char* a="dsa" "qwe";
printf("%s", a);
output: dsaqwe
My question is why does this thing work. If I give a space or nothing in between two string literals it concatenates the string literals.
How is this working?
It's defined by the ISO C standard, adjacent string literals are combined into a single one.
The language is a little dry (it is a standard after all) but section 6.4.5 String literals of C11 states:
In translation phase 6, the multibyte character sequences specified by any sequence of adjacent character and identically-prefixed wide string literal tokens are concatenated into a single multibyte character sequence.
This is also mentioned in 5.1.1.2 Translation phases, point 6 of the same standard, though a little more succinctly:
Adjacent string literal tokens are concatenated.
This basically means that "abc" "def" is no different to "abcdef".
It's often useful for making long strings while still having nice formatting, something like:
const char *myString = "This is a really long "
"string and I don't want "
"to make my lines in the "
"editor too long, because "
"I'm basically anal retentive :-)";
And to answer your unasked question, "What is this good for?"
For one thing, you can put constants in string literals. You can write
#define FIRST "John"
#define LAST "Doe"
const char* name = FIRST " " LAST;
const char* salutation = "Dear " FIRST ",";
and then if you'll need to change the name later, you'll only have to change it in one spot.
Things like that.
You answered your own question.
If I give a space or nothing in between two string literals it concatenates the string literals.
That's one of the features of the C syntax.
ISO C standard §5.1.1.2 says:-
Adjacent string literal tokens are concatenated.
White-space characters separating tokens are no longer significant.
Related
char str[] = "C:\Windows\system32"
auto raw_string = convert_to_raw(str);
std::cout << raw_string;
Desired output:
C:\Windows\system32
Is it possible? I am not a big fan of cluttering my path strings with extra backslash. Nor do I like an explicit R"()" notation.
Any other work-around of reading a backslash in a string literally?
That's not possible, \ has special meaning inside a non-raw string literal, and raw string literals exist precisely to give you a chance to avoid having to escape stuff. Give up, what you need is R"(...)".
Indeed, when you write something like
char const * str{"a\nb"};
you can verify yourself that strlen(str) is 3, not 4, which means that once you compile that line, in the binary/object file there's only one single character, the newline character, corresponding to \n; there's no \ nor n anywere in it, so there's no way you can retrieve them.
As a personal taste, I find raw string literals great! You can even put real Enter in there. Often just for the price of 3 characters - R, (, and ) - in addtion to those you would write anyway. Well, you would have to write more characters to escape anything needs escaping.
Look at
std::string s{R"(Hello
world!
This
is
Me!)"};
That's 28 keystrokes from R to last " included, and you can see in a glimpse it's 6 lines.
The equivalent non-raw string
std::string s{"Hello\nworld!\nThis\nis\nMe!"};
is 30 keystrokes from R to last " included, and you have to parse it carefully to count the lines.
A pretty short string, and you already see the advantage.
To answer the question, as asked, no it is not possible.
As an example of the impossibility, assume we have a path specified as "C:\a\b";
Now, str is actually represented in memory (in your program when running) using a statically allocated array of five characters with values {'C', ':', '\007', '\010', '\000'} where '\xyz' represents an OCTAL representation (so '\010' is a char equal to numerically to 8 in decimal).
The problem is that there is more than one way to produce that array of five characters using a string literal.
char str[] = "C:\a\b";
char str1[] = "C:\007\010";
char str2[] = "C:\a\010";
char str3[] = "C:\007\b";
char str4[] = "C:\x07\x08"; // \xmn uses hex coding
In the above, str1, str2, str3, and str4 are all initialised using equivalent arrays of five char.
That means convert_to_raw("C:\a\b") could quite legitimately assume it is passed ANY of the strings above AND
std::cout << convert_to_raw("C:\a\b") << '\n';
could quite legitimately produce output of
C:\007\010
(or any one of a number of other strings).
The practical problem with this, if you are working with windows paths, is that c:\a\b, C:\007\010, C:\a\010, C:\007\b, and C:\x07\x08 are all valid filenames under windows - that (unless they are hard links or junctions) name DIFFERENT files.
In the end, if you want to have string literals in your code representing filenames or paths, then use \\ or a raw string literal when you need a single backslash. Alternatively, write your paths as string literals in your code using all forward slashes (e.g. "C:/a/b") since windows API functions accept those too.
I get a code like this
virtual Qstring getEnergies() const {
return "estrain"
",eslip"
",edashpot";
}
Could you guys please explain the meaning of quotation and comma marks in that code? I am really thankful
Adjacent string literals are concatenated in C++. So
"foo" "bar"
Becomes
"foobar"
In your case the function will return a Qstring with the value of
"estrain,eslip,edashpot"
This behavior is defined in section 2.2.6 [lex.phases] of the C++ standard
Adjacent string literal tokens are concatenated.
The commas are a red herring; this is just concatenation of string literals.
In source code, "abc" "def" means the same thing as "abcdef".
The quotation mark here is because your function return type is string. So "" is for the string type return.
Now the second part as to why you have comma, then as others have answered it is for getting the adjacent string literal or different string literals. If you remove the comma then you will get a single string. If you don't have the comma then the return string will be equivalent to estraineslipedashpot
This is one usage I found in a open source software.And I don't understant how it works.
when I ouput it to the stdout,it was "version 0.8.0".
const char version[] = " version " "0" "." "8" "." "0";
It's called string concatenation -- when you put two (or more) quoted strings next to each other in the source code with nothing between them, the compiler puts them together into a single string. This is most often used for long strings -- anything more than one line long:
char whatever[] = "this is the first line of the string\n"
"this is the second line of the string\n"
"This is the third line of the string";
Before string concatenation was invented, you had to do that with a rather clumsy line continuation, putting a backslash at the end of each line (and making sure it was the end, because most compilers wouldn't treat it as line continuation if there was any whitespace after the backslash). There was also ugliness with it throwing off indentation, because any whitespace at the beginning of subsequent lines might be included in the string.
This can cause a minor problem if you intended to put a comma between the strings, such as when initializing an array of pointers to char. If you miss a comma, the compiler won't warn you about it -- you'll just get one string that includes what was intended to be two separate ones.
This is a basic feature of both C89 and C++98 called 'adjacent string concatenation' or thereabouts.
Basically, if two string literals are adjacent to each other with no punctuation in between, they are merged into a single string, as your output shows.
In the C++98 standard, section §2.1 'Phases of translation [lex.phases]' says:
6 Adjacent ordinary string literal tokens are concatenated. Adjacent wide string literal tokens are concatenated.
This is after the preprocessor has completed.
In the C99 standard, the corresponding section is §5.1.2.1 'Translation Phases' and it says:
6 Adjacent string literal tokens are concatenated.
The wording would be very similar in any other C or C++ standard you can lay hands on (and I do recognize that both C++98 and C99 are superseded by C++11 and C11; I just don't have electronic copies of the final standards, yet).
Part of the C++ standard implementation states that string literals that are beside each other will be concatenated together.
Quotes from C and C++ Standard:
For C (quoting C99, but C11 has something similar in 6.4.5p5):
(C99, 6.4.5p5) "In translation phase 6, the multibyte character
sequences specified by any sequence of adjacent character and
identically-prefixed string literal tokens are concatenated into a
single multibyte character sequence."
For C++:
(C++11, 2.14.5p13) "In translation phase 6 (2.2), adjacent string
literals are concatenated."
const char version[] = " version " "0" "." "8" "." "0";
is same as:
const char version[] = " version 0.8.0";
Compiler concatenates the adjacent pieces of string-literals, making one bigger piece of string-literal.
As a sidenote, const char* (which is in your title) is not same as char char[] (which is in your posted code).
The compiler automatically concatenates string literals written after each other (separated by white-space only).. It is as if you have written
const char version[] = "version 0.8.0";
EDIT: corrected pre-processor to compiler
Adjacent string literals are concatenated:
When specifying string literals, adjacent strings are concatenated.
Therefore, this declaration:
char szStr[] = "12" "34"; is identical to this declaration:
char szStr[] = "1234"; This concatenation of adjacent strings makes it
easy to specify long strings across multiple lines:
cout << "Four score and seven years "
"ago, our forefathers brought forth "
"upon this continent a new nation.";
Simply putting strings one after the other concatenates them at compile time, so:
"Hello" ", " "World!" => "Hello, World!"
This is a strange usage of the feature, usually it is to allow #define strings to be used:
#define FOO "World!"
puts("Hello, " FOO);
Will compile to the same as:
puts("Hello, World!");
What is the difference (if any) between this
_T("a string")
and
_T('a string')
?
First, _T isn't a standard part of C++. I've added the "windows" tag to your question.
Now, the difference between these is that the first is correct and the second is not. In C++, ' is for quoting single characters, and " is for quoting strings.
The second is wrong. You are placing a string literal in between single quotes.
'a string' is a so-called "multicharacter literal". It has type int, and an implementation-defined value. This is [lex.ccon] in the standard.
I don't know what values MSVC gives to multicharacter literals, and I don't know for sure what the MS-specific _T macro ends up doing with it, but I expect you get a narrow multicharacter literal on narrow builds, and a wide multicharacter literal on wide builds. The prefix L is the same for strings and character literals.
It's wrong, anyway: multicharacter literals are pretty much useless and certainly are no substitute for strings. "a string" is a string literal, which is what you want.
You use '' for single character and "" for strings. _T('a string') is wrong and its behaviour is compiler-specific.
In case of MSVC it uses first character only. Example:
#include <iostream>
#include <tchar.h>
int main()
{
if (_T('a string') == _T('a'))
std::cout << (int)'a' << " = " << _T('a');
}
output: 97 = 97
Single quotations are primarily used when denoting a single character:
char c = 'e' ;
Double quotations are used with strings and output statements:
string s = "This is a string";
cout << "Output where double quotations are used.";
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.