I have a homework assignment where part of the menu has to have "R\C" printed, but when I run the program the console just prints "RC". Does anyone know why is this happening and how I can fix it?
This is what I have in Visual Studio:
cout << "R\C" << endl;
The \C is being interpreted as an (invalid) escape sequence. You need to escape the \ character as \\ in order to print it as a single \, eg:
cout << "R\\C" << endl;
Alternatively, in C++11 and later, you can use a raw string literal instead, so you do not need to escape the \ character:
cout << R"(R\C)" << endl;
Escape \ with another \:
cout << "R\\C" << endl;
c++ reserve some characters, so you can't directly input them, usually you will have to put \ in front of them to signify that you want to use "\" as a string.
You have to use escape sequences for certain characters. For the character that you specified you would have to output as “\\” and your output would be \. Other escape sequences are:
\’
\t For Tab
\n For newline
\? For question marks
See this for more information.
You can use escape sequences.., like \t, \n, \a...
If you want to print ' \ ', you have to code like this
cout<<"\\";
Related
Take a look at the following example:
cout << "option 1:
\n option 2:
\n option 3";
I know,it's not the best way to output a string,but the question is why does this cause an error saying that a " character is missing?There is a single string that must go to stdout but it just consists of a lot of whitespace charcters.
What about this:
string x="
string_test";
One may interpret that string as: "\nxxxxxxxxxxxxstring_test" where x is a whitespace character.
Is it a convention?
That's called multiline string literal.
You need to escape the embedded newline. Otherwise, it will not compile:
std::cout << "Hello world \
and stackoverflow";
Note: Backslashes must be immediately before the line ends as they need to escape the newline in the source.
Also you can use the fun fact "Adjacent string literals are concatenated by the compiler" for your advantage by this:
std::cout << "Hello World"
"Stack overflow";
See this for raw string literals. In C++11, we have raw string literals. They are kind of like here-text.
Syntax:
prefix(optional) R"delimiter( raw_characters )delimiter"
It allows any character sequence, except that it must not contain the
closing sequence )delimiter". It is used to avoid escaping of any
character. Anything between the delimiters becomes part of the string.
const char* s1 = R"foo(
Hello
World
)foo";
Example taken from cppreference.
I have a QString with following content:
"MXTP24\\x00\\x00\\xF4\\xF9\\x80\r\n"
I want it to become:
"MXTP24\x00\x00\xF4\xF9\x80\r\n"
I need to replace the "\x" to "\x" so that I can start parsing the values. But the following code, which I think should do the job is not doing anything as I get the same string before and after:
qDebug() << "BEFORE: " << data;
data = data.replace("\\\\x", "\\x", Qt::CaseSensitivity::CaseInsensitive);
qDebug() << "AFTER: " << data;
Here, no change!
Then I tried like this:
data = data.replace("\\x", "\x", Qt::CaseSensitivity::CaseInsensitive);
Then compiler complaines that \x used with no following hex digits!
any ideas?
First let's look at what this piece of code does:
data.replace("\\\\x", "\\x", ....
First string becomes \\x in compiled code, and is used as regular expression. In reqular expression, backslash is special, and needs to be escaped with another backslash to mean actual single backslash character, and your regexp does just this. 4 backslashes in C+n string literal regexp means matching single literal backslash in target text. So your reqular expression matches literal 2-character string \x.
Then you replace it. Replacement isn't a reqular expression, so backslash doesn't need double escaping here, so you end up using literal 2-char replacement string \x, which is same as what you matched, so even if there is a match, nothing changes.
However, this is not your problem, your problem is how qDebug() prints strings. It prints them escaped. That \" at start of output means just plain double quote, 1 char, in the actual string because double quote is escaped. And those \\ also are single backslash char, because literal backslash is also escaped (because it is the escape char and has special meaning for the next char).
So it seems you don't need to do any search replace at all, just remove it.
Try printing the QString in one of these ways to get is shown literally:
std::cout << data << std::endl;
qDebug() << data.toLatin1().constData();
how can print escape characters without further processing and as \t or \n or ... in std::cout?
I dont want to process text manually before sending it to output?
Is there any switch to std::cout for this purpose?
Basically a raw string literal is a string in which the escape characters (like \n \t or \" ) of C++ are not processed. A raw string literal starts with R"( and ends in )", let's see it in an example the difference between a normal string and a raw string in C++:
string raw_str=R"(First line.\nSecond line.\nEnd of message.\n)";
cout<<raw_str<<endl;
result:
~$ ./a.out
First line.\nSecond line.\nEnd of message.\n
If you add one extra slash there as \\t you can see \t in the output of std::cout
For example: cout<<"\\t hello" will print \t hello.
I hope this helps
For a function I am making, I take a string in as a parameter and do things with it. However I treat characters in the string specially if there is a backslash before it. However I am having problems even seeing the blackslash!
std::string s = "01234\6";
std::cout << s << std::endl;
std::cout << s.at(5) << std::endl;
if(s.at(5)== '\\')
std::cout << "It's a backslash" << std::endl;
else
std::cout << "It's not a backslash" << std::endl;
outputs
01234
It's not a backslash
How am I supposed to check if mystring.at(i) == '\\' if it isn't showing up at all?
The input will be coming from another file (which I can't modify) like
myfunc("% \% %");
If I read the string I count 3 '%' characters (so its not ignored by the backslash), and 0 '\' characters
edit: Code how I count
char percent = '%';
int current_index = 0;
int percent_count = 0;
int ret = str.find(percent, current_index);
while(ret != std::string::npos)
{
percent_count++;
current_index = ret +1;
ret = str.find(percent, current_index);
}
return percent_count;
C++ supports three kinds of escape sequences:
simple-escape-sequence. It is one of:
\’ \" \? \\
\a \b \f \n \r \t \v
octal-escape-sequence. It is one of:
\ octal-digit
\ octal-digit octal-digit
\ octal-digit octal-digit octal-digit
\0 is the most well known octal escape sequence that represents the null character.
hexadecimal-escape-sequence. It is one of:
\x hexadecimal-digit
hexadecimal-escape-sequence hexadecimal-digit
When you use:
std::string s = "01234\6";
the \6 part represents an octal escape sequence. It does not represent two characters.
It is the same as
std::string s = "01234?";
where ? is the character represented by the octal number 6.
In order to have \ as an element of the string, you'll need to use:
std::string s = "01234\\6";
The checking method is right, but \ escape 6, so \6 is counted once, you can check sizeof("12345\6"), which 7, or strlen("12345\6"), which is 6.
Change "12345\6" to "12345\\6".
The C++ compiler would have already treated it specially if you have backslash in the string:
std::string s = "01234\6"; //\6 is treated differently already, as unicode character \6, not as backslash + 6
Unless what you mean is you want to have a text with backslash (say, from I/O). In that case, you should put \\ to make your compiler understand that you mean it as real backslash not a unicode character:
std::string s = "01234\\6"; //double backslash here
Then you can test your program.
No compiler C++ will interpret \ as a backslash, since its the escape character. You will have to use \\ to denote a backslash in a string.
Aim: to read a string in the form First\nSecond from a file and to print it as
First
Second
Problem: if the string is defined in the code, as in line = "First\nSecond";, then it is printed on two lines; if instead I read it from a file then is printed as
First\nSecond
Short program illustrating the problem:
#include "stdafx.h" // I'm using Visual Studio 2008
#include <fstream>
#include <string>
#include <iostream>
void main() {
std::ifstream ParameterFile( "parameters.par" ) ;
std::string line ;
getline (ParameterFile, line) ;
std::cout << line << std::endl ;
line = "First\nSecond";
std::cout << line << std::endl ;
return;
}
The parameters.par file contains only the line
First\nSecond
The Win32 console output is
C:\blabla>SOtest.exe
First\nSecond
First
Second
Any suggestion?
In C/C++ string literals ("...") the backslash is used to mark so called "escape sequences" for special characters. The compiler translates (replaces) the two characters '\' (ASCII code 92) followed by 'n' (ASCII code 110) by the new-line character (ASCII code 10). In a text file one would normally just hit the [RETURN] key to insert a newline character. If you really need to process input containing the two characters '\' and 'n' and want to handle them like a C/C++ compiler then you must explicitely replace them by the newline character:
replace(line, "\\n", "\n");
where you have to supply a replace function like this:
Replace part of a string with another string (Standard C++ does not supply such a replace function by itself.)
Other escape sequences supported by C/C++ and similar compilers:
\t -> [TAB]
\" -> " (to distinguish from a plain ", which marks the end of a string literal, but is not part of the string itself!)
\\ -> \ (to allow having a backslash in a string literal; a single backslash starts an escape sequence)
The character indicated in a string literal by the escape sequence \n is not the same as the sequence of characters that looks like \n!
When you think you're assigning First\nSecond, you're not. In your source code, \n in a string literal is a "shortcut" for the invisible newline character. The string does not contain \n - it contains the newline character. It's automatically converted for you.
Whereas what you're reading from your file is the actual characters \ and n.