Is there an easy helper function in common lisp to convert from hex to decimal?
(parse-integer "ff" :radix 16)
Never mind #xff = 255.....
Related
I am learning dart programming language for Flutter. In the integer class what does the word radix means ? Please explain me this. Thanks
Sometimes we have to work with string in radix number format. Dart int parse() method also supports convert string into a number with radix in the range 2..36:
For example, we convert a Hex string into int:
var n_16 = int.parse('FF', radix: 16);
The output of the code = 255
Using radix function we can also convert Binary numbers into decimal numbers like this
var decimal = int.parse('1001001', radix:2)'
Pretty straightforward, but I can't seem to find an answer. I have a string of 1s and 0s such as "01001010" - how would I parse that into a number?
Use string-to-number, which optionally accepts the base:
(string-to-number "01001010" 2)
;; 74
As explained by #sds in a comment, string-to-number returns 0 if the conversion fails. This is unfortunate, since a return value of 0 could also means that the parsing succeeded.
I'd rather use the Common Lisp version of this function, cl-parse-integer. The standard function is described in the Hyperspec, whereas the one in Emacs Lisp is slightly different (in particular, there is no secondary return value):
(cl-parse-integer STRING &key START END RADIX JUNK-ALLOWED)
Parse integer from the substring of STRING from START to END. STRING
may be surrounded by whitespace chars (chars with syntax ‘ ’). Other
non-digit chars are considered junk. RADIX is an integer between 2 and
36, the default is 10. Signal an error if the substring between START
and END cannot be parsed as an integer unless JUNK-ALLOWED is non-nil.
(cl-parse-integer "001010" :radix 2)
=> 10
(cl-parse-integer "0" :radix 2)
=> 0
;; exception on parse error
(cl-parse-integer "no" :radix 2)
=> Debugger: (error "Not an integer string: ‘no’")
;; no exception, but nil in case of errors
(cl-parse-integer "no" :radix 2 :junk-allowed t)
=> nil
;; no exception, parse as much as possible
(cl-parse-integer "010no" :radix 2 :junk-allowed t)
=> 2
This thread has an elisp tag. Because it also has a lisp tag, I would like to show standard Common Lisp versions of two solutions. I checked these on LispWorks only. If my solutions are not standard Common Lisp, maybe someone will correct and improve my solutions.
For solutions
(string-to-number "01001010" 2)
and
(cl-parse-integer "001010" :radix 2)
LispWorks does not have string-to-number and does not have cl-parse-integer.
In LispWorks, you can use:
(parse-integer "01001010" :radix 2)
For the solution
(read (concat "#2r" STRING))
LispWorks does not have concat. You can use concatenate instead. read won't work on strings in LispWorks. You have to give read a stream.
In LispWorks, you can do this:
(read (make-string-input-stream (concatenate 'string "#2r" "01001010")))
You can also use format like this:
(read (make-string-input-stream (format nil "#2r~a" "01001010")))
This seems hacky by comparison, but FWIW you could also do this:
(read (concat "#2r" STRING))
i.e. read a single expression from STRING as a binary number.
This method will signal an error if the expression isn't valid.
Hi I am trying to convert an octal number to decimal in swift. What would be the easiest way to do this?
From Octal to Decimal
There is a specific Int initializer for this
let octal = 10
if let decimal = Int(String(octal), radix: 8) {
print(decimal) // 8
}
From Decimal to Octal
let decimal = 8
if let octal = Int(String(decimal, radix: 8)) {
print(octal) // 10
}
Note 1: Please pay attention: parenthesis are different in the 2 code snippets.
Note 2: Int initializer can fail for string representations of number with more exotic radixes. Please read the comment by #AMomchilov below.
you can convert from octal to decimal easily. Swift supports octal syntax natively. You have to write "0o" before the octal number.
let number = 0o10
print(number) // it prints the number 8 in decimal
Integer Literals
Integer literals represent integer values of unspecified precision. By
default, integer literals are expressed in decimal; you can specify an
alternate base using a prefix. Binary literals begin with 0b, octal
literals begin with 0o, and hexadecimal literals begin with 0x.
Here is the documentation's reference.
I hope it helps you
In visual lisp, you can use (atoi "123") to convert "123" to 123. It seems there is no "atoi" like function in clisp ?
any suggestion is appreciated !
Now i want to convert '(1 2 3 20 30) to "1 2 3 20 30", then what's the best way to do it ?
parse-interger can convert string to integer, and how to convert integer to string ? Do i need to use format function ?
(map 'list #'(lambda (x) (format nil "~D" x)) '(1 2 3)) => ("1" "2" "3")
But i donot know how to cnovert it to "1 2 3" as haskell does:
concat $ intersperse " " ["1","2","3","4","5"] => "1 2 3 4 5"
Sincerely!
In Common Lisp, you can use the read-from-string function for this purpose:
> (read-from-string "123")
123 ;
3
As you can see, the primary return value is the object read, which in this case happens to be an integer. The second value—the position—is harder to explain, but here it indicates the next would-be character in the string that would need to be read next on a subsequent call to a reading function consuming the same input.
Note that read-from-string is obviously not tailored just for reading integers. For that, you can turn to the parse-integer function. Its interface is similar to read-from-string:
> (parse-integer "123")
123 ;
3
Given that you were asking for an analogue to atoi, the parse-integer function is the more appropriate choice.
Addressing the second part of your question, post-editing, you can interleave (or "intersperse") a string with the format function. This example hard-codes a single space character as the separating string, using the format iteration control directives ~{ (start), ~} (end), and ~^ (terminate if remaining input is empty):
> (format nil "Interleaved: ~{~S~^ ~}." '(1 2 3))
"Interleaved: 1 2 3."
Loosely translated, the format string says,
For each item in the input list (~{), print the item by its normal conversion (~S). If no items remain, stop the iteration (~^). Otherwise, print a space, and then repeat the process with the next item (~}).
If you want to avoid hard-coding the single space there, and accept the separator string as a separately-supplied value, there are a few ways to do that. It's not clear whether you require that much flexibility here.
Does anyone have a code snippet or a class that will take a long long and turn it into a 16 byte Hex string?
I'm looking to turn data like this
long long decimalRepresentation = 1719886131591410351;
and turn it into this
//Base 16 Hex Output: 17DE435307A07300
The %x operator doesn't want to work for me
NSLog(#"Hex: %x",decimalRepresentation);
//console : "Hex: 7a072af"
As you can see that's not even close. Any help is truly appreciated!
%x prints an unsigned integer in hexadecimal representation and sizeof(long long) != sizeof(unsigned). See e.g. "Data Type Size and Alignment" in the 64bit transitioning guide.
Use the ll specifier (thats two lower-case L) to get the desired output:
NSLog(#"%llx", myLongLong);