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Re: Coding while drunk and stoned take 2. (Yes, I know this is a cross post)

From Ben Bacarisse <ben.usenet@bsb.me.uk>
Newsgroups comp.programming
Subject Re: Coding while drunk and stoned take 2. (Yes, I know this is a cross post)
Date 2020-02-27 21:00 +0000
Organization A noiseless patient Spider
Message-ID <87ftevvj7v.fsf@bsb.me.uk> (permalink)
References <0b8d1b58-3651-47ed-a2b8-63dc1d659c91@googlegroups.com> <87y2souis2.fsf@bsb.me.uk> <bcc61784-004c-4bf9-bc09-fa44d22f83c8@googlegroups.com>

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Julio Di Egidio <julio@diegidio.name> writes:

> On Thursday, 27 February 2020 16:55:29 UTC+1, Ben Bacarisse  wrote:
>> "Chad Altenburg, Ph.D." <cdalten@gmail.com> writes:
> <snip>
>> > My gut feeling says this isn't right. But I have no idea how this
>> > could be done in one loop.
>> 
>> Here's a C solution.  It's messy because the natural way to do string
>> things in C is to pass an array to take the result.  You then need to
>> pass a size so the function does not write out-of-bounds.  But to be
>> useful the function should keep going and return the size that's needed
>> even when there was no room this time:
>> 
>> #include <stdio.h>
>> 
>> size_t zipzap(const char *src, char *dst, size_t n)
>> {
>>      size_t i = 0;
>>      while (*src) {
>>           if (i < n)
>>                dst[i] = *src;
>>           i += 1;
>>           if (src[0] == 'z' && src[1] && src[2] == 'p') {
>>                if (i < n)
>>                     dst[i] = 'p';
>>                i += 1;
>>                src += 3;
>>           }
>>           else src += 1;
>>      }
>>      dst[i <= n ? i : n-1] = 0;
>>      return i + 1;
>> }
>> 
>> int main(int argc, char **argv)
>> {
>>      char z[10];
>>      for (int a = 1; a < argc; a++) {
>>           size_t required = zipzap(argv[a], z, sizeof z);
>>           printf("%s -> %s (%zu)\n", argv[a], z, required);
>>      }
>> }
>> 
>> The more "scripty" way -- allocating the result -- looks simpler:
>> 
>> char *zipzap_str(const char *src)
>> {
>>      char result[strlen(src) + 1], *dst = result;
>>      while (*src) {
>>           *dst++ = *src++;
>>           if (src[-1] == 'z' && src[0] && src[1] == 'p') {
>>                *dst++ = 'p';
>>                src += 2;
>>           }
>>      }
>>      *dst++ = 0;
>>      char *rp = malloc(dst - result);
>>      return rp ? strcpy(rp, result) : 0;
>> }
>
> Man, that hurts...

Yes, it's grim.  It's just three explicit patterns in Haskell:

zipzap s = case s of
  []               -> []
  'z':_:'p' : rest -> 'z':'p' : zipzap rest
  c         : rest -> c       : zipzap rest

-- 
Ben.

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Thread

Coding while drunk and stoned take 2. (Yes, I know this is a cross post) "Chad Altenburg, Ph.D." <cdalten@gmail.com> - 2020-02-27 06:04 -0800
  Re: Coding while drunk and stoned take 2. (Yes, I know this is a cross post) Julio Di Egidio <julio@diegidio.name> - 2020-02-27 07:24 -0800
  Re: Coding while drunk and stoned take 2. (Yes, I know this is a cross post) Ben Bacarisse <ben.usenet@bsb.me.uk> - 2020-02-27 15:55 +0000
    Re: Coding while drunk and stoned take 2. (Yes, I know this is a cross post) Julio Di Egidio <julio@diegidio.name> - 2020-02-27 08:06 -0800
      Re: Coding while drunk and stoned take 2. (Yes, I know this is a cross post) Ben Bacarisse <ben.usenet@bsb.me.uk> - 2020-02-27 21:00 +0000
        Re: Coding while drunk and stoned take 2. (Yes, I know this is a cross post) "Chad Altenburg, Ph.D." <cdalten@gmail.com> - 2020-02-27 15:18 -0800
  Re: Coding while drunk and stoned take 2. (Yes, I know this is a cross post) Ben Bacarisse <ben.usenet@bsb.me.uk> - 2020-02-27 21:06 +0000

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