Printing Sequences 1 to 10 and 10 to 1 Using a Single Loop in C

If you're tasked with printing the numbers 10 to 1 and then 1 to 10 within a single C loop, you need to be creative!

Let's break down the problem and explore a few different approaches to solve it. This post will guide you through combining the sequences into a single loop and using clever arithmetic to achieve the desired output.

The Problem Statement

Your homework requires you to print the numbers in a specific pattern. You need to print from 10 to 1, followed by printing from 1 to 10. However, your teacher expects a solution that does not use two separate loops. Instead, you can merge these two sequences into a single loop.

Naive Approach: Two Separate Loops

The most straightforward approach is to use two loops, but this does not meet the homework requirement. Here's how you might do it naively:

for (int i  10; i > 0; i--) {    printf("%d ", i);}printf("
");for (int i  1; i 

While this is clear, it violates the homework instructions by using two loops. Therefore, we need a more clever solution.

Using Modern C Ranges/Views

A more modern C approach involves using the views::iota, views::reverse, and views::concat functionalities to merge the ranges into a single loop. This approach is efficient and concise:

for (auto i : std::views::concat(std::views::iota(1, 11), std::views::reverse(1, 11))) {    std::cout 

This code snippet combines the range from 1 to 10 and the reversed range from 10 to 1 into a single concatenated range and prints it out.

Clever Arithmetic: A Simpler and More Traditional Approach

Your teacher might be looking for a more traditional approach using simple arithmetic. Here's a neat trick:

for (auto i : std::views::iota(0, 20)) {    if (i 

This code snippet leverages a simple if-else condition to print the desired sequence while adhering to the single loop requirement.

Even Simpler Arithmetic

Another approach, as suggested by Trausti Thor Johannsson, uses the absolute value of the difference between 10 and i, adding 10 to the result:

for (auto i : std::views::iota(0, 20)) {    std::cout 

This method is even more concise, but it may be more difficult to understand due to the use of absolute values and bitwise operations.

Conclusion

The key takeaway is that you can achieve the desired output using a single loop and a combination of arithmetic or modern C ranges/views. While the traditional approach with if-else statements works well, the use of absolute values and arithmetic operations can provide a more elegant solution, albeit with a potential trade-off in readability.