Convert the type modifiers from prefix to C-fix

Question

Given a base type T, this can be augmented with * or [], each meaning pointer and array. For example, []*T means "array of pointer to T" and *[]T means "pointer to array of T".

Let's call * and [] type modifiers. C has a way of ordering these modifiers tied to the evaluation order. []*T in C-style becomes T *[], and *[]T becomes T (*)[]. You may be able to understand how the conversion works by having a look at the examples below. An explanation is also given at the end of this post.

* -> *
[] -> []
[]* -> *[]
*[] -> (*)[]
*[]* -> *(*)[]
[]*[] -> (*[])[]
*[][]* -> *(*)[][]
[]**[] -> (**[])[]
[]*[]* -> *(*[])[]
*[]*[] -> (*(*)[])[]

Your program or function should process an input string to an output string as the examples above. The input will only contain * and [] without whitespaces.

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This challenge is this challenge in reverse, simplified.

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Rules for C-fix:

• All of the *s always come before all of the []s (in real C code, the variable name comes between the last * and the first []).
• Evaluation starts in the middle, where the variable name would be.
• If there is both a * on the left and a [] on the right, without any parentheses to determine the order, the [] is bound first, and comes first in prefix:
  • C-fix *[]
    = C *a[] (where a is the variable name) = "array of pointers" =
    prefix []*.
• In order to change the order of evaluation so that a * gets bound first, it must be put in parentheses (with where the variable name would be):
  • prefix *[]
    = "pointer to an array" = C (*a)[](where a is the variable name) =
    C-fix (*)[].

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From the last example, (*(*)[])[] in actual C code would be something like int (*(*a)[])[] (where a is the variable name). The evaluation starts from a and [] is bound first unless a set of parentheses blocks the binding.

If you put int (*(*a)[])[] to cdecl.org, the output is "declare a as pointer to array of pointer to array of int". This is because:

• The first * is bound first due to the parentheses.
• There are no parentheses between the variable name and the [], causing that to be bound before the next *.
• Then, the next * is bound because it is in the parentheses with the variable name, whereas the remaining [] is not.
• Finally, the remaining [] is bound.

This is *[]*[]int a in prefix. Hence, *[]*[] -> (*(*)[])[].

Answer 1

Curry (PAKCS), 90 bytes

f s@(""?"[]")=s
f(s++"*")='*':f s
f(s++"*[]")="(*"++f s++")[]"
f(s++"][]")=f(s++"]")++"[]"

Try it online!

Answer 2

Retina 0.8.2, 52 bytes

^((.])*)(\**)
$3$1¶
{G`.
(.+)¶((.])+)(\**)
$4($1)$2¶

Try it online! Link includes test cases. Explanation:

^((.])*)(\**)
$3$1¶

Process any initial []s or *s into an output line.

{`

Repeat until there is nothing left to process.

G`.

Delete the original input line once there is nothing left.

(.+)¶((.])+)(\**)
$4($1)$2¶

Process some more []s and *s.

Answer 3

Charcoal, 27 bytes

ＦＳ≔⎇⁼ι*⁺ιω⁺⎇∧ω⁼§ω⁰*⪫()ωωιωω

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

ＦＳ

Loop over all of the input characters.

≔⎇⁼ι*⁺ιω⁺⎇∧ω⁼§ω⁰*⪫()ωωιω

If the current character is a *, then prepend it. Otherwise, append it, but if the output string starts with a * then wrap it in () first.

ω

Output the final string.

Answer 4

Python 3.8+, ⁹² 83 bytes

f=lambda p,c="":f(p[1:],["*"+c,[f"({c})",c][c[:1]!="*"]+p[0]][p[0]>"*"])if p else c

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Answer 5

C (clang), 119 117 bytes

-2 bytes thanks to @ceilingcat

*b;f(char*a,l){return l&&asprintf(&b,a[--l]<43?"*%*s":a[l]<93?:--l&&a[l-1]<43?l--,"(*%*s)[]":"%*s[]",l,f(a,l))?b:"";}

Try it online! Takes input as a character array and its length; returns an allocated string.