without the for loop in order to have a head start. accordingly to the lecturer they said this is the most basic of all.
http://imgur.com/54HeYtn

[Naughty Albatross]

Hi,
Can you give us more details about your assignment?

Is this image your program output? Or a sample output? Do you bring your code with you?

[PattuX]

I'll go pretty much from scratch here.

Everything you need for now is a main() function. Main functions in C++ always return an int (usually 0, if everything went well).

First you need to declare two integer variables, simply by int n;.

To write on your console, use std::cout << "n is " << n;. std is the standard library, cout is the standard output stream and << is the operator to write on it. For a new line, use "\n" or std::endl.

To read from the command line, use std::cin >> n;. This will save the input on the variable n.

To make sure your input has 4 digits, you can just repeadetly check whether your variable is <1000 or >9999. Do this with a while loop. Repeat for second value.

To get the single digits, there's probably a fancy way by converting the number to a string and then splitting it up, but I prefer the mathematical way.

int n1 = n/1000 will give you the first digit, as n and 1000 are integers and / only does a whole number division. (Keep this in mind, when you say double x = 1/2;, x will be zero, not one half!)
Then you can continue by making your 4 digit number a 3 digit number, by subtracting the 1000s like this:
n = n - n1 * 1000
You can keep on going like this for n2, n3 and n4.

The other variables are pretty straight forward, just int sum1 = n1 + m1; etc.
For division you need to cast a variable on double: double div1 = (double) n1 / m1;

The output is pretty straight forward as well. You just need to adjust the number of spaces to (8 - number of digits).
From http://stackoverflow.com/questions/22648978/c-how-to-find-the-length-of-an-integer

int len = 1;

// and for numbers greater than 0:
if (i > 0) {
    // we count how many times it can be divided by 10:
    // (how many times we can cut off the last digit until we end up with 0)
    for (len = 0; i > 0; len++) {
        i = i / 10;
    }
}

gives you the length of an integer. You might want this in a seperate function. Don't forget to add 1, if your number is negative.

To output x spaces, do a for-loop that starts with i=0, runs while i<x and where you increase i by 1 in every iteration. In every iteration you output a " ".

For the second method with the for loop you need vectors. Those are just like lists or arrays mostly.

First you need to #include <vector> .
Declare a vector of integers: vector<int> digits_of_n;
Then, instead of saving n1, n2, n3, n4, you use a for loop that runs 4 times and does the same thing, but uses digits_of_n.push_back(digit);
You'll end up with a vector that contains 4 ints, which are the 4 digits of n. To call a digit use digits_of_n.at(position);. Note that vectors start counting at 0, so .at(0) gives the first digit, .at(1) gives the second etc.
Output is the same.