I found this macro in c source which I'm porting now:
#define Round256(p0, p1) \
X##p0 += X##p1;
There is no variable X in that code. Can anyone tell me what the symbol # do in this context?
## textually appends two strings together.
So in your example, if called as follows:
Round256(one, two)
will be translated to:
Xone += Xtwo;
The macro simply stringizes the values
Suppose it was called like this
Round256(1,2)
It would be expanded by the preprocessor as
X1 += X2
Which seems to resemble variable X1...n somewhere....
The ## concatenates the stringized values together.
The reason I used numerical values is in the name of the macro itself.
## is the pasting operator. It contatenates X (literally) and each value. So Round256(one, two) will be converted to Xone += Xtwo, for example.
Related
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C preprocessor: stringize macro and identity macro
(2 answers)
What does #x inside a C macro mean?
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Closed 6 years ago.
Please explain the code
#include <stdio.h>
#define A(a,b) a##b
#define B(a) #a
#define C(a) B(a)
main()
{
printf("%s\n",C(A(1,2)));
printf("%s\n",B(A(1,2)));
}
Output
12
A(1,2)
I don't understand, how the first printf evaluates to 12?
Isn't it similar to the second, as C macro is simply a wrapper to B macro?
As mentioned in Wikipedia in C-preprocessor :
The ## operator (known as the "Token Pasting Operator") concatenates
two tokens into one token.
The # operator (known as the "Stringification Operator") converts a
token into a string, escaping any quotes or backslashes appropriately.
If you want to stringify the expansion of a macro argument, you have
to use two levels of macros:
You cannot combine a macro argument with additional text and stringify
it all together. You can however write a series of adjacent string
constants and stringified arguments: the C compiler will then combine
all the adjacent string constants into one long string.
#define xstr(s) str(s)
#define str(s) #s
#define foo 4
str (foo) // outputs "foo"
xstr (foo) // outputs "4"
Also, from C-FAQ Question 11.17 :
It turns out that the definition of # says that it's supposed to
stringize a macro argument immediately, without further expanding it
(if the argument happens to be the name of another macro).
So, similarly, going along these lines :
you're doing C(A(1,2)),
which would roll to C(12), // since no #, so inner argument is expanded
and then to B(12)
// [since you've done two levels of macros in the code:
// 1. from C() to B(), and then, 2. B() to #a]
= 12 .
Whereas, in the first case, only 1 level of stringification is plainly done as per definition of B(a)(since it gets stringified immediately because of #)
macro-replacement of B(A(1,2))
= stringification of A(1,2)
= A(1,2).
The confusion here comes from a simple rule.
When evaluating a macro the pre-processor first resolves the macros in the arguments passed to the macro. However, as a special case, if an argument is right of # or adjacent to ##, it doesn't resolve macros within such arguments. Such are the rules.
Your first case
C(A(1,2))
The pre-processor first applies the C(a) macro, which is defined as B(a). There's no # or ## adjacent to the argument in the definition (none of them in B(a) at all), thus the pre-processor must resolve macros in the argument:
A(1,2)
The definition of A(a,b) is a##b which evaluates into 12.
After the macros in the arguments of the C(a) macro are evaluated, the C macro becomes:
C(12)
The pre-processor now resolves the C(a) macro, which according to its definition becomes
B(12)
Once this is done, the pre-processor evaluates macros inside the result once again and applies the B(a) macro, so the result becomes
"12"
Your second case
B(A(1,2))
Similar to the first case, the pre-processor first applies the B(a) macro. But this time, the definition of the macro is such that the argument is preceded by #. Therefore, the special rule applies and macros inside the argument are not evaluated. Therefore, the result immediately becomes:
"A(1,2)"
The preprocessor goes over the result again trying to find more macros to expand, but now everything is a part of the string, and macros don't get expanded within strings. So the final result is:
"A(1,2)"
C preprocessor has two operators # and ##. The # operator turns the argument of a function like macro to a quoted string where ## operator concatenates two identifiers.
#define A(a,b) a##b will concatenate a with b returning ab as string.
so A(1,2) will return 12
#define B(a) #a will return a as string
#define C(a) B(a) will call previous one and return a as string.
so C(A(1,2)) = C(12) = B(12) = 12 (as string)
B(A(1,2)) = A(1,2) because A(1,2) is taken as an argument and returned as string A(1,2)
There are two operators used in the function-like macros:
## causes a macro to concatenate two parameters.
# causes the input to be effectively turned into a string literal.
In A(a,b) ## causes a to be concatenated with b. In B(a), # effectively creates a string literal from the input. So the expansion runs as follows:
C(A(1,2)) -> C(12) -> B(12) -> "12"
B(A(1,2)) -> "A(1,2)"
Because for C(A(1,2)), the A(1,2) part is evaluated first to turn into 12, the two statements aren't equal like they would appear to be.
You can read more about these at cppreference.
So when looking into getting my define macro to work, I found the # and ## macro helpers, and used them to simplify my macro. The key part of the macro sets a variable to a string containing the name of the variable (but not the variable name alone). As a simplified example, let's take a macro called SET(X) that should expand SET(something) into something = "pre_something".
The only way I've found to do it so far is with two macros like #define QUOTE(X) #X and #define SET(X) X = QUOTE(pre_##X). However, using multiple macros seems excessive, and may cause problems with further macro expansion (I think). Is there a cleaner, one-line way of doing the same thing?
#define SET(x) x = "pre_"#x
C does string concatenation at compile time, so two string literals next to each other are concatenated.
"hello " "world" -> "hello world"
I need to map a macro to a right value depending on the option selected e.g
#define A { if (fel == 1) x; if(fel != 1) y; }
I want A to assume the value of x if fel is 1, else it must assume the value y.
Where am I missing it?
Macros are preprocessor directives, meaning - they are expanded before the compilation and before runtime.
So, the way you've written this, A will literally be replaced by { if (fel == 1) x; if(fel != 1) y; }, not by x or y.
Macros can be avoided most of the time. Why not just use A as variable, not as macro?
If fel is compile-time known constant OR a macro itself, there are options with #if, but it looks like it's a normal variable.
You can't. Macro substitution is done at preprocessing stage and you can't set it based on a runtime condition.
Instead you can define A as a variable:
A = (fel == 1)? x: y;
Define the macro using the ternary operator:
#define A ((fel == 1) ? x : y)
The macro will only work if you use it in a context where fel and x and y are all in scope.
if is a statement, not an expression, thus it doesn't return any value. You need to write it as an expression, like using ternary operator, as suggested by others, or use GCC extension
Let's say I have already defined 9 macros from
ABC_1 to ABC_9
If there is another macro XYZ(num) whose objective is to call one of the ABC_{i} based on the value of num, what is a good way to do this? i.e. XYZ(num) should call/return ABC_num.
This is what the concatenation operator ## is for:
#define XYZ(num) ABC_ ## num
Arguments to macros that use concatenation (and are used with the operator) are evaluated differently, however (they aren't evaluated before being used with ##, to allow name-pasting, only in the rescan pass), so if the number is stored in a second macro (or the result of any kind of expansion, rather than a plain literal) you'll need another layer of evaluation:
#define XYZ(num) XYZ_(num)
#define XYZ_(num) ABC_ ## num
In the comments you say that num should be a variable, not a constant. The preprocessor builds compile-time expressions, not dynamic ones, so a macro isn't really going to be very useful here.
If you really wanted XYZ to have a macro definition, you could use something like this:
#define XYZ(num) ((int[]){ \
0, ABC_1, ABC_2, ABC_3, ABC_4, ABC_5, ABC_6, ABC_7, ABC_8, ABC_9 \
}[num])
Assuming ABC_{i} are defined as int values (at any rate they must all be the same type - this applies to any method of dynamically selecting one of them), this selects one with a dynamic num by building a temporary array and selecting from it.
This has no obvious advantages over a completely non-macro solution, though. (Even if you wanted to use macro metaprogramming to generate the list of names, you could still do that in a function or array definition.)
Yes, that's possible, using concatenation. For example:
#define FOO(x, y) BAR ##x(y)
#define BAR1(y) "hello " #y
#define BAR2(y) int y()
#define BAR3(y) return y
FOO(2, main)
{
puts(FOO(1, world));
FOO(3, 0);
}
This becomes:
int main()
{
puts("hello " "world");
return 0;
}
I wrote the following code in plain C:
#define _cat(A, B) A ## _ ## B
#define cat(A, B) _cat(A, B)
#define plus(A, B) cat(cat(plus,__typeof__(A)),__typeof__(B))(A, B)
int main(int argc, const char * argv[])
{
double x = 1, y = 0.5;
double r = plus(x, y);
printf("%lf",r);
return 0;
}
Here, I would like the macro plus to be expanded becoming a function name which contains the types of the parameters. In this example I would like it to expand the following way
double r = plus(x, y)
...
/* First becomes*/
double r = cat(cat(plus,double),double)(x, y)
...
/* Then */
double r = cat(plus_double,double)(x, y)
...
/* And finally */
double r = plus_double_double(x, y)
However all I got from the preprocessor is
double r = plus___typeof__(x)___typeof(y)(x,y)
and gcc will obviously refuse to compile.
Now, I know that typeof evaluates at compile-time and it is my understanding that a macro is only prevented from being evaluated when it is contained in second macro which directly involves the stringify #and the concatenation ## tokens (here's the reason why I split cat in the way you see). If this is right, why doesn't __typeof__(x) get evaluated to double by the preprocessor? Seems to me that the behaviour should be perfectly clear at build time. Shouldn't __typeof__(x) evaluate to double before even going in _cat?
I searched and searched but I couldn't find anything... Am I doing something really really stupid?
I'm running Mac OS X Mountain Lion but I'm mostly interested in getting it work on any POSIX platform.
The reason this does not work is typeof is not a macro but a reserved word in the gcc's dialect of C and is thus handled after the preprocessor has finished its work. A good analogy would be the sizeof operator which is not a macro either and is not expanded by the preprocessor. To do (approximately) what you want (pick a different function based on the type of the arguments) try the _Generic construct (new in C11)
Macro expansion occurs before C token analysis (see https://stackoverflow.com/a/1479972/1583175 for a diagram of the phases of translation)
The macro preprocessor is unaware of the type information -- it merely does text processing
The preprocessor knows nothing about types, only about textual tokens. __typeof__() gets evaluated by the compiler pass, after the preprocessor has finished performing macro replacements.