[英]Racket Binary Addition
I have a question concerning Racket. 我有一个关于球拍的问题。 Given 2 string binary numbers, I would like to obtain their sum in a string binary number. 给定2个字符串二进制数,我想以字符串二进制数获取它们的总和。 The function should take into account carry bits and a possible overflow. 该功能应考虑进位和可能的溢出。 So the function should look like this: 因此该函数应如下所示:
(define (binary-sum a b) ...)
For example: 例如:
(binary-sum "010" "001")
"011"
Or: 要么:
(binary-sum "0010" "1011")
"1101"
Or in case of overflow: 或在发生溢出的情况下:
(binary-sum "1111" "0001")
"overflow"
Please identify any packages I need. 请确定我需要的任何包裹。
Thank you! 谢谢!
Binary values in Racket are specified using the prefix #b
. 使用前缀#b
指定Racket中的二进制值。 Using that, we can create the following function: 使用它,我们可以创建以下函数:
> (define (binary-sum a b)
(+ (string->number (string-append "#b" a))
(string->number (string-append "#b" b))))
> (binary-sum "010" "001")
3
To format the output as binary as well, we can use ~r
, like so: 要将输出格式也设置为二进制,我们可以使用~r
,如下所示:
> (define (binary-sum a b)
(~r (+ (string->number (string-append "#b" a))
(string->number (string-append "#b" b))) #:base 2))
> (binary-sum "010" "001")
011
And finally, to add your overflow feature, we can simply compare the length of our result as a string to the length of one of the parameters. 最后,要添加您的溢出功能,我们可以简单地将结果的字符串长度与参数之一的长度进行比较。
(define (binary-sum a b)
(let ((sum (~r (+ (string->number (string-append "#b" a))
(string->number (string-append "#b" b)))
#:base 2)))
(if (> (string-length sum)
(string-length a))
"overflow"
sum)))
Which behaves exactly as your question specifies. 它的行为与您的问题所指定的完全相同。 Let me know if you have any remaining doubts or confusion. 如果您还有其他疑问或困惑,请告诉我。 Good luck! 祝好运!
Edit: Just wanted to note that if you ever need to merely add two binary values and don't mind getting an integer back, you can always just do it like so: 编辑:只是要注意,如果您只需要添加两个二进制值并且不介意取回整数,则总是可以这样进行:
> (+ #b010 #b001)
3
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