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Q&A Why is boolean value f (false) defined as a parsing-word while t (true) is not in Factor?

I noticed when attempting to generate random booleans that t and f are not treated in the same way: t random ! Error f random ! returns a value In the factor 0.98 and 0.99 documentation, t is ...

1 answer  ·  posted 2y ago by Razetime‭  ·  edited 2y ago by Alexei‭

Question factor boolean
#2: Post edited by user avatar Alexei‭ · 2022-08-19T05:42:12Z (about 2 years ago)
cleanup up the title a little bit
  • Why is boolean value f (false) defined as a parsing-word while t (true) is not? [Factor programming language]
  • Why is boolean value f (false) defined as a parsing-word while t (true) is not in Factor?
#1: Initial revision by user avatar Razetime‭ · 2022-08-18T04:50:29Z (about 2 years ago)
Why is boolean value f (false) defined as a parsing-word while t (true) is not? [Factor programming language]
I noticed when attempting to generate random booleans that `t` and `f` are not treated in the same way:

```factor
t random ! Error
f random ! returns a value
```

In the factor 0.98 and 0.99 documentation, [`t`](https://docs.factorcode.org/content/word-t,syntax.html) is defined as a simple word, whereas [`f`](https://docs.factorcode.org/content/word-f,syntax.html) is given special status as a parsing word. Why?