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

Question

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__;                                          \
})

Answer 1

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;
}

Answer 2

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.

Answer 3

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