Swift 4 Conditional compile logic not working correctly

Question

Here's some simple code to show the problem I am having:

import UIKit

let TEST = true
print(TEST)
#if TEST
    print("1. TEST should be true. Value is : ", TEST)
#endif

#if !TEST
    print("2. TEST should be false.  Value is : ", TEST)
#endif

Regardless of whether TEST is 'true' or 'false', the '#if !TEST' is always executed. The '#if TEST' branch is never executed.

This happens in a Playground and in compiled code. Xcode 9 Beta 5

I changed 'let TEST = true' to 'var TEST = true', but I still get the same erroneous behaviour. — Simon Youens
– Simon Youens, Commented Aug 23, 2017 at 16:51

Code Different · Accepted Answer · 2017-08-24 00:06:38Z

1

You are confusing between TEST the local variable and TEST the compiler flag.

What you have is a local variable named TEST
What you are testing for is the existence of a compiler flag named TEST

To set a compiler flag, go to the target's Build Settings and search for Other Swift Flags, then add -DTEST to your Debug build.

Also, Swift compiler flags are much simpler compared to C compiler flags: a flag either exists or not. You can't associate an int value or anything like that. Consequently, there's no point in getting its value. It's a true/false situation: you know that it exists or not; there's no other value associated with it.

answered Aug 24, 2017 at 0:06

Code Different

93.4k16 gold badges154 silver badges175 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

Simon Youens Over a year ago

CD, thanks for your response I had suspected something like that, but i couldn't find any good description. If I want more than one flag, do they go on the same line?

Collectives™ on Stack Overflow

Swift 4 Conditional compile logic not working correctly

1 Answer 1

1 Comment

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

1 Comment

Your Answer

Sign up or log in

Post as a guest

Linked

Related