你如何遍历字符串的索引？答案

【问题标题】：How do you loop over the indexes of a string?你如何遍历字符串的索引？
【发布时间】：2018-01-08 05:09:24
【问题描述】：

考虑：

val example = "1234567"

fn digit(c: char): int =
  case- c of
  | '0' => 0 | '1' => 1 | '2' => 2 | '3' => 3 | '4' => 4
  | '5' => 5 | '6' => 6 | '7' => 7 | '8' => 8 | '9' => 9

fn f(): int = loop(0, 0) where {
  fun loop(i: int, acc: int): int =
    if example[i] = '\000' then acc else
    loop(i + 1, acc + digit(example[i]))
}

implement main0() = () where {
  val () = println!("f: ", f())
}

这（尝试）循环遍历字符串的索引，将字符串的字符求和为数字。我已经用.foldleft 和streamize_string_char 解决了几个类似的问题，但实际任务需要对索引本身进行数学运算（即，不是使用每个字符，它应该只使用一个字符，如果i+10 处的字符为偶数）。

实际上数学是相关的，因为它似乎强制$UNSAFE.cast2int，因为strlen(input)的结果没有除法运算符：

fn day2(): uint = loop(input, 0, 0) where {
  val len = $UNSAFE.cast2int(strlen(input))
  fn nextindex(i: int): int = (i + len/2) mod len
  fn get(i: int): char = input[i]  // <-- also broken at this point
  // this next line is just me slowly going mad
  fun loop{n:int}{i:nat | i <= n}(s: string(n), i: size_t(i), acc: uint): uint =
    if i >= len then acc else
    if s[i] = s[nextindex(i)] then loop(i+1, acc + digit(s[i])) else
    loop(i+1, acc)
}

f()上面应该怎么写？请给我一个函数示例，该函数循环遍历字符串的索引并按索引从字符串中获取字符。同样，我不需要像

这样的解决方案

typedef charint = (char, int)
fn day1(): int = sum where {
  val lastchar = input[strlen(input)-1]
  val res = input.foldleft(TYPE{charint})((lastchar, 0), (lam((last, sum): charint, c: char) =>
                if last = c then (c, sum + digit(c)) else (c, sum)))
  val sum = res.1
}

因为我需要根据索引测试属性。

编辑：

好吧，我终于想出了一些种解决方案，但看看它有多荒谬。必须有正确和适当的 ATS 方式来执行此操作。

#include "share/atspre_staload.hats"

val example = "1234567"

fn digit(c: char): int =
  case- c of
  | '0' => 0 | '1' => 1 | '2' => 2 | '3' => 3 | '4' => 4
  | '5' => 5 | '6' => 6 | '7' => 7 | '8' => 8 | '9' => 9

fn f(): int = loop(0, 0) where {
  fn get(i: int): char = loop(i, string2ptr(example)) where {
    fun loop(i: int, s: ptr): char =
      if i > 0 then loop(i-1, ptr0_succ<char>(s)) else
      $UNSAFE.ptr0_get<char>(s)
  }
  fun loop(i: int, acc: int): int =
    if get(i) = '\000' then acc else
    loop(i + 1, acc + digit(get(i)))
}

implement main0() = () where {
  val () = println!("f: ", f())
}

输出：

f: 28

EDIT2：

不那么荒谬：

...
  val p = string2ptr(example)
  fn get(i: int): char = $UNSAFE.ptr0_get<char>(add_ptr_bsz(p, g0int2uint(i) * sizeof<char>))
...

EDIT3：

我可以再次使用string[i]

overload + with add_ptr_bsz
fn string_get_at(str, i) = $UNSAFE.ptr0_get<charNZ>(string2ptr(str)+g0int2uint(i))
overload [] with string_get_at

这与我在 prelude/DATS/string.dats 中看到的几乎相同......有什么问题？

【问题讨论】：

已批准编辑，但我怀疑提高可见性是否会带来任何好处。如果您不了解 ATS，就无法回答这个问题。

标签： string loops indexing ats

【解决方案1】：

好的，day2 的以下实现是安全的：

fn
day2
(input: string): uint = let

val
[n:int]
input = g1ofg0(input)

val n0 = strlen(input)
val n0 = sz2i(n0) // int(n)

fun
nextindex
(
 i: natLt(n)
) : natLt(n) = nmod(i + n0/2, n0)

fun
loop(i: natLte(n), acc: uint): uint =
  if i >= n0 then acc else
  (
    if input[i] = input[nextindex(i)]
      then loop(i+1, acc + digit2uint(input[i]))
      else loop(i+1, acc)
  )

in
  loop(0, 0u)
end // end of [day2]

【讨论】：

【解决方案2】：

这里有几个问题。你的函数f()可以写成：

fn digit2int(c: char): int = (c - '0')
fn f(): int = loop(example, 0, 0) where
{
  fun
  loop
  {n:int}
  {i:nat|i <= n}
  (cs: string(n), i: int(i), acc: int): int =
  if
  string_is_atend(cs, i)
  then acc else loop(cs, i+1, acc+digit2int(cs[i]))
}

这种编程涉及依赖类型。它通常对程序员提出了更多要求。

【讨论】：

【解决方案3】：

我尝试重新实现你的函数day2：

fn digit2int(c: char): int = (c - '0')

fn
day2(input: string): int =
  loop(0, 0) where
{
  val n0 = strlen(input)
  val n0 =
  g0uint2int_size_int(n0)
  val p0 = string2ptr(input)
  fn nextindex(i: int): int = (i + n0/2) mod n0
  fun get(i: int): char = $UNSAFE.ptr0_get_at<char>(p0, i)
  fun loop(i: int, acc: int): int =
    if i >= n0 then acc else
    (
      if get(i) = get(nextindex(i))
        then loop(i+1, acc + digit2int(get(i))) else loop(i+1, acc)
    )
}

我不得不说上面的实现非常丑陋（而且风格非常不安全）。如果有时间，我稍后会尝试给出一个安全优雅的实现。

【讨论】：

【解决方案4】：

我能够在提供的帮助下完成这项工作，但只是为了还有一个完整的功能示例，这是Day 1, 2017 Advent of Code的解决方案：

/* compile with: patscc -O2 -D_GNU_SOURCE -DATS_MEMALLOC_LIBC day1.dats -o day1 */
#include "share/atspre_staload.hats"
#include "share/atspre_staload_libats_ML.hats"

val input: string = "12341234" /* actual value provided by contest */

fn digit(c: char): uint =
  case- c of
  | '0' => 0u | '1' => 1u | '2' => 2u | '3' => 3u | '4' => 4u
  | '5' => 5u | '6' => 6u | '7' => 7u | '8' => 8u | '9' => 9u

typedef charint = (char, uint)

fn part1(): uint = sum where {
  val lastchar = input[strlen(input)-1]
  val res = input.foldleft(TYPE{charint})((lastchar, 0u), (lam((last, sum): charint, c: char) =>
                if last = c then (c, sum + digit(c)) else (c, sum)))
  val sum = res.1
}

typedef natLT(n:int) = [i:nat | i < n] int(i)
typedef natLTe(n:int) = [i:nat | i <= n] int(i)

fn part2(): uint = loop(0, 0u) where {
  val [n:int] input = g1ofg0(input)
  val len = sz2i(strlen(input))
  fn nextindex(i: natLT(n)): natLT(n) = nmod(i + len/2, len)
  fun loop(i: natLTe(n), acc: uint): uint =
    if i >= len then acc else
    if input[i] = input[nextindex(i)] then loop(i+1, acc + digit(input[i])) else
    loop(i+1, acc)
}

extern fun reset_timer(): void = "ext#reset_timer"
extern fun elapsed_time(): double = "ext#elapsed_time"
%{
#include <sys/time.h>
#include <time.h>
struct timeval timer_timeval;
void reset_timer() { gettimeofday(&timer_timeval, NULL); }
double elapsed_time() {
  struct timeval now;
  gettimeofday(&now, NULL);
  int secs = now.tv_sec - timer_timeval.tv_sec;
  double ms = (now.tv_usec - timer_timeval.tv_usec) / ((double)1000000);
  return(secs + ms);
}
%}

fn bench(f: () -> void) = () where {
  val () = reset_timer()
  val () = f()
  val c = elapsed_time()
  val () = println!(" (timing: ", c, ")")
}

implement main0() =
  begin
    bench(lam() => print!("part1: ", part1()));
    bench(lam() => print!("part2: ", part2()));
  end

输出（对于假的“12341234”输入）：

part1: 0 (timing: 0.000137)
part2: 20 (timing: 0.000001)

【讨论】：