I came across this code snippet in C++17 draft n4713:
#define R "x"
const char* s = R"y"; // ill-formed raw string, not "x" "y"
What is a "raw string"? What does it do?
Raw string literals are string literals that are designed to make it easier to include nested characters like quotation marks and backslashes that normally have meanings as delimiters and escape sequence starts. They’re useful for, say, encoding text like HTML. For example, contrast
"C:\\Program Files\\"
which is a regular string literal, with
R"(C:\Program Files\)"
which is a raw string literal. Here, the use of parentheses in addition to quotes allows C++ to distinguish a nested quotation mark from the quotation marks delimiting the string itself.
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 which starts with R"( and ends in )" ,introduced in C++11
prefix(optional) R "delimiter( raw_characters )delimiter"
prefix - One of L, u8, u, U
Thanks to #Remy Lebeau,
delimiter is optional and is typically omitted, but there are corner cases where it is actually needed, in particular if the string content contains the character sequence )" in it, eg: R"(...)"...)", so you would need a delimiter to avoid an error, eg: R"x(...)"...)x".
See an example:
#include <iostream>
#include <string>
int main()
{
std::string normal_str = "First line.\nSecond line.\nEnd of message.\n";
std::string raw_str = R"(First line.\nSecond line.\nEnd of message.\n)";
std::string raw_str_delim = R"x("(First line.\nSecond line...)")x";
std::cout << normal_str << std::endl;
std::cout << raw_str << std::endl;
std::cout << raw_str_delim << std::endl;
return 0;
}
output:
First line.
Second line.
End of message.
First line.\nSecond line.\nEnd of message.\n
"(First line.\nSecond line...)"
Live on Godbolt
I will make an addition about a concern in one of the comments:
But here in the code the R is defined as "x" and after
expansion of the #define the code is const char* s = "x""y";
and there isn't any R"(.
The code fragment in the question is to show invalid uses of the Raw Strings. Let me get the actual 3-lines of code here:
#define R "x"
const char* s = R"y"; // ill-formed raw string literal, not "x" "y"
const char* s2 = R"(a)" "b)"; // a raw string literal followed by a normal string literal
The first line is there to not get confused by a macro. macros are preprocessed code fragments that replace parts in the source. Raw String, on the other hand, is a feature of the language that is "parsed" according to language rules.
The second line is to show the wrong use of it. Correct way would be R"(x)" where you need parenthesis in it.
And the last is to show how it can be a pain if not written carefully. The string inside parenthesis CANNOT include closing sequence of raw string. A correction might be R"_(a)" "b)_". _ can be replaced by any character (but not parentheses, backslash and spaces) and any number of them as long as closing sequence is not included inside: R"___(a)" "b)___" or R"anything(a)" "b)anything"
So if we wrap these correction within a simple C++ code:
#include <iostream>
using namespace std;
#define R "x" // This is just a macro, not Raw String nor definition of it
const char* s = R"(y)"; // R is part of language, not a macro
const char* s2 = R"_(a)" "b)_"; // Raw String shall not include closing sequence of characters; )_"
int main(){ cout << s <<endl << s2 <<endl << R <<endl; }
then the output will be
y
a)" "b
x
Raw string literal. Used to avoid escaping of any character. Anything between the delimiters becomes part of the string. prefix, if present, has the same meaning as described above.
C++Reference: string literal
a Raw string is defined like this:
string raw_str=R"(First line.\nSecond line.\nEnd of message.\n)";
and the difference is that a raw string ignores (escapes) all the special characters like \n ant \t and threats them like normal text.
So the above line would be just one line with 3 actual \n in it, instead of 3 separate lines.
You need to remove the define line and add parentheses around your string to be considered as a raw string.
Related
Why does the following work:
string input = "a long string of text pasted from a .txt file";
But this version does not?
string input =
"
some
large
string ";
I thought C++ doesn't care about whitespace.
You can do something like this. It's called a raw string literal:
string input =
R"(
some
large
string )";
This will include the endline characters as well. The format is R"(string-literal)"
For the most parts no, it does not care about whitespace. But there are exceptions and string literals are one of them.
The rule is string literals cannot span multiple lines. But adjacent literals are automatically concatenated so you can just do
const char string[] = "very "
"long "
"string";
and it will be equivalent to
const char string[] = "very long string";
I am not sure about the origin of the rule, I suspect it might have been done to prevent confusion whether the newline should be part of the string or not (it's not unless explicitly escaped). Or maybe just some grammar/parser thing. Compiling C/C++ is kind of complicated and happens in multiple phases, see cppreference - string literals already have plenty of special treatment.
I came across this code snippet in C++17 draft n4713:
#define R "x"
const char* s = R"y"; // ill-formed raw string, not "x" "y"
What is a "raw string"? What does it do?
Raw string literals are string literals that are designed to make it easier to include nested characters like quotation marks and backslashes that normally have meanings as delimiters and escape sequence starts. They’re useful for, say, encoding text like HTML. For example, contrast
"C:\\Program Files\\"
which is a regular string literal, with
R"(C:\Program Files\)"
which is a raw string literal. Here, the use of parentheses in addition to quotes allows C++ to distinguish a nested quotation mark from the quotation marks delimiting the string itself.
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 which starts with R"( and ends in )" ,introduced in C++11
prefix(optional) R "delimiter( raw_characters )delimiter"
prefix - One of L, u8, u, U
Thanks to #Remy Lebeau,
delimiter is optional and is typically omitted, but there are corner cases where it is actually needed, in particular if the string content contains the character sequence )" in it, eg: R"(...)"...)", so you would need a delimiter to avoid an error, eg: R"x(...)"...)x".
See an example:
#include <iostream>
#include <string>
int main()
{
std::string normal_str = "First line.\nSecond line.\nEnd of message.\n";
std::string raw_str = R"(First line.\nSecond line.\nEnd of message.\n)";
std::string raw_str_delim = R"x("(First line.\nSecond line...)")x";
std::cout << normal_str << std::endl;
std::cout << raw_str << std::endl;
std::cout << raw_str_delim << std::endl;
return 0;
}
output:
First line.
Second line.
End of message.
First line.\nSecond line.\nEnd of message.\n
"(First line.\nSecond line...)"
Live on Godbolt
I will make an addition about a concern in one of the comments:
But here in the code the R is defined as "x" and after
expansion of the #define the code is const char* s = "x""y";
and there isn't any R"(.
The code fragment in the question is to show invalid uses of the Raw Strings. Let me get the actual 3-lines of code here:
#define R "x"
const char* s = R"y"; // ill-formed raw string literal, not "x" "y"
const char* s2 = R"(a)" "b)"; // a raw string literal followed by a normal string literal
The first line is there to not get confused by a macro. macros are preprocessed code fragments that replace parts in the source. Raw String, on the other hand, is a feature of the language that is "parsed" according to language rules.
The second line is to show the wrong use of it. Correct way would be R"(x)" where you need parenthesis in it.
And the last is to show how it can be a pain if not written carefully. The string inside parenthesis CANNOT include closing sequence of raw string. A correction might be R"_(a)" "b)_". _ can be replaced by any character (but not parentheses, backslash and spaces) and any number of them as long as closing sequence is not included inside: R"___(a)" "b)___" or R"anything(a)" "b)anything"
So if we wrap these correction within a simple C++ code:
#include <iostream>
using namespace std;
#define R "x" // This is just a macro, not Raw String nor definition of it
const char* s = R"(y)"; // R is part of language, not a macro
const char* s2 = R"_(a)" "b)_"; // Raw String shall not include closing sequence of characters; )_"
int main(){ cout << s <<endl << s2 <<endl << R <<endl; }
then the output will be
y
a)" "b
x
Raw string literal. Used to avoid escaping of any character. Anything between the delimiters becomes part of the string. prefix, if present, has the same meaning as described above.
C++Reference: string literal
a Raw string is defined like this:
string raw_str=R"(First line.\nSecond line.\nEnd of message.\n)";
and the difference is that a raw string ignores (escapes) all the special characters like \n ant \t and threats them like normal text.
So the above line would be just one line with 3 actual \n in it, instead of 3 separate lines.
You need to remove the define line and add parentheses around your string to be considered as a raw string.
For example, in the command line this works (the 1st argument has quotes but the 2nd argument doesn't):
"test.bat" "a" b
i.e it know that "a" is the 1st argument and b is the second
but using system() it doesn't work:
system("test.bat" "a" b)
this also doesn't work:
system("test.bat" \"a\" b)
This is gonna be simplest if we use a raw string literal. A raw string literal is a way of writing a string in c++ where nothing gets escaped. Let's look at an example:
char const* myCommand = R"(test.bat "a" b)";
The R at the beginning indicates that it's a raw string literal, and if you call system(myCommand), it will be exactly equivalent to typing
$ test.bat "a" b
into the command line. Now, suppose you want to escape the quotes on the command line:
$ test.bat \"a\" b
With a raw string literal, this is simple:
char const* myCommand = R"(test.bat \"a\" b)";
system(myCommand);
Or, alternatively:
system(R"(test.bat \"a\" b)");
Hope this helps!
A bit more info on raw string literals: Raw string literals are a great feature, and they basically allow you to copy+paste any text directly into your program. They begin with R, followed by a quote and a parenthesis. Only the stuff inside the parenthesis gets included. Examples:
using std::string;
string a = R"(Hello)"; // a == "Hello"
Begin and end with "raw":
string b = R"raw(Hello)raw"; // b == "Hello"
Begin and end with "foo"
string c = R"foo(Hello)foo"; // c == "Hello"
Begin and end with "x"
string d = R"x(Hello)x"; // d == "Hello"
The important thing is that we begin and end the literal with the same string of letters (called the delimiter), followed by the parenthesis. This ensures we never have a reason to escape something inside the raw string literal, because we can always change the delimiter so that it's not something found inside the string.
I got it to work now:
system(R"(C:\"to erase\test.bat" "a")");
I found the answer: system("test.bat" ""a"" b);
or more precisely: system("\"test.bat\" ""a"" b");
So the answer is to escape the quotes with a double quote
Is there a simple way to escape all occurrences of \ in a string? I start with the following string:
#include <string>
#include <iostream>
std::string escapeSlashes(std::string str) {
// I have no idea what to do here
return str;
}
int main () {
std::string str = "a\b\c\d";
std::cout << escapeSlashes(str) << "\n";
// Desired output:
// a\\b\\c\\d
return 0;
}
Basically, I am looking for the inverse to this question. The problem is that I cannot search for \ in the string, because C++ already treats it as an escape sequence.
NOTE: I am not able to change the string str in the first place. It is parsed from a LaTeX file. Thus, this answers to a similar question does not apply. Edit: The parsing failed due to an unrelated problem, the question here is about string literals.
Edit: There are nice solutions to find and replace known escape sequences, such as this answer. Another option is to use boost::regex("\p{cntrl}"). However, I haven't found one that works for unknown (erroneous) escape sequences.
You can use raw string literal. See http://en.cppreference.com/w/cpp/language/string_literal
#include <string>
#include <iostream>
int main() {
std::string str = R"(a\b\c\d)";
std::cout << str << "\n";
return 0;
}
Output:
a\b\c\d
It is not possible to convert the string literal a\b\c\d to a\\b\\c\\d, i.e. escaping the backslashes.
Why? Because the compiler converts \c and \d directly to c and d, respectively, giving you a warning about Unknown escape sequence \c and Unknown escape sequence \d (\b is fine as it is a valid escape sequence). This happens directly to the string literal before you have any chance to work with it.
To see this, you can compile to assembler
gcc -S main.cpp
and you will find the following line somewhere in your assembler code:
.string "a\bcd"
Thus, your problem is either in your parsing function or you use string literals for experimenting and you should use raw strings R"(a\b\c\d)" instead.
string s="abcdefghijklmnopqrstuvwxyz"
char f[]=" " (s.substr(s.length()-10,9)).c_str() " ";
I want to get the last 9 characters of s and add " " to the beginning and the end of the substring, and store it as a char[]. I don't understand why this doesn't work even though char f[]=" " "a" " " does.
Is (s.substr(s.length()-10,9)).c_str() not a string literal?
No, it's not a string literal. String literals always have the form "<content>" or expand to that (macros, like __FILE__ for example).
Just use another std::string instead of char[].
std::string f = " " + s.substr(s.size()-10, 9) + " ";
First, consider whether you should be using cstrings. In C++, generally, use string.
However, if you want to use cstrings, the concatenation of "abc" "123" -> "abc123" is a preprocessor operation and so cannot be used with string::c_str(). Instead, the easiest way is to construct a new string and take the .c_str() of that:
string s="abcdefghijklmnopqrstuvwxyz"
char f[]= (string(" ") + s.substr(s.length()-10,9) + " ").c_str();
(EDIT: You know what, on second thought, that's a really bad idea. The cstring should be deallocated after the end of this statement, so using f can cause a segfault. Just don't use cstrings unless you're prepared to mess with strcpy and all that ugly stuff. Seriously.)
If you want to use strings instead, consider something like the following:
#include <sstream>
...
string s="abcdefghijklmnopqrstuvwxyz"
stringstream tmp;
tmp << " " << s.substr(s.length()-10,9) << " ";
string f = tmp.str();
#Xeo tells you how to solve your problem. Here's some complimentary background on how string literals are handled in the compilation process.
From section A.12 Preprocessing of The C Programming language:
Escape sequences in character constants and string literals (Pars. A.2.5.2, A.2.6) are
replaced by their equivalents; then adjacent string literals are concatenated.
It's the Preprocessor, not the compiler, who's responsible for the concatenation. (You asked for a C++ answer. I expect that C++ treats string literals the same way as C). The preprocessor has only a limited knowledge of the C/C++ language; the (s.substr(s.length()-10,9)).c_str() part is not evaluated at the preprocessor stage.