我想写与BrainFuck一个程序,可以读出两个数高达9,计算它们的总和,然后打印出结果,如3 & 5给出结果8。如何计算两个数字的总和与BrainFuck
我只是想了解BF语言,但它看起来比我想象的要困难得多。
我想写与BrainFuck一个程序,可以读出两个数高达9,计算它们的总和,然后打印出结果,如3 & 5给出结果8。如何计算两个数字的总和与BrainFuck
我只是想了解BF语言,但它看起来比我想象的要困难得多。
认为语言是一个巨大的胶带(30K字节),在那里你可以读,写,向前或向后移动和增量/在一个时间递减一个单元(每个单元为1个字节,所以你有效地具有30K细胞)。或者,您可以读入和写出字节流保存的内容(以ASCII形式)。假设你知道基本的运营商,程序总结两个数字应该大致如下走:
, ; 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
我看到这个帖子2-3天前,我已经在它的工作,现在我有一个解决方案多位数加法。首先,我认为这个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
,我会告诉你如何使用我的代码。
$ bf sum.bf
199+997=
1196
$
只有+ ve号码可以在我的代码中添加。并且请确保您在两个输入中使用相同的数字位数。即如果你想用3加57,那么给出57 + 03 =或03 + 57 =等输入。 现在的代码。我用一个例子记录下来。尽管如此,我还是不想看看自己的代码,因为自己设计比学习或解决bf中的代码更容易。首先你需要知道如何比较两个数字。我在this问题的答案是一个解决方案。 在文档中,我使用'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
我知道它有点冗长的代码,可能是有可能更好的解决方案。 但它仍值得一试。
这是我知道的
, ;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
你应该在10使用数字和结果在10
我也做了一个程序,只能处理单个数字输入和解答:如果你想要得到的结果与一个以上的数字
#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
.
,你有你的brainfuck-使用十进制转换器码。其余的是正常的计算:
, writing input char in cell(0)
>, writing input char in cell(1)
>>++++++++[<++++++>-] writing 48/0x30/ASCII('0') in cell(2)
< go to cell(2)
[-<-<->>] subtract cell(2) from cell(1) and cell(2)
< go to cell(1)
[<+>-] calculating the sum
cell(0) = cell(0) plus cell(1); cell(1) = 0
< go to cell(0)
>[-]++++++++[>[-]<[->+<]>-]<<<<<<<<< here begins the converter code
this first line sets all cells between cell(1) and cell(9) to 0 (including!)
[->+<]> moving cell(0) to cell(1)
[>+<-<+>]> moving cell(1) to cell(0) and cell(2)
this ends up in copying cell(0) to cell(2)
[ here begins a complex loop
it doesn't end at the same cell as it starts
>>>>>
[->+<]
>
+
<<<<<
++++++++++
<
[
-
>>
+
<
-
[>>>]
>
[
[<+>-]
>
+
>>
]
<<<<<
]
>
[-]
>
[-<<+>>]
>
[-<<+>>]
<<
]
>>>>>
[ this is the output loop
it also doesn't end at the same cell as it starts
<<<<
+++++++
[-<+++++++>]
<
-
[<+>-]
<
.
[-]
>>>>>>
[-<+>]
<
-
]
<<<<<<< this probably sets the pointer to 0
but I don't know if there are still values in some cells
可悲的是我不得不说,我不明白整个转换器代码,虽然我真的尝试过。我只是从德国维基百科网站复制它。也许有人可以解释它?:)
非常感谢。这对我现在确实有意义。 – Yahoo
我经历了一遍又一遍的代码后,我意识到它只能用于单个值。right!我的意思是,如果结果是单值例如3 + 2 = 5就没问题,但如果我们有类似4 + 6 = 10的东西,那么情况会怎样呢? – Yahoo
@Yahoo:是的,那是真的。这仅仅是一个例子。但是你总是可以扩展它以处理大于9的值。另外,请记住,打印一个多位数字本身需要一些思考,因为你需要实现'n' putchars,其中'n'是位数总和。 – dirkgently