Decode the constant/variable

Question

To save codepoints, a compute-er can share 0 and -, and share . with $(indicating a variable name).

A number matches RegEx /^-?(0|[1-9]\d*)(\.\d+)?$/ and a variable matches RegEx /^\$\d+$/.

Given input where 0 and - are represented as same char(your choice) and so do . and $, return as number or variable. You can assume that it's possible.

Test cases

0 => 0
10 => 10
01 => -1
00 => -0
001 => (undefined behavior)
1.3 => 1.3
.33 => 33ドル
0.33 => 0.33
00.33 => -0.33
42.0 => 42.0
3.4.5 => (undefined behavior)

Shortest code wins.

• tsh: Is 42 a valid input? If so, add it add another testcase.

Answer 1

Retina 0.8.2, 14 bytes

1T1`0.`-$`^.\d

Try it online! Link includes test cases. Explanation: Port of @tsh's JavaScript answer.

1T1`

Transliterate only the first character of the first match. (There will only be one match, but I still need to specify it so that I can only transliterate the first character.)

0.`-$`

Transliterate 0 to - and . to $.

^.\d

Match only the first two characters and only if the second is a digit. (Transliterate will ignore the first character if it's not a 0 or ..)

Answer 2

JavaScript (Node.js), 42 bytes

s=>s.replace(/^[0.](?=\d)/,x=>x<1?'-':'$')

Try it online!

Answer 3

Perl 5 -p, (削除) 25 (削除ここまで) 21 bytes

@Neil saved 4 bytes

s/^[.\$]/\$/;s;^0\B;-

Try it online!

• Neil: You can use \B instead of (?=\d) to save 4 bytes.

Answer 4

C (gcc), 45 bytes

f(char*x){*x=*x-48|x[1]%46<1?*x-46?*x:36:45;}

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Answer 5

05AB1E, (削除) 21 (削除ここまで) 16 bytes

¦нdić„-$ì¤0tskèì

Try it online or verify all test cases.

Explanation:

¦ # Remove the first character of the (implicit) input
 н # Pop and take the new first character (aka the second),
 # or an empty string if the input was 1 character long
 di # Pop, and if this second character is a digit:
 ć # Extract head; push first character and remainder-string
 „-$ì # Prepend "-$" before this first character
 ¤ # Push the last character of this triplet (without popping),
 # aka that same first character
 0t # Push 0.0 (square root of 0)
 s # Swap so the first character is at the top again
 k # Get the 0-based index of this character in "0.0",
 # or -1 if it's neither "0" nor "."
 è # Use that index to index into "-$ÿ",
 # where -1 indexes into this last character (`ÿ`)
 ì # Prepend that to the remainder-string
 # (after which it is output implicitly as result)
 # (implicit else: output the implicit input as is)

Answer 6

Charcoal, 21 bytes

Pθ¿‹.⊟KD2→§+-$KK⌕0.KK

Try it online! Link is to verbose version of code. Explanation:

Pθ

Output the input string without moving the cursor.

¿‹.⊟KD2→

If the second character exists and is a digit, then...

§+-$KK⌕0.KK

... transliterate the character under the cursor from 0 to - or . to $ or anything else to itself.

Answer 7

Java, (削除) 57 (削除ここまで) 53 bytes

s->s.replaceAll("^0\\B","-").replaceAll("^\\.","\\$")

-4 bytes thanks to @Neil.

Try it online.

Explanation:

s-> // Method with String as both parameter and return-type
 s.replaceAll("^0 // Replace a leading "0",
 \\B", // which has a (zero-width) digit after it
 "-") // with "-"
 .replaceAll("^\\.", // Then(/or) replace a leading "."
 "\\$") // with "$"

• Neil: replaceAll("^0\\B","-") saves 4 bytes.