你如何遍历字符串的索引？

Question

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] = '[=10=]0' then acc else
    loop(i + 1, acc + digit(example[i]))
}

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

这（尝试）遍历字符串的索引，将字符串的字符作为数字求和。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，如果 i+10 处的字符是偶数，它应该只使用字符。

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)
}

上面的f()应该怎么写呢？请给我一个循环遍历字符串索引并按索引从字符串中获取字符的函数示例。同样，我不需要

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，但看看它是多么荒谬。必须有一个正确和适当的 ATS 方法来做到这一点。

#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) = '[=13=]0' then acc else
    loop(i + 1, acc + digit(get(i)))
}

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

输出：

f: 28

编辑 2：

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>))
...

编辑 3：

我可以用

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？

Answer 1

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更多要求。

Answer 2

我试过重新实现你的函数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（而且是非常不安全的风格）。有空的话，后面会尽量给出一个安全优雅的实现。

Answer 3

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]

Answer 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，这里是 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)

你如何遍历字符串的索引？

How do you loop over the indexes of a string?

string

indexing

loops

ats