编写一个quine，该quine在运行时会在当前目录中创建一个名为自身的源文件。我们将在其中使用Windows，因此文件名（并因此是quine）必须具有以下限制：

• 这些字符都不是 \ / : ? * < > |
• 少于211255个字符

局限性和假设

• 您的代码必须是完整的程序（毕竟将要运行）。

• 不能从源文件复制。

• 您可以假设没有另一个以Quine为名称的文件（因为它将产生Quine（1））。

• 错误是允许的（只要它们不进入源代码即可）

这是 代码高尔夫球，以字节为单位的最短代码胜出！

编辑

也许我不清楚，但带有quine名称的文件实际上必须包含该quine。我的错。

Vitsy，10 27字节

'rddd++&"rdd8++a[v}v1-D);].

如果:允许，我可以将其缩短为11个字节。:(

说明：

'rddd++&"rdd8++a[v}v1-D);].
'                           Capture all instructions as a string until encountering
                              ' again, looping if necessary.
 r                          Reverse the current stack (output is top-down).
  ddd++                     Push char literal ' to the stack.
       &                    Push a new stack to the stack stack.
        "                   Same as ', but " specific.
         r                  Reverse the current stack.
          dd8++             Push char literal " to the stack.
               a            Push 10 to the stack.
                [        ]  Loop forever.
                 v          Capture the top value as a variable.
                  }         Take the bottom item of the stack and put it on the top.
                   v        Dump the variable to the stack.
                    1-      Subtract 1.
                      D     Duplicate the top item.
                       );   Pop n; if n is 0, exit the loop.
               a[v v1-D);]  This is a makeshift for loop with 10 iterations.
                          . Pop the top stack as n, and the second stack as o. 
                              Write a file called "n" with the contents of "o".

最后的两个堆栈相同。在某些Java版本上，FileNotFoundException由于FileInputStream类的不同实现，这可能会抛出a 。

编辑前的先前答案：

&'rddd++}.

说明：

&           Push a new stack to the stack stack.
 'rddd++    Modified standard quine.
        }   Move the ' to the right place.
         .  Pop the top stack as n, and the second stack as o. Write a file called
                 "n" with the contents of "o". (Quine name, no content)

因为我也是如此，所以下面是堆栈中发生的步骤的示意图（每个堆栈由表示[]）：

Initial state: [[]]

&              [[], []]
                  Push a new stack to the stack stack.

 '             [[], ["r", "d", "d", "d", "+", "+", "}", ".", "&"]]
                  This stack state occurs because ' will loop around the line
                    until finding the next '.

  r            [[], ["&", ".", "}", "+", "+", "d", "d", "d", "r"]]
                  Reverses the top (last) stack.

   ddd         [[], ["&", ".", "}", "+", "+", "d", "d", "d", "r", 13, 13, 13]
                  Push thirteen thrice.

      ++       [[], ["&", ".", "}", "+", "+", "d", "d", "d", "r", "'"]
                  Because char ' is 39 = 13 + 13 + 13.

        }      [[], [".", "}", "+", "+", "d", "d", "d", "r", "'", "&"]]
                  Takes the bottom item, then puts it on the top.

         .     []
                  Writes a file with content from the second-to-top stack and the
                    name as the concatenation of all elements in the top stack,
                    with top member priority (backwards from my representation).

我实际上不确定这是如何工作的。微小的变化会导致此中断。例如，放置&在不同的其他位置应该只是工作为好，但它会导致ClassCastExceptionS，IOExceptions和ArrayOutOfBoundsExceptions ^取决于你把它。我可能需要进行一些错误修正。

Node.js，56 52字节

function f(){require('fs').writeFile(f+='f()',f)}f()

这会打印警告

DeprecationWarning：不推荐使用不带回调的异步函数。

如果您想要全绿色，则需要4个字节才能更改writeFile为writeFileSync。

Lua，96个字节。

s="s=%qs=string.format(s,s)f=io.open(s)f.write(f,s)"s=string.format(s,s)f=io.open(s)f.write(f,s)

在我的手机上输入此字词即可正常使用，但是当我进入计算机时会对其进行测试。

C，134字节

s[255];fd;char p[255]="s[255];fd;char p[255]=%c%s%c;main(){sprintf(s,p,34,p,34);creat(s,0);}";main(){sprintf(s,p,34,p,34);creat(s,0);}