I came across this following code:
#include<stdio.h>
#define d(x) x(#x[3])
int main(){
d(putchar);
}
Which prints c as the output. I wonder what does the macro #define d(x) x(#x[3]) does? In C language is there an operator like #? I can see this inside the macro body i.e here x(#x[3]). According to my normal eye it looks something different I see in C language but actually What does this does?
Edit : Whats the real use of # in real world?
I'm a novice in C and it will be good if the explanation is in simple terms. Thanks in advance.
The character '#' is a stringizer -- it turns a symbol into a string. The code becomes
putchar("putchar"[3]);
The hash sign means "stringify", so d(x) expands to putchar("putchar"[3]), whence the c.
From here:
Function macro definitions accept two special operators (# and ##) in the replacement sequence:
If the operator # is used before a parameter is used in the replacement sequence, that parameter is replaced by a string literal (as if it were enclosed between double quotes)
#define str(x) #x
cout << str(test);
Put simply, it changes the "x" parameter into a string. In this case test becomes a char array containing 't', 'e', 's', 't', '\0'.
The # is a pre-processor operator which turns a literal into a string. In fact your d macro prints the fourth char of the converted string of your literal.
Related
How do you escape the % sign when using printf in C?
printf("hello\%"); /* not like this */
You can escape it by posting a double '%' like this: %%
Using your example:
printf("hello%%");
Escaping the '%' sign is only for printf. If you do:
char a[5];
strcpy(a, "%%");
printf("This is a's value: %s\n", a);
It will print: This is a's value: %%
As others have said, %% will escape the %.
Note, however, that you should never do this:
char c[100];
char *c2;
...
printf(c); /* OR */
printf(c2);
Whenever you have to print a string, always, always, always print it using
printf("%s", c)
to prevent an embedded % from causing problems (memory violations, segmentation faults, etc.).
If there are no formats in the string, you can use puts (or fputs):
puts("hello%");
if there is a format in the string:
printf("%.2f%%", 53.2);
As noted in the comments, puts appends a \n to the output and fputs does not.
With itself...
printf("hello%%"); /* like this */
Use a double %%:
printf("hello%%");
Nitpick:
You don't really escape the % in the string that specifies the format for the printf() (and scanf()) family of functions.
The %, in the printf() (and scanf()) family of functions, starts a conversion specification. One of the rules for conversion specification states that a % as a conversion specifier (immediately following the % that started the conversion specification) causes a '%' character to be written with no argument converted.
The string really has 2 '%' characters inside (as opposed to escaping characters: "a\bc" is a string with 3 non null characters; "a%%b" is a string with 4 non null characters).
Like this:
printf("hello%%");
//-----------^^ inside printf, use two percent signs together
You can use %%:
printf("100%%");
The result is:
100%
You are using the incorrect format specifier. You should use %% for printing %. Your code should be:
printf("hello%%");
Read more all format specifiers used in C.
The backslash in C is used to escape characters in strings. Strings would not recognize % as a special character, and therefore no escape would be necessary. printf is another matter: use %% to print one %.
You can simply use % twice, that is "%%"
Example:
printf("You gave me 12.3 %% of profit");
Yup, use printf("hello%%"); and it's done.
The double '%' works also in ".Format(…).
Example (with iDrawApertureMask == 87, fCornerRadMask == 0.05):
csCurrentLine.Format("\%ADD%2d%C,%6.4f*\%",iDrawApertureMask,fCornerRadMask) ;
gives the desired and expected value of (string contents in) csCurrentLine;
"%ADD87C, 0.0500*%"
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.
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.";
The preprocessor can be used to replace certain keywords with other words using #define. For example I could do #define name "George" and every time the preprocessor finds 'name' in the program it will replace it with "George".
However, this only seems to work with code. How could I do this with strings and text? For example if I print "Hello I am name" to the screen, I want 'name' to be replaced with "George" even though it is in a string and not code.
I do not want to manually search the string for keywords and then replace them, but instead want to use the preprocessor to just switch the words.
Is this possible? If so how?
I am using C++ but C solutions are also acceptable.
#define name "George"
printf("Hello I am " name "\n");
Adjacent string literals are concatenated in C and C++.
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."
EDIT: as requested, add quotes from C and C++ Standard. Thanks to #MatteoItalia for the C++11 quote.
#define name "George"
printf("Hello I am %s\n", name);
Here name will be replaced by "George"
Your issue is that the preprocessor will (wisely) not replace tokens that are inside string literals.
So you must either use a function like printf or a variable rather than the preprocessor, or pull the token out of the string like so:
#include <iostream>
#define name "George"
int main(int argc, char** argv) {
std::cout << "Hello I am " << name << std::endl;
}
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 ").