Using macros to name functions in C - c

The other day I was trying to use a macro to define a function like so (simplified, obviously):
#define DEF_ADD(name) \
int add_name(int x, int y) { \
return x + y; \
} \
Now, I expected the snippet DEF_ADD(hello) to define the function add_hello. However, the macro call instead defines a function called add_name. I expect that this has something to do with scanning, but I couldn't find a way to recreate the behavior I'm looking for. Any ideas how to do this?

You need
#define DEF_ADD(name) \
int add_##name(int x, int y) { \
return x + y; \
}
Note well the ##. I've also dropped the final newline character. Else the parameter name is not used and you create add_name as the function name.

Related

Extract a function name inside a macro

In C, we often have to run such code
if (! somefun(x, y, z)) {
perror("somefun")
}
Is it possible to create a macro which, used as follows:
#define chkerr ...
chkerr(somefun(x, y, z));
would compile to the above?
I already know I can use __VA_ARGS__ macro, but this would require me to call it like
chkerr(somefun, x, y, z)
Short variant (you spotted already):
#define chkErr(FUNCTION, ...) \
if(!FUNCTION(__VA_ARGS__)) \
{ \
perror(#FUNCTION); \
}
Be aware that this can impose big problems in nested if/else or similar constructs:
if(x)
chkErr(f, 10, 12) //;
//^ semicolon forgotten!
else
chkErr(f, 12, 10);
would compile to code equivalent to the following:
if(x)
{
if(!f(10, 12))
perror("f");
else if(!f, 12, 10))
perror("f");
}
Quite obviously not what was intended with the if/else written with the macros... So you really should prefer to let it look like a real function (requiring a semicolon):
#define chkErr(FUNCTION, ...) \
do \
{ \
if(!FUNCTION(__VA_ARGS__)) \
{ \
perror(#FUNCTION); \
} \
} \
while(0)
You would call it like this:
chkErr(someFunction, 10, 12);
In case of error, output would be:
someFunction: <error text>
However, this hides the fact that a function actually gets called, making it more difficult to understand for "outsiders". Same output, not hiding the function call, but requiring one additional comma in between function and arguments (compared to a normal function call):
#define chkErr(FUNCTION, ARGUMENTS) \
do \
{ \
if(!FUNCTION ARGUMENTS) \
{ \
perror(#FUNCTION); \
} \
} \
while(0)
chkErr(someFunction,(12, 10));
// ^ (!)
Another variant with the charm of retaining the function call would print out this entire function call:
#define chkErr(FUNCTION_CALL) \
do \
{ \
if(!FUNCTION_CALL) \
{ \
perror(#FUNCTION_CALL); \
} \
} \
while(0)
chkErr(someFunction(10, 12));
In case of error, output would be:
someFunction(10, 12): <error text>
Addendum: If you really want exactly the output as shown in the question and still have the function call retained (without comma in between), you are a little in trouble. Actually, it is possible, but it requires some extra work:
Problem is how the preprocessor operates on macro arguments: Each argument is a token. It can easily combine tokens, but cannot split them.
Leaving out any commas results in the macro accepting one single token, just as in my second variant. Sure, you can stringify it, as I did, but you get the function arguments with. This is a string literal, and as the pre-processor cannot modify string literals, you have to operate on them at runtime.
Next problem then is, though, string literals are unmodifiable. So you need to modify a copy!
The following variant would do all this work for you:
#define chkErr(FUNCTION_CALL) \
do \
{ \
if(!FUNCTION_CALL) \
{ \
char function_name[] = #FUNCTION_CALL; \
char* function_name_end = strchr(function_name, '('); \
if(function_name_end) \
*function_name_end = 0; \
perror(function_name); \
} \
} \
while(0)
Well, decide you if it is worth the effort...
By the way - whitespace between function name and opening parenthesis is not eliminated. If you want to be perfect:
unsigned char* end = (unsigned char*) function_name;
while(*end && *end != '(' && !isspace(*end))
++end;
*end = 0;
Or, much nicer (thanks chqrlie for the hint):
function_name[strcspn(function_name, "( \t")] = 0;
Anything else I can think of would require an additional pre-processing step:
#define CAT(X, Y) CAT_(X, Y)
#define CAT_(X, Y) X ## Y
#define chkErr(FUNCTION_CALL) \
do \
{ \
if(!FUNCTION_CALL) \
{ \
perror(CAT(CHK_ERR_TEXT_, __LINE__)); \
} \
} \
while 0
chkErr(function(10, 12));
Ah, huh, this would result in code like this:
if(!function(10, 12))
{
perror(CHK_ERR_TEXT_42);
}
And now, where to get these macros from? Well, the pre-processing, remember? Possibly a perl or python script, e. g. generating an additional header file you'd have to include. You would have to make sure this pre-processing is done every time before the compiler's pre-processor runs.
Well, all not impossible to solve, but I'll leave this to the masochists among us...
C11 6.4.2.2 Predefined identifiers
The identifier __func__ shall be implicitly declared by the translator as if, immediately following the opening brace of each function definition, the declaration
static const char __func__[] = "function-name";
appeared, where function-name is the name of the lexically-enclosing function.
You can used it this way:
#define chkErr(exp) do { if (!(exp)) perror(__func__); } while (0)
chkerr(somefun(x, y, z));
Unfortunately, this would produce an error message with the name of the calling function, not somefun. Here is a simple variant that should work and even produce more informative error messages:
#define chkErr(exp) do { if (!(exp)) perror(#exp); } while (0)
chkerr(somefun(x, y, z));
In case somefun(x, y, z) returns a non zero value, the error message will contain the string "somefun(x, y, z)".
You can combine both techniques to give both the offending call and the location:
#include <errno.h>
#include <stdio.h>
#include <string.h>
#define chkErr(exp) \
do { if (!(exp)) \
fprintf(stderr, "%s:%d: in function %s, %s failed: %s\n",\
__FILE__, __LINE__, __func__, #exp, strerror(errno)); \
} while (0)
chkerr(somefun(x, y, z));
This assumes somefun() returns 0 or NULL in case of error and set errno accordingly. Note however that most system calls return non zero in case of error.
You can use the original call format:
chkerr(somefun(x, y, z));
With a macro and a helper function:
#define chkerr(fcall) \
if (!fcall) { \
perror(extract_fname(#fcall)); \
}
const char *extract_fname(const char *fcall);
The extract_fname function would get text and return everything until the open parenthesis.
Yes it is possible with an ugly, unsafe variadic macro:
#define chkerr(func, ...) \
if(!func(__VA_ARGS__)) \
{ \
perror(#func); \
}
...
chkerr(somefunc, 1, 2, 3);
But it is a very bad idea.
Call for sanity:
If there was just the original code with the plain if statement, the reader would think "Here they call a function and do some basic error control. Okay, basic stuff. Moving on...". But after the changes, anyone who reads the code will instead freeze and think "WTF is this???".
You can never write a macro that is clearer than the if statement - which makes the if statement superior to the macro.
Some rules to follow:
Function-like macros are dangerous and unreadable. They should only be used as the very last resort.
Avoid inventing your own secret macro language with function-like macros. C programmers who read your code know C. They don't know your secret macro language.
"To avoid typing" is often a poor rationale for program design decisions. Avoiding code repetition is a good rationale, but taking it to the extremes will affect code readability. If you avoid code repetition and make the code more readable at the same time, it is a good thing. If you do it but the code turns less readable, it is hard to justify.
It's not possible to extract just the function name. The C processor sees the literals you pass as single tokens, which can't be manipulated. Your only options are to print the function with arguments like Aconcague suggests or pass the name as a separate parameter:
#define chkErr(FUNCTION_NAME, FUNCTION_CALL) \
if(!FUNCTION_CALL) \
{ \
perror(#FUNCTION_NAME); \
}
chkErr(someFunction, someFunction(10, 12));

Using typeof to convert a variable declaration to a type?

Currently, I have a scenario much like this:
#define my_macro(var) __builtin_types_compatible_p(typeof(var), foo) ? do_something : do_something_else
However, inadvertently the macro gets passed this parameter:
my_macro(int x);
Which of course fails, because typeof(int x) isn't valid.
Is there a transformation I can apply inside the my_macro which will allow typeof to work on that expression? I cannot change what is passed into the macro, only what happens inside the macro.
So, something like this:
#define my_macro(var) typeof(?? var ??)
Or, is there another expression I should be using here?
Well, I found a way to do it, using yet another GCC extension, this time the statement expression:
#define my_typeof(definition) typeof(({ definition, _def; _def; }))
Which, of course, expands to:
typeof(({ int x, _def; _def; }))
Pretty ugly, but why do I care? It works.
You could handle x or int x separately but to handle both with one macro you would need the ability to parse/separate a space-delimited argument within the C preprocessor. To my knowledge, no such support exists in the C preprocessor. Without such parsing capabilities you must find some other clever way to write a macro that works around this limitation, for example, something in the spirit of my_macro2() in the following code sample:
#include <stdio.h>
#define my_macro1(var) \
do { \
typeof(var) blah; \
printf("sizeof(var)=%d\n", sizeof(blah)); \
} while(0)
#define my_macro2(var) \
do { \
var, newvar_sametype; \
typeof(newvar_sametype) blah; \
printf("sizeof(newvar_sametype)=%d\n", sizeof(blah)); \
} while(0)
int
main()
{
int x;
my_macro1(x);
my_macro2(char y);
return 0;
}

How to check `typeof` for void value at compile time?

let's say that I want to have C macro that works on any type.
I'm using GCC compiler (>= 4.6) and can use GNU99 macros.
//code...
any_type_t *retVal = function_that_runs_very_long_time(a, b, &&c, **d, &e, *f);
//other code...
usage of macro for TIMER can look for example like this
//code...
any_type_t *retVal =
TIMER(
function_that_runs_very_long_time(a, b, &&c, **d, &e, *f),
"TIMING FOR VALUE <%d, %d>", a, b
);
//other code...
So TIMER has to return value of given function and print duration of its run.
There is problem with functions that have void return type.
I can obviously have two macros like TIMER_TYPE and TIMER_VOID, but I want to use single one to time function with any return value.
Thank you for suggestions.
Edited example of this TIMER macro
#define TIMER(expr, fmt_msg, ...) \
({ \
struct timeval before, after; \
uint64_t time_span; \
int time_span_sec, time_span_usec; \
gettimeofday(&before, NULL); \
typeof(expr) _timer_expr__ = (expr); \ // <- static if?
gettimeofday(&after, NULL); \
time_span = (after.tv_sec * 1000000 + after.tv_usec) \
- (before.tv_sec * 1000000 + before.tv_usec); \
time_span_sec = time_span / 1000000; \
time_span_usec = time_span % 1000000; \
TRACE(fmt_msg "\n%s : %d.%d seconds", \
#expr, time_span_sec, time_span_usec, ...); \
_timer_expr__; \
})
What an interesting question, kudos!
After few experiments, I found a solution which uses __builtin_types_compatible_p and __builtin_choose_expr intrinsics of GCC.
__builtin_types_compatible_p
Quoting GCC manual:
Built-in Function: int __builtin_types_compatible_p (type1, type2)
You can use the built-in function __builtin_types_compatible_p to determine whether two types are the same.
This built-in function returns 1 if the unqualified versions of the types type1 and type2 (which are types, not expressions) are compatible, 0 otherwise. The result of this built-in function can be used in integer constant expressions.
This built-in function ignores top level qualifiers (e.g., const, volatile). For example, int is equivalent to const int.
So here is how we can check for "voidness".
#define __type_is_void(expr) __builtin_types_compatible_p(typeof(expr), void)
__builtin_choose_expr
Built-in Function: type __builtin_choose_expr (const_exp, exp1, exp2)
You can use the built-in function __builtin_choose_expr to evaluate code depending on the value of a constant expression. This built-in function returns exp1 if const_exp, which is an integer constant expression, is nonzero. Otherwise it returns exp2.
This built-in function is analogous to the ? : operator in C, except that the expression returned has its type unaltered by promotion rules. Also, the built-in function does not evaluate the expression that is not chosen. For example, if const_exp evaluates to true, exp2 is not evaluated even if it has side-effects.
If exp1 is returned, the return type is the same as exp1's type. Similarly, if exp2 is returned, its return type is the same as exp2.
So __builtin_choose_expr intrinsic is something like a "static switch" evaluated at compile-time.
Preparation
I don't paste here your TIMER macro, but I assume it is able to split it into two versions: one for void expr and one for the rest. Here are just stubs which evaluate the expression and yield the result of the same type.
#define __DO(expr) \
({ typeof(expr) __ret; __ret = (expr); __ret; })
#define __DO_VOID(expr) \
(void) (expr)
Naive solution
Now we can statically switch between two implementations, depending on the actual type of the expression. But in fact the naive solution doesn't work, see below.
#define DO(expr) \
__builtin_choose_expr(__type_is_void(expr), \
__DO_VOID(expr), \
__DO(expr)) # won't work
Attempt to compile this code passing a void expression gives the following error:
test.c:28:9: error: variable or field ‘__ret’ declared void
test.c:28:9: error: void value not ignored as it ought to be
Although __DO_VOID is chosen, __DO generates errors. This behavior is described in manual:
... the unused expression (exp1 or exp2 depending on the value of const_exp) may still generate syntax errors. This may change in future revisions.
Working solution
The trick is to substitute the original void expr with some non-void value to be able to compile the __DO case (which is anyway a dead code when expr is void).
#define __expr_or_zero(expr) __builtin_choose_expr(__type_is_void(expr), 0, (expr))
#define DO(expr) \
__builtin_choose_expr(__type_is_void(expr), \
__DO_VOID(expr), \
__DO(__expr_or_zero(expr))) # works fine!
That's it! Here is the complete source code on Ideone: http://ideone.com/EFy4pE
can you accept an answer of "this isn't really possible" ?
not the part about returning from a macro. but the part about conditionally testing expr for its return type.
in effect, you're asking for something like the following:
let's say instead of some magical check called "is_expr_type_void(expr)", you instead simply pass a 1 or a 0 at the time of the call to indicate is_void or !is_void in the following variation of your macro:
#define TIMER(is_void, expr, fmt_msg, ...) \
({ \
struct timeval before, after; \
uint64_t time_span; \
int time_span_sec, time_span_usec; \
gettimeofday(&before, NULL); \
if (is_void) \
(expr) \
else \
typeof(expr) _timer_expr__ = (expr); \ // <- static if?
gettimeofday(&after, NULL); \
time_span = (after.tv_sec * 1000000 + after.tv_usec) \
- (before.tv_sec * 1000000 + before.tv_usec); \
time_span_sec = time_span / 1000000; \
time_span_usec = time_span % 1000000; \
TRACE(fmt_msg "\n%s : %d.%d seconds", \
#expr, time_span_sec, time_span_usec, ...); \
if (!is_void) \
_timer_expr__; \
})
this simply cannot work. the preprocessor would create code for that if-else conditional in all cases, both void and non-void function calls. and both sides would compile fine for non-void functions. but the compiler would always choke on the "else" part of the conditional when TIMER is invoked with a void function … despite the fact that the code would never be called.
(now if there existed a really smart compiler that could both identify that it would be dead code and dead-strip it prior to flagging it as a compile time error, you'd be in luck! but i don't think gcc 4.6 is that smart … )
this leaves you with what would be a preferred option of a #if (is_void) conditional inside the #define. but that's simply not allowed. since, as this answer points out in attempting to answer a similar question about conditional preprocessing, the preprocessor is not turing-complete.
so … despite your desire to have a single macro, i think your simplest answer is to create one for void functions, and one for functions with return values.
As long as you've got typeof and _Generic, you can do also do it without __builtin_types_compatible_p or __builtin_choose_expr.
The caveat is that _Generic won't let you match void so instead of matching Expr against void, match (typeof(Expr)*){0} against void*.
Below is eldar-abusalimov example modified to use _Generic instead of __builtin_types_compatible_p and __builtin_choose_expr:
#include <stdio.h>
#define __type_is_void(expr) _Generic((typeof(expr)*){0}, void*:1, default:0)
#define __expr_or_zero(expr) _Generic((typeof(expr)*){0}, void*:0, default:(expr))
#define DO(expr) \
_Generic((typeof(expr)*){0}, \
void*:__DO_VOID(expr), \
default:__DO(__expr_or_zero(expr)))
#define __DO(expr) \
({ typeof(expr) __ret; puts("do nonvoid"); __ret = (expr); __ret; })
#define __DO_VOID(expr) \
(void)({ puts("do void"); (void)(expr); })
void foo(void) { }
int bar(void) { return 1; }
int main(void)
{
DO(foo());
DO(bar());
return 0;
}
If you really need to return from macro, use inline function instead.

Using and returning output in C macro

I'm trying to instrument some code to catch and print error messages. Currently I'm using a macro somethng like this:
#define my_function(x) \
switch(function(x)) { \
case ERROR: \
fprintf(stderr, "Error!\n"); \
break; \
}
Normally, I never capture the function output and this works fine. But I've found a couple cases where I also need the return value of function(). I tried something like the following, but this produces a syntax error.
#define my_function(x) \
do { \
int __err = function(x); \
switch(__err) { \
case ERROR: \
fprintf(stderr, "Error!\n"); \
break; \
} \
__err; \
} while(0)
I could declare a global variable to hold the return value of the function, but that looks ugly and my program is multithreaded, so that's likely to cause problems. I'm hoping there's a better solution out there.
GCC has a feature called statement expressions
So if define macro like
#define FOO(A) ({int retval; retval = do_something(A); retval;})
then you will be able to use it like
foo = FOO(bar);
This is relatively complicated code, there is not much reason to have it in a macro. Make it inline (C99) or static (C89) or both if you really want to place it in a header file. With any reasonable compiler this then should result in the same efficiency as a macro.
A very late reply. But none the less. I agree inline functions are better but MACROs do offer some pretty printing fun you can't get with inline functions. I agree with #qrdl that you can indeed use statement expressions had you restructured your statements a bit. Here is how it would work with a macro -
#define my_function(x, y) ({ \
int __err = 0; \
do { \
__err = function(x, y); \
switch(__err) { \
case ERROR: \
fprintf(stderr, "Error!\n"); \
break; \
} \
} while(0); \
__err; \
})
Sorry, this is an edit...
I think you just need the curly braces. No need for the do..while keywords
Make sure that the backslashes are the last characters on each line (no space after).
If you need to get the err value out of the macro, you can just add a parameter
Like so:
#define my_function(x, out) \
{ \
int __err = function(x); \
switch(__err) { \
case ERROR: \
fprintf(stderr, "Error!\n"); \
break; \
} \
__err; \
(*(out)) = _err; \
}
To preserve the pass-by-reference C paradigm, you should call my_function this way:
int output_err;
my_function(num, &output_err);
This way, later, if you decide to make my_function a real function, you don't need to change the call references.
Btw, qrdl's "Statement Expressions" is also a good way to do it.
there is no need to declare variable if your function is returning something then you can directly get that value. For example:
#define FOO(A) do_something(A)
Here do_something returns some integer. Then you can easily use it like:
int a = FOO(a);

Function-like macro definition in C

I'd like to define a function like MACRO . i.e.
#define foo(x)\
#if x>32\
x\
#else\
(2*x)\
#endif
that is,
if x>32, then foo(x) present x
else, foo(x) present (2*x)
but my GCC complains about:
int a = foo(31);
I think C preprocessor should be handle this correctly. since at compile time, it knows x=33. it could replace foo(33) with (2*33)
You can as follows
#define foo(x) ((x) > 32 ? (x) : (2 * (x)))
But that evaluates x multiple times. You can instead create a static function, which is cleaner
static int foo(int x) {
if(x > 32)
return x;
return 2 * x;
}
Then you are also able to pass things to foo that have side effects, and have the side effect happen only one time.
What you have written is using the #if, #else and #endif preprocessor directives, but you need to use language constructs if you pass variables to the macro and want to evaluate their values. Using if, and else statements as in the actual language constructs don't work either, because control flow statements don't evaluate to values. In other words, an if statement is steering control flow only ("if A, then execute B, else execute C"), not evaluating to any values.
#define \
foo(x) \
({ \
int xx = (x); \
int result = (xx > 32) ? xx : (2*xx); \
result; \
})
int a = foo(31);
Expands out to
int a = if 31>32
31
else
(2*31)
endif;
That's how C macros work, via simple, dumb substitution. If you expect gcc to do anything more complex or intelligent with them, then your expectation is erroneous.
Given that, it's easy to see why your code won't work. An alternative that would suffice for this example would be:
#define foo(x) (x > 32 ? x : 2*x)
On the other hand, I would question whether macros are really the appropriate tool for such a thing to begin with. Just put it in the function and the compiler will inline the code if it thinks it will speed it up.
Consider:
int x = rand()
int y = foo( x );
x is not known at compile time.
The problem is not about the theory: provided that you, for some reason, want to have a macro that expands differently according to the value of a parameter passed to it, and this parameter is a constant, known to the macro preprocessor, there's no reason why it couldn't work... for a generic macro processor... But cpp unluckly does not allow the presence of other macro processor "commands" into a macro definition...
So your
#define foo(x) \
#if x>32 \
x \
#else \
2*x \
#endif
does not expand to
#if X>32
X
#else
2*X
#endif
where X is the known parameter (so change X to e.g. 31), which requires another pass by the preprocessor.
Moreover newlines are ignored, while they are important for such an use; otherwise, the following could be considered as a trick (that need another preprocessing pass however)
#define foo(x,y) \
y if x>32 \
x \
y else \
2*x \
y endif
that with foo(20,#) produces
# if 20>32 20 # else 2*20 # endif
which would work, if it would be
# if 20>32
20
# else
2*20
# endif
... but it is not (and as said, the output of the preprocessor must be feeded to the preprocessor again...)
So my answer is that if you need these things, you can't use the C preprocessor; you should use a uncommon (not standard?) C preprocessor, or just another macro processor, and if you need the sort of things that "cpp" has to "integrate" itself with C, then you can't use a generic one (like M4) so easily...

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