How do I add numbers in a different order?

Question

This is a simple question don't look for difficulties.

I have such a cycle

int arr[]{ 111, 222, 333 };
    
for (size_t i = 0; i < 3; i++) {
   test = 1000 * test + arr[i];    
}
std::cout << test;

it returns the result 111222333

the question is how to make it output the result 333222111 in short add numbers from the other side

you need to change this line to something else to add values in a different way

test = 1000 * test + arr[i];

I will update the question a little bit, that would be interesting, what if there is no array and you get the data in this form.

test = 1000 * test + X;

X is an unknown number and not an array

Well needless to say I love the accepted answer on the duplicate ;-) — Bathsheba, Aug 24 '20 at 13:57
If you're interested in c++20, you can use `std::ranges::views::reverse` as mentioned in the second dupe target. — cigien, Aug 24 '20 at 14:08
An update shouldn't invalidate answers. You should instead ask a new question. — Thomas Sablik, Aug 24 '20 at 21:40

Ted Lyngmo · Answer 1 · 2020-08-24T14:01:08.310

3

You could do the loop from the end using reverse iterators:

#include <iostream>
#include <iterator>

int main() {
    int arr[]{ 111, 222, 333 };
    int test = 0;        
    for(auto it = std::rbegin(arr); it != std::rend(arr); ++it) {
        test = 1000 * test + *it;
    }
    std::cout << test;
}

Here's an alternative using std::accumulate and a lambda:

#include <iostream>
#include <iterator>
#include <numeric>

int main() {
    int arr[]{ 111, 222, 333 };
    int test = std::accumulate(std::rbegin(arr), std::rend(arr), 0,
        [](const auto a, const auto b){
            return 1000 * a + b;
        }
    );
    std::cout << test;
}

edited Aug 24 '20 at 14:01

answered Aug 24 '20 at 13:52

Ted Lyngmo

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Yup sensible. You'll get another vote if you can use a C++11 lamda in another answer? – Bathsheba Aug 24 '20 at 13:54
@Bathsheba You mean as in using `std::accumulate` or `std::for-each` or something like that? – Ted Lyngmo Aug 24 '20 at 13:55
Absolutely, yes! – Bathsheba Aug 24 '20 at 13:56
1

@Bathsheba Added :) – Ted Lyngmo Aug 24 '20 at 14:01

MikeCAT · Accepted Answer · 2020-08-24T14:16:26.390

2

You can use

test = 1000 * test + arr[2 - i];

instead of the line.

To handle input other than arrays, you can use "delta" to keep track the place to insert new values.

int delta = 1;
for (size_t i = 0; i < 3; i++) {
    test += delta * X;
    delta *= 1000;
}

edited Aug 24 '20 at 14:16

answered Aug 24 '20 at 13:45

MikeCAT

73,922
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You were first to get the index correct. Oops. +1. – Bathsheba Aug 24 '20 at 13:45
@Bathsheba Actually I prefer `arr[3 - 1 - i]` because we can use same `3` and therefore it is easy to replace this to constant value. – MikeCAT Aug 24 '20 at 13:47
@MikeCAT: I can't resist putting in my preferred way `-->`. – Bathsheba Aug 24 '20 at 13:47
updated the question and your example won't work now) – buster Aug 24 '20 at 14:13
@buster added answer for the new question. – MikeCAT Aug 24 '20 at 14:16
Yes, it's a better fit for me) – buster Aug 24 '20 at 14:22

Bathsheba · Answer 3 · 2020-08-24T13:56:05.097

1

Assuming you can't reverse the elements in arr, perhaps the simplest way is to write

test = 1000 * test + arr[2 - i];

perhaps not hardcoding the 2. This is considerably simpler than running the loop the other way: size_t is unsigned so getting the stopping conditional correct takes some thought. Personally I use

for (size_t i = 3; i --> 0;) {

for this kind of thing. Note well the slide operator -->. It's not really an operator at all but it consists of the postfix -- followed by the comparison. It's not to everyone's taste and some software houses ban its use.

edited Aug 24 '20 at 13:56

answered Aug 24 '20 at 13:45

Bathsheba

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thank, but I don't think it's worth writing this code `-->` :) – buster Aug 24 '20 at 13:50
@buster: you don't like `for (size_t i = 3; i --> 0;) {`? – Bathsheba Aug 24 '20 at 13:51
Another variant is `for (size_t i = 2; i <= 2; --i) {` but I don't like the repeat. – Bathsheba Aug 24 '20 at 13:51
@Bathsheba it looks confusing to me, although it seems simple. – buster Aug 24 '20 at 13:52
@buster: perhaps I've been doing this for too long now, but I find it natural, and no worse than function pointer declarations. – Bathsheba Aug 24 '20 at 13:53
updated the question and your example won't work now) – buster Aug 24 '20 at 14:13

sumit · Answer 4 · 2020-08-24T13:53:01.603

1

Iterate the loop from i=2 to 0 i.e for(size_t i = 2; i >= 0; i --)

edited Aug 24 '20 at 13:53

answered Aug 24 '20 at 13:48

sumit

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2

1

This works if you use an `int`, but will not work for a `size_t` type. – Bathsheba Aug 24 '20 at 13:49
updated the question and your example won't work now) – buster Aug 24 '20 at 14:13

How do I add numbers in a different order?

4 Answers4