我正在尝试使用 BrainFuck 编写一个程序，该程序可以读取最多为 9 的两个数字，计算它们的总和，然后将结果打印出来，例如 3 & 5 给出结果 8 。
我只是想理解 BF 语言，但它看起来比我想象的要困难得多。
问问题
13210 次
7 回答
33
将语言想象成一个巨大的磁带（30K 字节长），您可以在其中读取、写入和向前或向后移动，并一次增加/减少一个单元格（每个单元格为 1 个字节，因此您实际上有 30K 个单元格）。或者，您可以读入和写出字节流包含的内容（以 ASCII 形式）。假设您了解基本运算符，则对两个数字求和的程序应遵循以下几行：
, ; read character and store it in p1
> ; move pointer to p2 (second byte)
, ; read character and store it in p2
[ ; enter loop
< ; move to p1
+ ; increment p1
> ; move to p2
- ; decrement p2
] ; we exit the loop when the last cell is empty
< ; go back to p1
------------------------------------------------ ; subtract 48 (ie ASCII char code of '0')
. ; print p1
于 2012-05-20T03:24:43.510 回答
14
我在 2-3 天前看到了这篇文章，我已经在研究它，现在我有一个关于多位数加法的解决方案。一开始我觉得这个 PL 的名字有点冒犯，但现在我知道，如果我被授权命名这种编程语言，我会选择相同的。
现在，我将告诉您如何使用我的代码。
$ bf sum.bf
199+997=
1196
$
我的代码中只能添加 +ve 数字。并确保在两个输入中使用相同的位数。即，如果您想将 57 与 3 相加，则输入 57+03= 或 03+57= 之类的输入。现在是代码。我已经记录了一个例子。我仍然不喜欢查看我的代码，因为自己设计比在 bf 中学习或排除代码更容易。首先，您需要知道如何比较两个数字。我对这个问题的回答是一种解决方案。在文档中，我使用了 'plus' 而不是 + ，因为 + 是 bf 中的有效操作。
>> +
[- >,>+<
----- ----- ----- ----- ; checking with ascii 43 ie plus symbol
----- ----- ----- -----
---
[
+++++ +++++ +++++ +++++
+++++ +++++ +++++ +++++
+++
< ] >>
]
; first input is over and terminated by a 'plus' symbol
<->>>>>+
[- >,>+<
----- ----- ----- ----- ; checking with ascii 61 ie = symbol
----- ----- ----- -----
----- ----- ----- ------
[
+++++ +++++ +++++ +++++
+++++ +++++ +++++ +++++
+++++ +++++ +++++ ++++++
< ] >>
]
; second input is over and terminated by an = symbol
; now the array looks like 0 0 0 49 0 50 0 0 0 0 0 0 0 0 49 0 53 0 0 1 0
; for an input 12'plus'15=
<<<<
[<+<]
; filled with 1's in between
+ [<+>-<<[>-]>] ; This is a special loop to traverse LEFT through indefinite no of 0s
; Lets call it left traverse
<<
[<+<]
>[>]<
; now the array looks like
; 0 0 1 49 1 50 0 0 0 0 0 0 0 1 49 1 53 0 0 1 for eg:12plus15
[
[->+> + [>+<->>[<-]<] ; Right traverse
>>[>]<+ [<]
+ [<+>-<<[>-]>] ; Left traverse
<<-<
]
+ [>+<->>[<-]<]
>> [>] <<-<[<]
+ [<+>-<<[>-]>]
<<-<
]
; now actual addition took place
; ie array is 00000000000000 98 0 103 0 0 1
+ [>+<->>[<-]<]
>>
[
----- ----- ----- -----
----- ----- ----- -----
----- ---
>>]
; minus 48 to get the addition correct as we add 2 ascii numbers
>-< ; well an undesired 1 was there 2 place after 103 right ? just to kill it
; now the array is 00000 00000 0000 50 0 55
; now comes the biggest task Carry shifting
<<
[<<]
+++++ +++++ +++++ +++++
+++++ +++++ +++++ +++++
+++++ +++
[>>]
; we added a 48 before all the digits in case there is an overall carry
; to make the size n plus 1
; array : 00000 00000 00 48 0 50 0 55
<<
<<
[
[>>->[>]>+>>>> >>>+<<<< <<<<<[<]><<]
>+[>]>-
[-<<[<]>+[>]>]
>>>>>+>>>
+++++ +++++ +++++ +++++ +++++
+++++ +++++ +++++ +++++ +++++
+++++ +++
<
; comparison loop: 0 1 0 a b 0
; (q) (p) (num) (58)
[->-[>]<<] ; comparison loop to check each digit with 58: greater means
; we need to minus 10 and add 1 to next significant digit
<[-
; n greater than or equal to 58 (at p)
<<<< <<<
[<]+
>
----- ----- ; minus 10 to that digit
<<+ ; plus 1 to next digit
>
[>]
>>>>>>
]
< [-<
; n less than 58 (at q)
<<<<<<
[<]+
[>]
>>>>>
]
; at (q)
>>>[-]>[-]
<<<<< <<<<<
[<]>
<<
]
; Its all over now : something like 0 48 0 52 0 66 ( ie 0 4 18 )
; will turn into 0 48 0 53 0 56 (ie 0 5 8)
>>
----- ----- ----- -----
----- ----- ----- -----
----- ---
; here we are just checking first digit is 48 or not
; its weird to print 0 ahead but it is defenitely needed
; if it is 49 ie 1
[
+++++ +++++ +++++ +++++
+++++ +++++ +++++ +++++
+++++ +++
.
[-]
]
>>
[.>>]
+++++ +++++
. ; to print nextline : ascii 10
我知道它的代码有点长，可能有更好的解决方案。但仍然值得一试。
于 2013-10-16T20:14:26.020 回答
3
这就是我所知道的
, ;read character and store it in p1
------------------------------------------------ ;return ascii to Dec
< ;move pointer to p2 (second byte)
, ;read character and store it in p2
------------------------------------------------ ;return ascii to Dec
[ ; enter loop
- ; decrement p2
> ; move to p1
+ ; increment p1
< ; move to p2
] ; we exit the loop when the last cell is empty
> ;go back to p1
++++++++++++++++++++++++++++++++++++++++++++++++ ;return Dec to ascii
. ;print p1
输入：
12
输出：
3
你应该使用小于 10 的数字，结果小于 10
于 2014-02-10T14:19:19.523 回答
2
我创建了一个解决方案，可以将任意数字与任意数字相加：
,[+>,]>++++++++[<<[------<]>[>]>-]<<[++++[----[<]<+[<+<+>>-]<<[>>+<<
-]>[<<<[<<]+[>>]>-]>>>>[>]<-[[<]<<<<<[<<]>>>+>[>>]>>>>[>]<-]<[<]<<<<<[<<]>>[[-]>
>]>>>>[>]>+<]>-[<<[<]<[-]>>[>]>+]<<]<[-]<->>>>>->+[[-]>+>[<]<]<->>+>->+[>>+]<<[+
[--[>+>+<<-]>>[<<+>>-]>+<<[>++++++++++<[->-[>>]<<<]>>>>[<<<----------<<+>>>>]<]>
>[-]<<<[[-]<[-]<<[->>>>>+<<<<<]]<]<]>>>>>>>>>>[<<]>>[+[->++++++++[<++++++>-]<.[-
]]>>]
例如：
input: 13611+85399+35943
output: 134953
确保“内存溢出行为”设置为换行。
于 2019-07-10T10:01:52.933 回答
0
我构建了一个可以处理无限数量的输入数字的代码，但它只能输出 1 或 2 位数字，因此数字总和的范围可以是 0-99
>, go to cell #1 and read input
[ loop while input is not null
>+++++++[<------->-] subtract 49 from input
<+ add 1 to input so we have the value of the input numeral in cell #1
[-<+>] add cell #1 to cell #0
, read next numeral
] repeat until no number is being input (the input number can be longer than two digits)
>- go to cell #2 and set it to minus 1 so the loop will run (i will explain that at the end of the loop)
[ loop while number in cell #0 is greater than 9
+ set the value of #2 to 0
<[->+>+<<] copy the value of #1 to #2 and #3
>>[-<<+>>] copy the value of #3 back to #1
<<< go to #0
[->>>+>+<<<<] copy it to #3 and #4
>>>>[-<<<<+>>>>] restore it in #0
< go to 3
[-[-[-[-[-[-[-[-[-[ check if number is greater than 9
<<+<----- ----->>> if yes increment number in cell #1 and decrease number in cell #0 by 10
[-]]]]]]]]]]] set #3 to 0 so we can leave the loop
<< go to #1
we need to check if the number in cell #1 was incremented
cell #2 will store the most recent value of cell #1
[->->+<<] so we subtract the value of cell #2 by the value of cell #1 and
store the value of cell #1 in cell #3
>>[-<<+>>] restore the value of #1
< go back to cell #2
] if cell #1 was increased then cell #2 will be minus 1 and the loop will restart
<[>+++++++[<+++++++>-] if cell #1 is greater than 0 then add 49 to it
<-.[-]] subtract 1 so we have the ascii code of the value in #1 then print it and set it to zero
+++++++[<+++++++>-]<-. do the same with the value in cell #0
或简而言之：
>,[>+++++++[<------->-]<+[-<+>],]>-[+<[->+>+<<]>>[-<<+>>]<<<[->>>+>+<<<<]>>>>[-<<<<+>>>>]<[-[-[-[-[-[-[-[-[-[<<+<---------->>>[-]]]]]]]]]]]<<[->->+<<]>>[-<<+>>]<]<[>+++++++[<+++++++>-]<-.[-]]+++++++[<+++++++>-]<-.
于 2018-06-19T09:26:02.610 回答
0
许多人已经回答了这个问题，但是由于每个解决方案都有所不同，我也将添加我的。
我的解决方案会进行个位数加法（如果结果大于 9）。因此，例如对于输入“89”，它会返回“17”。我添加了很多评论，所以它应该相对容易理解。
[
A brainfuck program for doing a single digit addition.
Ex. input: '13' -> output: '4'
input: '99' -> output: '18'
Author: Florian Baierl
]
initialize #0 with 48 (ASCII char for '0')
>++++ ++++
[
<++++ ++
>-
]
save input to #1 and #2
,>,
substract value from #0 from #1 and #2
<<
[
>-
>-
<<-
]
move to #1
>
substract from #1 and add to #2; now the answer is in #2
[
->+<
]
since we need to modify the answer afterwards write it to #3 and #6
as well
>
[>+>>>+<<<<-]
Is the answer bigger than 9?
to continue the memory tape should look like this:
0 1 0 (a) b 0
with the pointer pointing to a
<+
>>>+++++ +++++ b is 10
< point to a
+>+< necessary in case a and b are zero
loop to determine whether a or b reach 0 first
[->-[>]<<]
<[-
a was bigger or equals b
the answer is still stored in #6
subtract 10 from #6 (the answer)
>>>>> ----- -----
write 48 to #4
<++++ ++++
[
<++++ ++
>-
]
add 48 to #5 and #6
<
[->+>+<<]
print out the results
>+.>.
leave loop (back to #2)
<<<<
]
<[-
a was samller so we can simply print the answer out after adding 48 to it
the answer is still stored in #6
>>
++++ ++++
[
<++++ ++
>-
]
<
[
->>>>+<<<<
]
print #2
>>>>.
we want to leave the loop so go somewhere with the value '0'
>
]
于 2018-05-06T17:45:40.240 回答
0
我还制作了一个只能处理单个数字条目和答案的程序：
#Make the first cell (Cell 0) hold a value of 48
>++++ ++++
[
<++++ ++
>-
]
#Get inputs and minus 48 from each to get Decimal
,>,
<<
[
>-
>-
<<-
]
#Adds the contents of Cells 1 and 2
>
[
>+
<-
]
#Moves answer to Cell 0
>
[
<+
>-
]
<
[
<+
>-
]
#Converts answer to ASCII
>++++ ++++
[
<++++ ++
>-
]
<
[
<+
>-
]
<
#Print answer
.
于 2016-11-08T00:25:01.797 回答