CODEWITHSUNDEEP
java
Alexandru
Alexandru

Reputation: 25840

How do I eliminate conditional from code?

How do I eliminate the ternary operator from the code below? I'm looking for something using only integer operators (shifts, additions, etc).

a = (a < 0) ? (-a * 2) - 1 : (a * 2)

Upvotes: 1

Views: 300

Answers (6)

Peter Lawrey
Peter Lawrey

Reputation: 533880

A simpler answer is to use 

(a ^ a >> 31) * 2 + (a >>> 31)

Upvotes: 0

Jeffrey L Whitledge
Jeffrey L Whitledge

Reputation: 59553

It should be something like this:

a = ((a ^ (a >> 31)) + 1) * 2 - ~(a >>> 31) - 1;

(I haven’t actually tried it, though—not being a Java programmer.)

Upvotes: 0

Thorbj&#248;rn Ravn Andersen
Thorbj&#248;rn Ravn Andersen

Reputation: 75426

if (a < 0) 
  a = (-a * 2) - 1 
else 
  a = (a * 2)

Upvotes: 2

Skip Head
Skip Head

Reputation: 7760

This seems to work:

a= (2 * Math.abs(a)) - (a >>> 31);

Upvotes: 1

T.J. Crowder
T.J. Crowder

Reputation: 1075785

What you have is several answers demonstrating that getting rid of the ternary is not a good idea. It means your code is hard to read, hard to debug, hard to maintain. You haven't flagged this as homework or an intellectual exercise, and so the right answer is: Don't. Leave the ternary in place. It's clear, it's expressive, and it does the job.

Upvotes: 3

Tom Hawtin - tackline
Tom Hawtin - tackline

Reputation: 147164

Off the top of my head, assuming a is an int:

a = 2*a*(1+2*(a>>31)) - (a>>>31);

>> is arithmetic (signed) shift left. So a>>31 replaces all the bits with the sign bit. All bits set gives -1. Multiple by two and add one, negatives go to -1 and positives (and zero) to 1.

>>> is logical (unsigned) shift left. So a>>>31 clears the top 31 bits and places the sign bit in bit 0 (0 for positive numbers and zero; 1 for negative numbers).

The "proper" way of doing something vaguely like this is:

a ^= a>>31;
a *= 2;

Upvotes: 7

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