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What are the applications of the ## preprocessor operator and gotchas to consider?
(13 answers)
The ## operator in C
(7 answers)
Closed 9 years ago.
What's the meaning of "##" in the following?
#define CC_SYNTHESIZE(varType, varName, funName)\
protected: varType varName;\
public: inline varType get##funName(void) const { return varName; }\
public: inline void set##funName(varType var){ varName = var; }
The operator ## concatenates two arguments leaving no blank spaces between them:
e.g.
#define glue(a,b) a ## b
glue(c,out) << "test";
This would also be translated into:
cout << "test";
It concatenates tokens without leaving blanks between them. Basically, if you didn't have the ## there
public: inline varType getfunName(void) const { return varName; }\
the precompiler would not replace funName with the parameter value. With ##, get and funName are separate tokens, which means the precompiler can replace funName and then concatenate the results.
This is called token pasting or token concatenation.
The ## (double number sign) operator concatenates two tokens in a macro invocation (text and/or arguments) given in a macro definition.
Take a look here at the official GNU GCC compiler documentation for more information.
Related
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# and ## in macros
(3 answers)
Stringification - how does it work?
(2 answers)
Closed 5 years ago.
What does the '#' symbol do after the second define? And isn't the second line enough? Why the first one?
#define MAKESTRING(n) STRING(n)
#define STRING(n) #n
This is stringize operation, it will produce a string literal from macro parameter, e.g. "n". Two lines are required to allow extra expantion of macro parameter, for example:
// prints __LINE__ (not expanded)
std::cout << STRING(__LINE__) << std::endl;
// prints 42 (line number)
std::cout << MAKESTRING(__LINE__) << std::endl;
Hash symbol takes macro argument into a c-string.
For example
#define MAKESTRING(x) #x
printf(MAKESTRING(text));
will print text
And first line is only alternative name for this macro.
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Creating C macro with ## and __LINE__ (token concatenation with positioning macro)
(3 answers)
Closed 6 years ago.
Consider the following macro:
#define CAT(X, Y) X ## Y
#define CMB(A, B) CAT(A, B)
#define SLB_LOGGING_ALGORITHM CMB(Logging, SLB_ALGORITHM)
where SLB_ALGORITHM is a defined pre-processor symbol.
If I just use CAT directly instead of CMB, SLB_ALGORITHM does not get expanded. Why is that the case and how exactly does the indirection help?
## is a string concatenator, so if you call CAT(Logging, SLB_ALGORITHM) from SLB_LOGGING_ALGORITHM macro, this will result in concatenation of string Logging with string SLB_ALGORITHM, that is: LoggingSLB_ALGORITHM which is likely not what you want.
When calling CMB(Logging, SLB_ALGORITHM) from SLB_LOGGING_ALGORITHM macro, instead, preprocessor first expands Logging and SLB_ALGORITHM (call to CMB()) then concatenate the expanded strings (call to CAT()).
To quote this answer:
When you have a macro replacement, the preprocessor will only expand the macros recursively if neither the stringizing operator # nor the token-pasting operator ## are applied to it.
So the preprocessor does not expand a given macro when ## is applied to it. This is why it is exapnded in the CMB(A, B) level but not when directly using CAT(X, Y) .
So, it's been a while since I have written anything in C++ and now I'm working on a project using C++11 and macros.
I know that by using the stringify operator I can do this:
#define TEXT(a) #a //expands to "a"
How am I supposed to use the preprocessor for recognizing the tokens like + and * to do this:
#define TEXT(a)+ ??? //want to expand to "a+"
#define TEXT(a)* ??? //want to expand to "a*"
when the input has to be in that syntax?
I have tried doing that:
#define + "+"
but of course it doesn't work. How can I make the preprocessor recognize those tokens?
NOTE:
This is actually part of a project for a small language that defines and uses regular expressions, where the resulting string of the macros is to be used in a regex. The syntax is given and we have to use it as it is without making any changes to it.
eg
TEXT(a)+ is to be used to make the regular expression: std::regex("a+")
without changing the fact that TEXT(a) expands to "a"
First,
#define TEXT(a) #a
doesn't “convert to "a"”. a is just a name for a parameter. The macro expands to a string that contains whatever TEXT was called with. So TEXT(42 + rand()) will expand to "42 + rand()". Note that, if you pass a macro as parameter, the macro will not be expanded. TEXT(EXIT_SUCCESS) will expand to "EXIT_SUCCESS", not "0". If you want full expansion, add an additional layer of indirection and pass the argument to TEXT to another macro TEXT_R that does the stringification.
#define TEXT_R(STUFF) # STUFF
#define TEXT(STUFF) TEXT_R(STUFF)
Second, I'm not quite sure what you mean with TEXT(a)+ and TEXT(a)*. Do you want, say, TEXT(foo) to expand to "foo+"? I think the simplest solution in this case would be to use the implicit string literal concatenation.
#define TEXT_PLUS(STUFF) # STUFF "+"
#define TEXT_STAR(STUFF) # STUFF "*"
Or, if you want full expansion.
#define TEXT_R(STUFF) # STUFF
#define TEXT_PLUS(STUFF) TEXT_R(STUFF+)
#define TEXT_STAR(STUFF) TEXT_R(STUFF*)
Your assignment is impossible to solve in C++. You either misunderstood something or there’s an error in the project specification. At any rate, we’ve got a problem here:
TEXT(a)+ is to be used to make the regular expression: std::regex("a+") without changing the fact that TEXT(a) expands to "a" [my emphasis]
TEXT(a) expands to "a" — meaning, we can just replace TEXT(a) everywhere in your example; after all, that’s exactly what the preprocessor does. In other words, you want the compiler to transform this C++ code
"a"+
into
std::regex("a+")
And that’s simply impossible, because the C++ preprocess does not allow expanding the + token.
The best we can do in C++ is use operator overloading to generate the desired code. However, there are two obstacles:
You can only overload operators on custom types, and "a" isn’t a custom type; its type is char const[2] (why 2? Null termination!).
Postfix-+ is not a valid C++ operator and cannot be overloaded.
If your assignment had just been a little different, it would work. In fact, if your assignment had said that TEXT(a)++ should produce the desired result, and that you are allowed to change the definition of TEXT to output something other than "a", then we’d be in business:
#include <string>
#include <regex>
#define TEXT(a) my_regex_token(#a)
struct my_regex_token {
std::string value;
my_regex_token(std::string value) : value{value} {}
// Implicit conversion to `std::regex` — to be handled with care.
operator std::regex() const {
return std::regex{value};
}
// Operators
my_regex_token operator ++(int) const {
return my_regex_token{value + "+"};
}
// more operators …
};
int main() {
std::regex x = TEXT(a)++;
}
You don't want to jab characters onto the end of macros.
Maybe you simply want something like this:
#define TEXT(a, b) #a #b
that way TEXT(a, +) gets expanded to "a" "+" and TEXT(a, *) to "a" "*"
If you need that exact syntax, then use a helper macro, like:
#define TEXT(a) #a
#define ADDTEXT(x, y) TEXT(x ## y)
that way, ADDTEXT(a, +) gets expanded to "a+" and ADDTEXT(a, *) gets expanded to "a*"
You can do it this way too:
#define TEXT(a) "+" // "a" "+" -> "a+"
#define TEXT(a) "*" // "a" "*" -> "a*"
Two string literals in C/C++ will be joined into single literal by specification.
This question already has answers here:
Closed 11 years ago.
Possible Duplicate:
How to convert concatenated strings to wide-char with the C preprocessor?
I have a string literal defined using a #define:
#define B "1234\0"
How do I use this definition to get this wide string literal at compile time?:
L"1234\0"
(just the #defined string literal with L prepended to make it into a wide string).
I tried this:
#define MAKEWIDE(s) L##s
but this generates LB.
Token pasting needs an additional level of indirection to deal properly with macros used as operands. Try something like:
#define PASTE(x, y) x##y
#define MAKEWIDE(x) PASTE(L,x)
This will work just fine:
#define B "1234\0"
#define LB L"" B
is it possible to concatenate strings during preprocessing?
I found this example
#define H "Hello "
#define W "World!"
#define HW H W
printf(HW); // Prints "Hello World!"
However it does not work for me - prints out "Hello" when I use gcc -std=c99
UPD This example looks like working now. However, is it a normal feature of c preprocessor?
Concatenation of adjacent string litterals isn't a feature of the preprocessor, it is a feature of the core languages (both C and C++). You could write:
printf("Hello "
" world\n");
You can indeed concatenate tokens in the preprocessor, but be careful because it's tricky. The key is the ## operator. If you were to throw this at the top of your code:
#define myexample(x,y,z) int example_##x##_##y##_##z## = x##y##z
then basically, what this does, is that during preprocessing, it will take any call to that macro, such as the following:
myexample(1,2,3);
and it will literally turn into
int example_1_2_3 = 123;
This allows you a ton of flexibility while coding if you use it correctly, but it doesn't exactly apply how you are trying to use it. With a little massaging, you could get it to work though.
One possible solution for your example might be:
#define H "Hello "
#define W "World!"
#define concat_and_print(a, b) cout << a << b << endl
and then do something like
concat_and_print(H,W);
From gcc online docs:
The '##' preprocessing operator performs token pasting. When a macro is expanded, the two tokens on either side of each '##' operator are combined into a single token, which then replaces the '##' and the two original tokens in the macro expansion.
Consider a C program that interprets named commands. There probably needs to be a table of commands, perhaps an array of structures declared as follows:
struct command
{
char *name;
void (*function) (void);
};
struct command commands[] =
{
{ "quit", quit_command },
{ "help", help_command },
...
};
It would be cleaner not to have to give each command name twice, once in the string constant and once in the function name. A macro which takes the name of a command as an argument can make this unnecessary. The string constant can be created with stringification, and the function name by concatenating the argument with _command. Here is how it is done:
#define COMMAND(NAME) { #NAME, NAME ## _command }
struct command commands[] =
{
COMMAND (quit),
COMMAND (help),
...
};
I just thought I would add an answer that cites the source as to why this works.
The C99 standard §5.1.1.2 defines translation phases for C code. Subsection 6 states:
Adjacent string literal tokens are concatenated.
Similarly, in the C++ standards (ISO 14882) §2.1 defines the Phases of translation. Here Subsection 6 states:
6 Adjacent ordinary string literal tokens are concatenated. Adjacent wide string literal tokens are concatenated.
This is why you can concatenate strings simply by placing them adjacent to one another:
printf("string"" one\n");
>> ./a.out
>> string one
The preprocessing part of the question is simply the usage of the #define preprocessing directive which does the substitution from identifier (H) to string ("Hello ").