An efficient way to write this code snippet?

Question

Ways of rewriting this code more efficiently and/or shortly to save bytes and appear less verbose?

if(N % 2 == 0 && (N == 6 || N == 8 || N == 10 || N == 12 || N == 14 || N== 16 || N == 18 || N == 20 ))
{
    printf("Weird\n");
}

Given the language is C, although doesn't matter if its java or php or any other language :-).

Why not `N%2 == 0 && N > 5 && N < 21`? – skypjack Apr 30 '16 at 19:58 — skypjack, Apr 30 '16 at 19:58

score 10 · Answer 1 · answered Apr 30 '16 at 20:00

10

Why won't you just do :

if (N % 2 == 0 && (N >= 6 && N <= 20))
{
     /* ... */
}

answered Apr 30 '16 at 20:00

Damn, I wasn't fast enough :o – Apr 30 '16 at 20:01
That's nice code. +1 – raddevus Apr 30 '16 at 20:06
This is nothing more than my comment to the OP question... – skypjack Apr 30 '16 at 20:37
Almost the same, yes, with a pair of parenthesis :) – Apr 30 '16 at 20:38

Anton Angelov · Answer 2 · 2016-04-30T20:08:08.207

5

Given N is unsigned integer.

if(N % 2 == 0 && (N-6) <= 14)
{
    printf("Weird\n");
}

edited Apr 30 '16 at 20:08

answered Apr 30 '16 at 19:59

Anton Angelov

1,223
7
19

!(N % 2) - very bad style. Mixing boolean logic with arithmetic – SomeWittyUsername Apr 30 '16 at 20:03
Yes, of course. It does hide the actual logic, but saves bytes in source code. – Anton Angelov Apr 30 '16 at 20:04
1

Efficiency isn't about saving source code. (N % 2) == 0 will yield exactly the same binary – SomeWittyUsername Apr 30 '16 at 20:05
1

Yes, efficienty is about performance and less resource consumption. And both variants most likely have same efficiency, since most likely will be resolved to the same assembly code. N % 2 == 0 just has more readability. Anyway, I edited it back to how it was before. – Anton Angelov Apr 30 '16 at 20:12

John Bollinger · Accepted Answer · 2016-04-30T20:41:09.333

5

Supposing that N has an unsigned integral type, and you are more interested minimal source code than in clarity, then you might write

if (((N - 6) & 14) + 6 == N) {
    printf("Weird\n");
}

Explanation:

The first operation, N - 6, shifts the range of the acceptable values from 6 - 20 to 0 - 14. The latter range can be expressed with four bits.

The second operation, & 14, is a bitwise and with bit pattern 1110. That will leave even numbers in the range 0 - 14 unchanged, but it will convert odd numbers to even and it will clear any higher bits. The only values that survive unchanged are the ones derived from the original target values in the first step.

The third operation reverses the first, so that the result can be compared directly with the original number. If the combined sequence of operations reproduces the original number, then that number is one of the ones you were looking for.

edited Apr 30 '16 at 20:41

answered Apr 30 '16 at 20:27

John Bollinger

160,171
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I would really appreciate it if you could explain the logic behind Thai for me. I see some bit wise operations, right? – yousafe007 Apr 30 '16 at 20:29
@YousufEjazi007, done. – John Bollinger Apr 30 '16 at 20:41
You're a god! I'm thanks bro!!! – yousafe007 Apr 30 '16 at 20:46

An efficient way to write this code snippet?

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