I'm writing strings which contain backslashes (\) to a file:
x1 = "\\str"
x2 = "\\\str"
# Error: '\s' is an unrecognized escape in character string starting "\\\s"
x2="\\\\str"
write(file = 'test', c(x1, x2))
When I open the file named test, I see this:
\str
\\str
If I want to get a string containing 5 backslashes, should I write 10 backslashes, like this?
x = "\\\\\\\\\\str"
[...] If I want to get a string containing 5 \ ,should i write 10 \ [...]
Yes, you should. To write a single \ in a string, you write it as "\\".
This is because the \ is a special character, reserved to escape the character that follows it. (Perhaps you recognize \n as newline.) It's also useful if you want to write a string containing a single ". You write it as "\"".
The reason why \\\str is invalid, is because it's interpreted as \\ (which corresponds to a single \) followed by \s, which is not valid, since "escaped s" has no meaning.
Have a read of this section about character vectors.
In essence, it says that when you enter character string literals you enclose them in a pair of quotes (" or '). Inside those quotes, you can create special characters using \ as an escape character.
For example, \n denotes new line or \" can be used to enter a " without R thinking it's the end of the string. Since \ is an escape character, you need a way to enter an actual . This is done by using \\. Escaping the escape!
Note that the doubling of backslashes is because you are entering the string at the command line and the string is first parsed by the R parser. You can enter strings in different ways, some of which don't need the doubling. For example:
> tmp <- scan(what='')
1: \\\\\str
2:
Read 1 item
> print(tmp)
[1] "\\\\\\\\\\str"
> cat(tmp, '\n')
\\\\\str
>
Related
I am trying to send two commands at once with sprintf. Commands should be separated with 0x0A (LF). I thought I could enter special characters using two slashes, so I am writing:
sprintf(tmpstr,"VSET1:%ld.%3.3d\\x0AVSET2:%ld.%3.3d",mv/1000, AbsVal((int)mv%1000), mv / 1000, AbsVal((int)mv % 1000));
and it seems only the second command (VSET2) is recognized.
What am I doing wrong?
Use \n in the format string. Also, use a single backslash not \\.
If you are writing your buffer to a file, open the file in binary mode.
Whether you use \n or \x0A, you have to open the file in binary mode to avoid non-portable translations.
See Escape sequences.
When you use \\x0A in a string literal, the first backslash escapes the second backslash. As a result, the string contains a backslash character, '\\', followed by characters 'x', '0', and 'A'.
To use the character represented by 0x0A, you need to use \x0A.
You should be using a single backslash instead of two backslashes. Try the statement given below:
sprintf(tmpstr,"VSET1:%ld.%3.3d\x0AVSET2:%ld.%3.3d",mv/1000, AbsVal((int)mv%1000), mv / 1000, AbsVal((int)mv % 1000));
However, what you have done in your program will print a string "\x0A", rather than an ASCII character (0xAA (Line Feed)).
In C, all escape sequences consist of two or more characters, the first of which is the backslash, \ (called the "Escape character"); the remaining characters determine the interpretation of the escape sequence.
C deal with backslashes as escape sequences by default. However, in your program, you have told C compiler to not use your backslash as an escape sequence by adding an extra backslash to your string.
This works perfect. You not only get to insert \n but looks correct in the code. No need for a \ at the end of lines either. I use this for big paragraphs. Personal data has been obfuscated.
enter code here
wchar_t msg[200];
swprintf(msg, L"XYZ%d: ABCD Limit set to %d%%. %d times it has abcd and xyz rstu %d%%\n"
"Do you want to fix it?\n"
"An Yes will fix it\n"
"No will ignore it and continue\n"
"Cancel will abort the run\n", xxx, yyy, zzz, aaa);
I want to replace all _ and % in a string with \_ and \% respectively.
I tried
String.replace("_foo%_bar", ~r/_|%/, "\\\\0")
But this just produces "\\0foo\\0\\0bar".
How to properly escape the first backslash not to affect the back reference syntax?
You need to use
String.replace("_foo%_bar", ~r/_|%/, "\\\\\\0")
Here, "\\\\" defines 2 literal \ chars that are parsed as a single literal \ char in the replacement, and "\\0" is parsed as a \0, the backreference to the whole match value.
You may also use
String.replace("_foo%_bar", ~r/_|%/, ~S(\\\0))
to avoid overescaping, as ~S sigil does not allow escape sequences and backslashes have literal meanings inside them.
You need one more pair of backslashes:
iex(1)> IO.puts String.replace("_foo%_bar", ~r/_|%/, "\\\\\\0")
\_foo\%\_bar
But I'd suggest using Regex.replace/3 with a function as a callback here:
iex(2)> IO.puts Regex.replace(~r/_|%/, "_foo%_bar", &("\\" <> &1))
\_foo\%\_bar
I have a line in my file which contains the following string
$print = "SM_sdo_debugss_cxct6_CSCTM_4 \csctm_gen[4]_ctm_i_nctm_I_csctm (4+5)";
$my_meta = '\csctm_gen[4]_ctm_i_nctm_I_csctm';
print "I got this\n" if($print =~ /\Q$my_meta\E/);
But it's not able to find the $my_meta string in $print. Why?
Your first string is in double quotes, so backslash escape sequences are processed.
\cs stands for Ctrl-S, which can also be written chr(19) or "\x13".
Your second string is in single quotes, which ignores backslash escapes (apart from \\ and \').
So your regex ends up looking for a 3-character sequence \ c s, but your target string contains a single byte 0x13.
To fix this, either write "... \\cs ..." in your first string (the first backslash escapes the second one), or use single quotes for your first string ('... \cs ...').
This question already has answers here:
using \ in a string as literal instead of an escape
(2 answers)
Closed 7 years ago.
I want the following printout to appear in the my screen when I run my code:
\begin{tabular}
\hline \\
For that, I am using the following command on my code:
std::cout<<"\begin{tabular}<< std::endl;
std::cout<<"\hline \\"<< std::endl;
And I'm getting the following compiler message (regarding the second line of code):
unknown escape sequence: '\h'
and the following incomplete printout:
egin{tabular}
hline\
Where in the first one the "\b" is missing and the first and last \ are missing for the second sentence.
The question is: does anyone know how I can print the \ symbol as text, such that it will get printed and not be interpreted as a command, etc?
The backslash forms escape sequences in C++. You have two options:
Double all the backslashes, a la "\\hline \\\\" (the backslash will escape itself, just as in TeX).
Use C++11 raw strings, which look like R"(\hline \\)" (the text is enclosed by the parens inside the quotes).
Just escape it wiht '\'. So if you want to print out '\' character you must do:
cout<<'\\'<<endl;
Use double backslash (\) if you would like to print \ as a character in your output, otherwise, single \ followed by some character has inherent meaning of some special character, e.g. \n for newline, \r for carriage return \t for tab etc
Double all the backslashes.
e.g.
std::cout<<"\\hline \\\\"<< std::endl;
backslash is an escape code in C++.
As in, below, the "\"escape sequence and n is escape code. Which means a newline character.
"hello\n"
so if you want to print \ as well, you need to escape it too.
"hello\\hi"
I want to create a string that contains all possible special chars.
However, the compiler gives me the warning "Unrecognized escape sequence" in this line:
wstring s=L".,;*:-_⁊#‘⁂‡…–«»¤¤¡=„+-¶~´:№\¯/?‽!¡-¢–”¥—†¿»¤{}«[-]()·^°$§%&«|⸗<´>²³£€™℗#©®~µ´`'" + wstring(1,34);
Can anybody please tell me which one of the characters I may not add to this string the way I did?
You have to escape \ as \\, otherwise \¯ will be interpreted as an (invalid) escape sequence:
wstring s=L".,;*:-_⁊#‘⁂‡…–«»¤¤¡=„+-¶~´:№\\¯/?‽!¡-¢–”¥—†¿»¤{}«[-]()·^°$§%&«|⸗<´>²³£€™℗#©®~µ´`'" + wstring(1,34);
Escape sequence is a character string that has a different meaning than the literal characters themselves. In C and C++ the sequence begins with \ so if your string contains a double quote or backslash it must be escaped properly using \" and \\
In long copy-pasted strings it may be difficult to spot those characters and it's also less maintainable in the future so it's recommended to use raw string literals with the prefix R so you don't need any escapes at all
wstring s = LR"(.,;*:-_⁊#‘⁂‡…–«»¤¤¡=„+-¶~´:№\¯/?‽!¡-¢–”¥—†¿»¤{}«[-]()·^°$§%&«|⸗<´>²³£€™℗#©®~µ´`')"
+ wstring(1,34);
A special delimiter string may be inserted outside the parentheses like this LR"delim(special string)delim" in case your raw string contains a )" sequence