Increment a variable in C++ with a void function

Question

I wanted to make one of my code more concise so I wrote this:

while(m <= r) 
    nums[m] ? nums[m] == 1 ? m++ : swap(nums[m], nums[r--]) : swap(nums[m++], nums[l++]);

But it's not working because 'swap' is a void function but 'm++' returns int. ('right operand to ? is void, but left operand is of type int' error). So I'd like to know how can I replace m++ so it's of type void.

I know that I can create a new void function (for example void increase(int &x){x++;}) but I want to keep my code as an one liner.

The best working variant I made is 'swap(nums[m], nums[m++])', which does nothing to my array, but it looks awful. What other functions can I use?

A simple cast to `(void)` could fix your issue: [demo on coliru](http://coliru.stacked-crooked.com/a/e105271bf6154f06). However, the whole expression looks a bit scaring... — Scheff's Cat, Aug 21 '20 at 10:22
.... arrr ... concise doesn't always mean better. Always prefer readability. This code would fail any decent code review, even if it would compile. — bolov, Aug 21 '20 at 10:44
@Scheff: I think using `operator ,` after `swap` to also return `int` would be more in the scaring spirit ;) — Jarod42, Aug 21 '20 at 11:13
You're worried about the wrong thing. This code is absolutely unreadable and, worse, unmaintainable. It should be rewritten with `if` statements. — Pete Becker, Aug 21 '20 at 13:46
That's a complete abuse of the ternary operator. It's not meant to replace the `if` statement, it's meant to choose between two values. — Mark Ransom, Aug 21 '20 at 16:09

dfrib · Answer 1 · 2020-08-21T10:46:02.387

I wanted to make one of my code more concise

Baking in a number of side effects into a single expression of nested non-parenthesized ternary operators (making use of implicit conversion to bool) does, if anything, make your code more complex, more bug-prone and may hide the fact that the original code should actually have been broken up and re-factored into something simpler.

Why not favour clarity over over-complex brevity? E.g. starting with the straight-forward approach:

while(m <= r) {
    if (nums[m] != 0) {
        if (nums[m] == 1) {
            ++m;
        }
        else {
            swap(nums[m], nums[r--]); 
        }
    }
    else {
        swap(nums[m++], nums[l++]);
    }
}

which can be re-factored into:

while(m <= r) {
    if (nums[m] != 0) {
        if (nums[m] == 1) {
            ++m;
        }
        else {
            swap(nums[m], nums[r]); 
            --r;
        }
    }
    else {
        swap(nums[m], nums[l]);
        ++m;
        ++l;
    }
}

which can be re-factored into:

while(m <= r) {
    const std::size_t swap_from_idx = m;
    std::size_t swap_with_idx = m;  // default: no swapping.

    if (nums[m] == 1) {
        ++m;
        continue;
    }
    else if (nums[m] == 0) {
        swap_with_idx = l;
        ++l;
    }
    else {
        swap_with_idx = r;
        --r;
        ++m;
    }
    swap(nums[swap_from_idx], nums[swap_with_idx]);
}

or e.g.:

while(m <= r) {
    // No swapping.
    if (nums[m] == 1) {
        ++m;
    }
    // Swapping.
    else {
        const std::size_t swap_from_idx = m;
        std::size_t swap_with_idx = l;
        
        if (nums[m] == 0) {
            ++l;
        }
        else {
            swap_with_idx = r;
            --r;
            ++m;
        }
        swap(nums[swap_from_idx], nums[swap_with_idx]);
    }
}

At this point you may ask yourself is the original loop design is overly complex, and/or if you should break out part of the loop body into a separate utility function.

If your if/else if/else logic is reaching a too high cyclomatic complexity, the answer is seldom to try to hide it by means of a highly complex tenary operator expression, but rather by re-factoring and, if applicable, breaking out some parts in separate functions.

Yeah, thanks. I initially did the same. I just had a challenge to write it as an one liner. — Brega Stanislav, Aug 21 '20 at 20:19

score 0 · Accepted Answer · answered Aug 21 '20 at 15:47

0

If you want the expression m++ to have a side-effect, and have a void type, you can simply cast the expression like this:

(void)m++

which will evaluate the m++ first, and then cast it to void.

Here's a demo.

answered Aug 21 '20 at 15:47

cigien

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Woah, it was so simple. C++ is truly amazing. I found that you can also write it as void(m++). I think this one looks better. – Brega Stanislav Aug 21 '20 at 20:23
@BregaStanislav Yeah, they're both equivalent. Pick whichever one looks better :) Make sure to read the advice in other answers as well regarding your code. – cigien Aug 21 '20 at 20:27

Increment a variable in C++ with a void function

2 Answers2