When encountering a problem whose input is meant to be read sequentially, but you may also want to push back onto the input like a stack, which languages have the optimal boilerplates for doing so, and how?

For example, from Create a Boolean Calculator, we have a string like:


We want to "pop" 5 chars 1AND0, compute it to be 0, then "push" it back to form:


Then, repeat until there's only 1 char left. In this case:


It's trivial to write while loops and such, but this seems common enough in codegolf problems that there must be canonical forms. I wasn't able to find a question specifically about this setup though, and going through each question by language was difficult.

  • \$\begingroup\$ I think this question is very difficult to answer without a specific language tag. The answer will vary wildly based on the length of keywords and the language paradigm. I haven't VTC yet, but I think this is probably lacking focus unless you specify a language. \$\endgroup\$ Commented Jul 15, 2020 at 22:05
  • \$\begingroup\$ Hm, I guess I was hoping to pick my golfing language for these types of questions by knowing which was best at it. \$\endgroup\$ Commented Jul 15, 2020 at 22:09
  • \$\begingroup\$ I see, then it may be reasonable if you rephrase this as "in which language is doing task X the shortest" (that way it has a precise answer). I think that would probably make it the first of its kind though, so I'd advise asking something on meta. I'm wrong often enough about what the majority wants here so I wouldn't trust just me! \$\endgroup\$ Commented Jul 15, 2020 at 22:19
  • \$\begingroup\$ Why not write a challenge that requires this trick to be used? You basically want to sequentially replace a string prefix so if you require each step in the output you should find the best way to do that among the competing languages. \$\endgroup\$
    – Sanchises
    Commented Jul 15, 2020 at 22:21
  • 2
    \$\begingroup\$ This looks totally on-topic for a tips question. Selecting a language is part of golfing, and the feature being asked for is pretty specific. \$\endgroup\$
    – xnor
    Commented Jul 16, 2020 at 2:44

1 Answer 1


The approach you're asking seems like reduce/left fold in general. Many languages have this, such as Python (reduce(f,seq) or functools.reduce(f,seq)), APL (f⍨/⌽seq), Jelly (f/seq), and Haskell (foldl f start seq).

As a Python example, let's assume we already have the input parsed as a list seq=[1, 'AND', 0, 'OR', 1, 'XOR', 1]. Then reduce(f,seq) is equivalent to

f(f(f(f(f(f(1, 'AND'), 0), 'OR'), 1), 'XOR'), 1)

The trouble here is that we need to take 3 arguments at a time. A way this could be done is by grouping most of the sequence into pairs seq2=[1, ['AND',0], ['OR',1], ['XOR',1]], so reduce(f,seq) would be equivalent to

f(f(f(1, ['AND',0]), ['OR',1]), ['XOR',1])

This could work well in Jelly because it has a builtin s that could help split into pairs (output looks funny strings are internally lists of chars).

However, a loop-based approach would work better in Python by assigning to a slice of an array:

seq=[1, 'AND', 0, 'OR', 1, 'XOR', 1]
while len(seq)>1:
  seq[1:3] = [f(*seq[1:3])]

This would output f(f(f(1, 'AND', 0), 'OR', 1), 'XOR', 1).

As @AdHocGarfHunter notes in the comments, recursion is a good idea too:

# (ungolfed)
def r(s):
  if len(s)>1:
    return r(f(*s[:3]) + s[3:])
    return s[0]

APL has little boilerplate for this: {1=⍴⍵:⊃⍵⋄∇(3↓⍵),f3↑⍵} ( is the recursion). Jelly does too, with 1 byte recursion.


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