如何在达特char型工作？ （打印字母） [英] How to work with char types in Dart? (Print alphabet)

问题描述

我想学习飞镖语言，通过转由我校C编程给出的exercices。

I am trying to learn the Dart language, by transposing the exercices given by my school for C programming.

在我们的C池中的第一个exercice是的编写一个函数 print_alphabet（），打印字母小写的;它是被禁止直接打印的字母的

The very first exercice in our C pool is to write a function print_alphabet() that prints the alphabet in lowercase; it is forbidden to print the alphabet directly.

在POSIX C，直截了当的解决办法是：

In POSIX C, the straightforward solution would be:

#include <unistd.h>

void    print_alphabet(void)
{
    char    c;

    c = 'a';
    while (c <= 'z')
    {
        write(STDOUT_FILENO, &c, 1);
        c++;
    }
}

int     main(void)
{
    print_alphabet();
    return (0);
}

不过，据我所知，飞镖（1.1.1）的当前版本没有处理字符的简单方法。我想出了（我的第一个版本），最远的是这样的：

However, as far as I know, the current version of Dart (1.1.1) does not have an easy way of dealing with characters. The farthest I came up with (for my very first version) is this:

void  print_alphabet()
{
  var c = "a".codeUnits.first;
  var i = 0;

  while (++i <= 26)
  {
    print(c.toString());
    c++;
  }
}

void main() {
  print_alphabet();
}

它打印每一个字符，每行一个的ASCII值，作为一个字符串（97......122）。不是真的是我的本意......

Which prints the ASCII value of each character, one per line, as a string ("97" ... "122"). Not really what I intended…

我试图寻找这样做的正确方法。但缺少了字符键入像在C的给我一点很难，作为一个初学者！

I am trying to search for a proper way of doing this. But the lack of a char type like the one in C is giving me a bit of a hard time, as a beginner!

推荐答案

飞镖不具有的性格类型。

Dart does not have character types.

要一个code点转换为字符串，使用String构造String.fromChar code：

To convert a code point to a string, you use the String constructor String.fromCharCode:

int c = "a".codeUnitAt(0);
int end = "z".codeUnitAt(0);
while (c <= end) {
  print(new String.fromCharCode(c));
  c++;
}

对于这样简单的东西，我会使用打印，而不是标准输出，如果你不介意的换行。

For simple stuff like this, I'd use "print" instead of "stdout", if you don't mind the newlines.

还有：

int char_a = 'a'.codeUnitAt(0);
print(new String.fromCharCodes(new Iterable.generate(26, (x) => char_a + x)));

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