The word ." prints a string. More precisely it compiles the (.") and the string up to the next " in the currently compiled word.
But how can I print
That's the "question".
with Forth?
In a Forth-2012 System (e.g. Gforth) you can use string literals with escaping via the word s\" as:
: foo ( -- ) s\" That's the \"question\"." type ;
In a Forth-94 system (majority of standard systems) you can use arbitrary parsing and the word sliteral as:
: foo ( -- ) [ char | parse That's the "question".| ] sliteral type ;
A string can be also extracted up to the end of the line (without printable delimiter); a multi-line string can be extracted too.
Specific helpers for particular cases can be easily defined.
For example, see the word s$ for string literals that are delimited by any arbitrary printable character, e.g.:
s$ `"test" 'passed'` type
Old school:
34 emit
Output:
"
Using gforth:
: d 34 emit ;
cr ." That's the " d ." question" d ." ." cr
Output:
That's the "question".
Related
I am writing a simple scanner in flex. I want my scanner to print out "integer type seen" when it sees the keyword "int". Is there any difference between the following two ways?
1st way:
%%
int printf("integer type seen");
%%
2nd way:
%%
"int" printf("integer type seen");
%%
So, is there a difference between writing if or "if"? Also, for example when we see a == operator, we print something. Is there a difference between writing == or "==" in the flex file?
There's no difference in these specific cases -- the quotes(") just tell lex to NOT interpret any special characters (eg, for regular expressions) in the quoted string, but if there are no special characters involved, they don't matter:
[a-z] printf("matched a single letter\n");
"[a-z]" printf("matched the 5-character string '[a-z]'\n");
0* printf("matched zero or more zero characters\n");
"0*" printf("matched a zero followed by an asterisk\n");
Characters that are special and mean something different outside of quotes include . * + ? | ^ $ < > [ ] ( ) { } /. Some of those only have special meaning if they appear at certain places, but its generally clearer to quote them regardless of where they appear if you want to match the literal characters.
I am attempting to use this string:
var passwordRegex = "^[A-Za-z0-9 !\"#$%&'()*+,-./:;<=>?#[\\]^_`{|}~].{8,}$"
As my regular expression pattern, but it keeps failing saying the pattern is invalid. I used the \ character to escape for the characters: " and \, but it throws the error: invalid escape sequence in literal for regex key characters like ^ & [ ] | . etc.
What am I missing in order to allow the characters:
! " # $ % & ' ( ) * + , - . / : ; < = > ? # [ \ ] ^ _ ` { | } ~
(including space) in my regex? I assume it's something with how I am escaping, but I can't find anything anywhere for these characters in regards to SWIFT's regular expression.
The problem is that characters [, \, and ] need to be escaped because they have special meaning in a regular expression.
So you need \[, \\, and \] in the regular expression. But since this is inside a Swift string, each \ needs to be escaped with a \.
So [\] becomes \[\\\] in the regular expression which becomes \\[\\\\\\] in the Swift string.
The final valid string is:
var passwordRegex = "^[A-Za-z0-9 !\"#$%&'()*+,-./:;<=>?#\\[\\\\\\]^_`{|}~].{8,}$"
I have following type of arrays
["def, \" \""]
["abc,\" \",def"]
How can i check if there's a \ in value and convert them correctly as follows?
["def", " "]
["abc", " ", "def"]
I have been trying to correct them using
join(',')
delete('\"') split(', ')
but no luck
When you write:
input = "{ \"foo\": \"bar\", \"num\": 3}"
The actual string stored in input is:
{ "foo": "bar", "num": 3}
The escape \" here is interpreted by Ruby parser, so that it can distinguish between the boundary of a string (the left most and the right most "), and a normal character " in a string (the escaped ones).
String#delete deletes a character set specified the first parameter, rather than a pattern. All characters that is in the first parameter will be removed. So by writing
input.delete('\\"')
You got a string with all \ and " removed from input, rather than a string with all \" sequence removed from input. This is wrong for your case. It may cause unexpected behavior some time later.
This is probably a rather trivial question for the Erlang experts - I'm trying to have my ejabberd server store offline messages (in a Riak db) which inherently do contain double quotes (") around various values, etc. I get a format error when I try to create a Riak database object from them, and testing of replacing the double quotes with an escape character (\") corrects the issue. The question is how can I do this replacement manually?
I tried the following code but somehow doesn't work.
(ejabberd#xxx-xx-xx-xxx)4> re:replace(""hello"", """, "\"", [{return, list}, global]).
* 1: syntax error before: hello
So essentially I'm trying to replace the embedded " around the hello word with \".
I don't know Erlang, but you probably need something like this:
"\"hello\"", "\"", "\\\""
You must escape both " and \ in replacement string.
The Erlang literal syntax for strings uses the "\" (backslash)
character as an escape code. You need to escape backslashes in literal
strings, both in your code and in the shell, with an additional
backslash, i.e.: "\".
Example:
Let's make an example. I use $ Erlang symbol which will be substituted with ascii integer of a character to show what is happening behind each string which basically is a list of integer.
Subject = [$"] ++ "hello" ++ [$"] = "\"hello\"".
Target = [$"] = "\"".
Replacement = [$\\, $\\, $"] = "\\\\\"".
Result = re:replace(Subject, Target, Replacement, [{return, list}, global]).
Now with getting the length of Subject and Result we can find the difference:
7 = length(Subject). %% => 7 characters: " h e l l o "
9 = length(Result). %% => 9 characters: \ " h e l l o \ "
English isn't my mother tongue,so it's a little hard to describe the question.
I wanna to get 'd=40' in str by lua string.gsub(),but there's some problem.
------code below---
local str =
[==[
-- a=10
- -b=20
--c=30
d=40
]==]
local pat1 = [=[%s[%s]]=]
local pat2 = [=[\n[%s]]=]
str:gsub(pat1, function(s) print("pat1>>" .. s) end) --pat1>>d=40
str:gsub(pat2, function(s) print("pat2<<" .. s) end) --not match
local re1,_ = str:gsub("\n","$")
local re2,_ = str:gsub("%s","$")
print(re1) --a=10$- -b=20$ --c=30$d=40$
print(re2) --$a=10$-$-b=20$$ --c=30$d=40$
As Lua 5.1 Reference Manual Say
%s: represents all space characters.
I Think it equal to '\n',' 'and'\t'.
Question : Why pat2 can't match?
But I think pat2 is right,there's a '\n'befor'd=40' ,
so I think It can match ,but it can't work,why?
When you use [[]] notation for strings, that's a special string literal that takes the string exactly as you provide it. No character escaping is done. You can put some number of = characters in the brackets, to make it a bit easier to let you use [ characters in the string.
The string literal "\n" is one character, representing the newline. That's because of the use of the escape character \. The escape character applied to the 'n' character means "the newline character."
The string literal [[\n]] is exactly what it says: the character '\' followed by the character 'n'. Because no escaping is done, \n is not treated specially. It's exactly what it looks like.
Therefore, when you say local pat2 = [=[\n[%s]]=] You're saying "the first character should be '\' followed by 'n' followed by a space. That's not what you want; you want the escaping to work. So you should use a regular string literal: local pat2 = "\n[%s]".