How do you say something like this?
static const string message = "This is a message.\n
It continues in the next line"
The problem is, the next line isn't being recognized as part of the string..
How to fix that? Or is the only solution to create an array of strings and then initialize the array to hold each line?
Enclose each line in its own set of quotes:
static const string message = "This is a message.\n"
"It continues in the next line";
The compiler will combine them into a single string.
You can use a trailing slash or quote each line, thus
"This is a message.\n \
It continues in the next line"
or
"This is a message."
"It continues in the next line"
In C++ as in C, string litterals separated by whitespace are implicitly concatenated, so
"foo" "bar"
is equivalent to:
"foobar"
So you want:
static const string message = "This is a message.\n"
"It continues in the next line";
Related
I am working to convert multiline strings into a list of tokens that might be easier for me to work with.
In accordance with the specific needs of my project, I'm padding any carat symbol that appears in my input with spaces, so that "^" gets turned into " ^ ". I'm using something like the following function to do so:
let bad_function string = Str.global_replace (Str.regexp "^") " ^ " (string)
I then use something like the below function to then turn this multiline string into a list of tokens (ignoring whitespace).
let string_to_tokens string = (Str.split (Str.regexp "[ \n\r\x0c\t]+") (string));;
For some reason, bad_function adds carats to places where they shouldn't be. Take the following line of code:
(bad_function " This is some
multiline input
with newline characters
and tabs. When I convert this string
into a list of tokens I get ^s showing up where
they shouldn't. ")
The first line of the string turns into:
^ This is some \n ^
When I feed the output from bad_function into string_to_tokens I get the following list:
string_to_tokens (bad_function " This is some
multiline input
with newline characters
and tabs. When I convert this string
into a list of tokens I get ^s showing up where
they shouldn't. ")
["^"; "This"; "is"; "some"; "^"; "multiline"; "input"; "^"; "with";
"newline"; "characters"; "^"; "and"; "tabs."; "When"; "I"; "convert";
"this"; "string"; "^"; "into"; "a"; "list"; "of"; "tokens"; "I"; "get";
"^s"; "showing"; "up"; "where"; "^"; "they"; "shouldn't."]
Why is this happening, and how can I fix so these functions behave like I want them to?
As explained in the Str module.
^ Matches at beginning of line: either at the beginning of the
matched string, or just after a '\n' character.
So you have to quote the '^' character using the escape character "\".
However, note that (also from the doc)
any backslash character in the regular expression must be doubled to
make it past the OCaml string parser.
This means you have to put a double '\' to do what you want without getting a warning.
This should do the job:
let bad_function string = Str.global_replace (Str.regexp "\\^") " ^ " (string);;
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();
I am trying to read data from a text file and split the read line based on quotes. For example
"Hi how" "are you" "thanks"
Expected output
Hi how
are you
thanks
My code:
getline(infile, line);
ch = strdup(line.c_str());
ch1 = strtok(ch, " ");
while (ch1 != NULL)
{
a3[i] = ch1;
ch1 = strtok(NULL, " ");
i++;
}
I don't know what to specify as delimiter string. I am using strtok() to split, but it failed. Can any one help me?
Please have a look at the example code here. You should provide "\"" as delimiter string to strtok.
For example,
ch1 = strtok (ch,"\"");
Probably your problem is related with representing escape sequences. Please have a look here for a list of escape sequences for characters.
Given your input: "Hi how" "are you" "thanks", if you use strtok with "\"" as the delimiter, it'll treat the spaces between the quoted strings as if they were also strings, so if (for example) you printed out the result strings, one per line, surrounded by square brackets, you'd get:
[Hi how]
[ ]
[are you]
[ ]
[thanks]
I.e., the blank character between each quoted string is, itself, being treated as a string. If the delimiter you supplied to strtok was " \"" (i.e., included both a quote and a space) that wouldn't happen, but then it would also break on the spaces inside the quoted strings.
Assuming you can depend on every item you care about being quoted, you want to skip anything until you get to a quote, ignore the quote, then read data into your input string until you get to another quote, then repeat the whole process.
I'm attempting to initialise a string variable in C++, and the value is so long that it's going to exceed the 80 character per line limit I'm working to, so I'd like to split it to the next line, but I'm not sure how to do that.
I know that when splitting the contents of a stream across multiple lines, the syntax goes like
cout << "This is a string"
<< "This is another string";
Is there an equivalent for variable assignment, or do I have to declare multiple variables and concatenate them?
Edit: I misspoke when I wrote the initial question. When I say 'next line', I'm just meaning the next line of the script. When it is printed upon execution, I would like it to be on the same line.
You can simply break the line like this:
string longText("This is a "
"very very very "
"long text");
In the C family, whitespaces are insignificant, so you can freely use character literals spanning multiple lines this way.
It can also simply be
cout << "This is a string"
"This is another string";
You can write this:
const char * str = "First phrase, "
"Second phrase, "
"Third phrase";