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I need to write a C macro that checks to ensure all parameters passed to it are unsigned and of the same integer type. Ex: all input params are uint8_t, or all uint16_t, or all uint32_t, or all uint64_t.
Here is how this type of checking can be done in C++: Use static_assert to check types passed to macro
Does something similar exist in C, even if only by way of a gcc extension?
Note that static asserts are available in gcc via _Static_assert. (See my answer here: Static assert in C).
This fails to work:
int a = 1;
int b = 2;
_Static_assert(__typeof__ a == __typeof__ b, "types don't match");
Error:
main.c: In function ‘main’:
main.c:23:20: error: expected expression before ‘__typeof__’
_Static_assert(__typeof__ a == __typeof__ b, "types don't match");
UPDATE:
Here's precisely how to do what I want in C++ (using a function template, static_assert, and the <type_traits> header file). I needed to learn this anyway, for comparison purposes, so I just did. Run this code for yourself here: https://onlinegdb.com/r1k-L3HSL.
#include <stdint.h>
#include <stdio.h>
#include <type_traits> // std::is_same()
// Templates: https://www.tutorialspoint.com/cplusplus/cpp_templates.htm
// Goal: test the inputs to a "C macro" (Templated function in this case in C++) to ensure
// they are 1) all the same type, and 2) an unsigned integer type
// 1. This template forces all input parameters to be of the *exact same type*, even
// though that type isn't fixed to one type! This is because all 4 inputs to test_func()
// are of type `T`.
template <typename T>
void test_func(T a, T b, T c, T d)
{
printf("test_func: a = %u; b = %u; c = %u; d = %u\n", a, b, c, d);
// 2. The 2nd half of the check:
// check to see if the type being passed in is uint8_t OR uint16_t OR uint32_t OR uint64_t!
static_assert(std::is_same<decltype(a), uint8_t>::value ||
std::is_same<decltype(a), uint16_t>::value ||
std::is_same<decltype(a), uint32_t>::value ||
std::is_same<decltype(a), uint64_t>::value,
"This code expects the type to be an unsigned integer type\n"
"only (uint8_t, uint16_t, uint32_t, or uint64_t).");
// EVEN BETTER, DO THIS FOR THE static_assert INSTEAD!
// IE: USE THE TEMPLATE TYPE `T` DIRECTLY!
static_assert(std::is_same<T, uint8_t>::value ||
std::is_same<T, uint16_t>::value ||
std::is_same<T, uint32_t>::value ||
std::is_same<T, uint64_t>::value,
"This code expects the type to be an unsigned integer type\n"
"only (uint8_t, uint16_t, uint32_t, or uint64_t).");
}
int main()
{
printf("Begin\n");
// TEST A: This FAILS the static assert since they aren't unsigned
int i1 = 10;
test_func(i1, i1, i1, i1);
// TEST B: This FAILS to find a valid function from the template since
// they aren't all the same type
uint8_t i2 = 11;
uint8_t i3 = 12;
uint32_t i4 = 13;
uint32_t i5 = 14;
test_func(i2, i3, i4, i5);
// TEST C: this works!
uint16_t i6 = 15;
uint16_t i7 = 16;
uint16_t i8 = 17;
uint16_t i9 = 18;
test_func(i6, i7, i8, i9);
return 0;
}
With just TEST A uncommented, you get this failure in the static assert since the inputs aren't unsigned:
main.cpp: In instantiation of ‘void test_func(T, T, T, T) [with T = int]’:
<span class="error_line" onclick="ide.gotoLine('main.cpp',46)">main.cpp:46:29</span>: required from here
main.cpp:32:5: error: static assertion failed: This code expects the type to be an unsigned integer type
only (uint8_t, uint16_t, uint32_t, or uint64_t).
static_assert(std::is_same<decltype(a), uint8_t>::value ||
^~~~~~~~~~~~~
with just TEST B uncommented, you get this failure to find a valid function from the template since the template expects all inputs to be the same type T:
main.cpp: In function ‘int main()’:
main.cpp:54:29: error: no matching function for call to ‘test_func(uint8_t&, uint8_t&, uint32_t&, uint32_t&)’
test_func(i2, i3, i4, i5);
^
main.cpp:26:6: note: candidate: template void test_func(T, T, T, T)
void test_func(T a, T b, T c, T d)
^~~~~~~~~
main.cpp:26:6: note: template argument deduction/substitution failed:
main.cpp:54:29: note: deduced conflicting types for parameter ‘T’ (‘unsigned char’ and ‘unsigned int’)
test_func(i2, i3, i4, i5);
^
And with just TEST C uncommented, it passes and looks like this!
Begin
test_func: a = 15; b = 16; c = 17; d = 18
References:
http://www.cplusplus.com/reference/type_traits/is_same/
https://en.cppreference.com/w/cpp/types/is_same
https://en.cppreference.com/w/cpp/language/decltype
How do I restrict a template class to certain built-in types?
Related:
Use static_assert to check types passed to macro [my own answer]
Static assert in C [my own answer]
If the most important aspect here is that you want it to fail to compile if a and b are different types, you can make use of C11's _Generic along with GCC's __typeof__ extension to manage this.
A generic example:
#include <stdio.h>
#define TYPE_ASSERT(X,Y) _Generic ((Y), \
__typeof__(X): _Generic ((X), \
__typeof__(Y): (void)NULL \
) \
)
int main(void)
{
int a = 1;
int b = 2;
TYPE_ASSERT(a,b);
printf("a = %d, b = %d\n", a, b);
}
Now if we try to compile this code, it will compile fine and everybody is happy.
If we change the type of b to unsigned int, however, it will fail to compile.
This works because _Generic selection uses the type of a controlling expression ((Y) in this case) to select a rule to follow and insert code corresponding to the rule. In this case, we only provided a rule for __typeof__(X), thus if (X) is not a compatible type for (Y), there is no suitable rule to select and therefore cannot compile. To handle arrays, which have a controlling expression that will decay to a pointer, I added another _Generic that goes the other way ensuring they must both be compatible with one another rather than accepting one-way compatibility. And since--as far as I particularly cared--we only wanted to make sure it would fail to compile on a mismatch, rather than execute something particular upon a match, I gave the corresponding rule the task of doing nothing: (void)NULL
There is a corner case where this technique stumbles: _Generic does not handle Variably Modifiable types since it is handled at compile time. So if you attempt to do this with a Variable Length Array, it will fail to compile.
To handle your specific use-case for fixed-width unsigned types, we can modify the nested _Generic to handle that rather than handling the pecularities of an array:
#define TYPE_ASSERT(X,Y) _Generic ((Y), \
__typeof__(X): _Generic ((Y), \
uint8_t: (void)NULL, \
uint16_t: (void)NULL, \
uint32_t: (void)NULL, \
uint64_t: (void)NULL \
) \
)
Example GCC error when passing non-compatible types:
main.c: In function 'main':
main.c:7:34: error: '_Generic' selector of type 'signed char' is not compatible with any association
7 | __typeof__(X): _Generic ((Y), \
| ^
It is worth mentioning that __typeof__, being a GCC extension, will not be a solution that is portable to all compilers. It does seem to work with Clang, though, so that's another major compiler supporting it.
What you want is doable in standard C11, no extensions or GCC required.
We'll build up to the final answer, so all can follow.
According to the C11 standard [6.7.10], static_assert-declaration: _Static_assert( constant-expression , string-literal ) is a Declaration. Thus if we are going to use a macro, we had best provide a scope for a declaration, to keep things tidy. Typically of the usual form:
#define MY_AMAZING_MACRO() do {_Static_assert(...some magic...);} while(0)
Next, so that our _Static_assert within the macro at least repeats via stdio the actual issue if the assert fails, well use familiar stringification setup:
#define STATIC_ASSERT_H(x) _Static_assert(x, #x)
#define STATIC_ASSERT(x) STATIC_ASSERT_H(x)
Next, we'll use C11's Generic selection feature to declare a macro that returns a constant 1 if the object is of the type we're looking for, and zero otherwise:
#define OBJ_IS_OF_TYPE(Type, Obj) _Generic(Obj, Type: 1, default: 0)
Next we''l make a macro to test if all four of your inputs are of the same type:
#define ALL_OBJS_ARE_OF_TYPE(Type, Obj_0, Obj_1, Obj_2, Obj_3) \
(OBJ_IS_OF_TYPE(Type, Obj_0) && \
OBJ_IS_OF_TYPE(Type, Obj_1) && \
OBJ_IS_OF_TYPE(Type, Obj_2) && \
OBJ_IS_OF_TYPE(Type, Obj_3))
Next, using the above, well make a macro to test if all four of your inputs are further one of the four types:
#define IS_ACCEPTABLE(Type_0, Type_1, Type_2, Type_3, Obj_0, Obj_1, Obj_2, Obj_3) \
(ALL_OBJS_ARE_OF_TYPE(Type_0, Obj_0, Obj_1, Obj_2, Obj_3) || \
ALL_OBJS_ARE_OF_TYPE(Type_1, Obj_0, Obj_1, Obj_2, Obj_3) || \
ALL_OBJS_ARE_OF_TYPE(Type_2, Obj_0, Obj_1, Obj_2, Obj_3) || \
ALL_OBJS_ARE_OF_TYPE(Type_3, Obj_0, Obj_1, Obj_2, Obj_3))
And FINALLY, putting it all together:
#define TEST_FUNC(a,b,c,d) \
do \
{ \
STATIC_ASSERT(IS_ACCEPTABLE(uint8_t, uint16_t, uint32_t, uint64_t, \
a, b, c, d)); \
} while(0)
Of course, you could separate the above into more distinct, individual STATIC_ASSERTs, as you wish, if you want more verbose error output if any of the _Static_asserts fail.
To make some code compile in C and C++ I use this in a few places:
#ifdef __cplusplus
#define typeof(x) decltype(x) // works ok in most cases, except C++ reference types
#endif
char* a = (typeof(a)) malloc(4);
In C, this compiles to char* a = (char *) malloc(4) where the cast is totally unecessary, but in C++ void * is not implicitly promoted to char * and an error is issued if a cast is not present.
This is just as well when I can compile with -std=gnu11 on GCC or Clang, but what when I want to make my code compile as ISO C11? I thought I could use C11's _Generic to implement typeof(x) to cast some types:
#define gettype(x) _Generic((x), \
short: (short ), \
char: (char ), \
char*: (char *), \
default: (void *) )
int main (void) {
short a = (gettype(a)) 1;
return a;
}
But no matter what type defined in gettype(x) is given in a's declaration,
typeof.h: In function ‘main’:
typeof.h:2:24: error: expected expression before ‘,’ token
short: (short ), \
^
typeof.h:8:13: note: in expansion of macro ‘gettype’
char a = (gettype(a)) 1;
^~~~~~~
typeof.h:8:25: error: expected ‘,’ or ‘;’ before numeric constant
char a = (gettype(a)) 1;
gcc -E says that line expands just fine:
short a = (_Generic((a), short: (short ), char: (char ), char*: (char *), default: (void *) )) 1; ^
Is there some syntax I am missing, or is it simply not possible in C to generate cast code using _Generic?
The problem is that you can't have a partial expression inside the generic selection. A possible work-around could be to put a full expression inside it:
#define cast(from, to) _Generic((from), \
short: (short) (to), \
char: (char) (to), \
char*: (char*) (to), \
default: (void*) (to))
int main (void) {
short a = cast(a, 1);
return 0;
}
No, it's not possible. (Now watch someone prove me wrong!)
In a _Generic expression, each generic-association is either
type-name : assignment-expression
or
default : assignment-expression
It can't be a type name or something that expands to a type name. In particular, though a _Generic expression is resolved at compile time, it is not a macro. The final result is always an expression.
And I don't believe there's any other way to do what you want in standard C.
I just figured out..if Visual C++, in C, not C++, if you have two unrelated non-void pointer result types in a ternary expression, the type of the ternary expression is the first.
This can be useful.
So, in a corner I am painted into where I have a bunch of C code, I need to cast a void* to the type of something else, within a macro, that should not double evaluate...
typedef struct DesiredType { ... } DesiredType;
typedef struct ArbitraryType { ... } ArbitraryType;
ArbitraryType/*void*/* function_to_void_double_eval (void* a)
{
...
}
#if defined(_MSC_VER) && !defined(__cplusplus)
#define MACRO(x) (0 ? (DesiredType*)0 : function_to_avoid_double_eval(x))
#else // assume gcc
use typeof and temporaries in a macro
#endif
I have a function that I need to macro'ize. The function contains temp variables and I can't remember if there are any rules about use of temporary variables in macro substitutions.
long fooAlloc(struct foo *f, long size)
{
long i1, i2;
double *data[7];
/* do something */
return 42;
}
MACRO Form:
#define ALLOC_FOO(f, size) \
{\
long i1, i2;\
double *data[7];\
\
/* do something */ \
}
Is this ok? (i.e. no nasty side effect - other than the usual ones : not "type safe" etc). BTW, I know "macros are evil" - I simply have to use it in this case - not much choice.
There are only two conditions under which it works in any "reasonable" way.
The macro doesn't have a return statement. You can use the do while trick.
#define macro(x) do { int y = x; func(&y); } while (0)
You only target GCC.
#define min(x,y) ({ int _x = (x), _y = (y); _x < _y ? _x : _y; })
It would help if you explain why you have to use a macro (does your office have "macro mondays" or something?). Otherwise we can't really help.
C macros are only (relatively simple) textual substitutions.
So the question you are maybe asking is: can I create blocks (also called compound statements) in a function like in the example below?
void foo(void)
{
int a = 42;
{
int b = 42;
{
int c = 42;
}
}
}
and the answer is yes.
Now as #DietrichEpp mentioned it in his answer, if the macro is a compound statement like in your example, it is a good practice to enclose the macro statements with do { ... } while (0) rather than just { ... }. The link below explains what situation the do { ... } while (0) in a macro tries to prevent:
http://gcc.gnu.org/onlinedocs/cpp/Swallowing-the-Semicolon.html
Also when you write a function-like macro always ask yourself if you have a real advantage of doing so because most often writing a function instead is better.
First, I strongly recommend inline functions. There are very few things macros can do and they can't, and they're much more likely to do what you expect.
One pitfall of macros, which I didn't see in other answers, is shadowing of variable names.
Suppose you defined:
#define A(x) { int temp = x*2; printf("%d\n", temp); }
And someone used it this way:
int temp = 3;
A(temp);
After preprocessing, the code is:
int temp = 3;
{ int temp = temp*2; printf("%d\n", temp); }
This doesn't work, because the internal temp shadows the external.
The common solution is to call the variable __temp, assuming nobody will define a variable using this name (which is a strange assumption, given that you just did it).
This is mostly OK, except that macros are usually enclosed with do { ... } while(0) (take a look at this question for explanations):
#define ALLOC_FOO(f, size) \
do { \
long i1, i2;\
double *data[7];\
/* do something */ \
} while(0)
Also, as far as your original fooAlloc function returns long you have to change your macro to store the result somehow else. Or, if you use GCC, you can try compound statement extension:
#define ALLOC_FOO(f, size) \
({ \
long i1, i2;\
double *data[7];\
/* do something */ \
result; \
})
Finally you should care of possible side effects of expanding macro argument. The usual pattern is defining a temporary variable for each argument inside a block and using them instead:
#define ALLOC_FOO(f, size) \
({ \
typeof(f) _f = (f);\
typeof(size) _size = (size);\
long i1, i2;\
double *data[7];\
/* do something */ \
result; \
})
Eldar's answer shows you most of the pitfalls of macro programming and some useful (but non standard) gcc extension.
If you want to stick to the standard, a combination of macros (for genericity) and inline functions (for the local variables) can be useful.
inline
long fooAlloc(void *f, size_t size)
{
size_t i1, i2;
double *data[7];
/* do something */
return 42;
}
#define ALLOC_FOO(T) fooAlloc(malloc(sizeof(T)), sizeof(T))
In such a case using sizeof only evaluates the expression for the type at compile time and not for its value, so this wouldn't evaluate F twice.
BTW, "sizes" should usually be typed with size_t and not with long or similar.
Edit: As to Jonathan's question about inline functions, I've written up something about the inline model of C99, here.
Yes it should work as you use a block structure and the temp variables are declared in the inner scope of this block.
Note the last \ after the } is redundant.
A not perfect solution: (does not work with recursive macros, for example multiple loops inside each other)
#define JOIN_(X,Y) X##Y
#define JOIN(X,Y) JOIN_(X,Y)
#define TMP JOIN(tmp,__LINE__)
#define switch(x,y) int TMP = x; x=y;y=TMP
int main(){
int x = 5,y=6;
switch(x,y);
switch(x,y);
}
will become after running the preprocessor:
int main(){
int x=5,y=6;
int tmp9 = x; x=y; y=tmp9;
int tmp10 = x; x=y; y=tmp10;
}
They can. They often shouldn't.
Why does this function need to be a macro? Could you inline it instead?
If you're using c++ use inline, or use -o3 with gcc it will inline all functions for you.
I still don't understand why you need to macroize this function.
In C (not C++/C#) how do I check if a variable is of a certain type?
For example, something like this:
double doubleVar;
if( typeof(doubleVar) == double ) {
printf("doubleVar is of type double!");
}
Or more general: How do I compare two types so that compare(double1,double2) will evaluate to true, and compare(int,double) will evaluate to false. Also I'd like to compare structs of different composition as well.
Basically, I have a function that operates on variables of type "struct a" and "struct b". I want to do one thing with the "struct a" variables and the other with the "struct b" variables. Since C doesn't support overloading and the void pointer losses its type information I need to check for type. BTW, what would be the sense in having a typeof operator, if you can't compare types?
The sizeof method seems to be a practical workaround solution for me. Thanks for your help. I still find it a bit strange since the types are known at compile time, but if I imagine the processes in the machine I can see, why the information is not stored in terms of types, but rather in terms of byte size. Size is the only thing really relevant besides addresses.
Getting the type of a variable is, as of now, possible in C11 with the _Generic generic selection. It works at compile-time.
The syntax is a bit like that for switch. Here's a sample (from this answer):
#define typename(x) _Generic((x), \
_Bool: "_Bool", unsigned char: "unsigned char", \
char: "char", signed char: "signed char", \
short int: "short int", unsigned short int: "unsigned short int", \
int: "int", unsigned int: "unsigned int", \
long int: "long int", unsigned long int: "unsigned long int", \
long long int: "long long int", unsigned long long int: "unsigned long long int", \
float: "float", double: "double", \
long double: "long double", char *: "pointer to char", \
void *: "pointer to void", int *: "pointer to int", \
default: "other")
To actually use it for compile-time manual type checking, you can define an enum with all of the types you expect, something like this:
enum t_typename {
TYPENAME_BOOL,
TYPENAME_UNSIGNED_CHAR,
TYPENAME_CHAR,
TYPENAME_SIGNED_CHAR,
TYPENAME_SHORT_INT,
TYPENAME_UNSIGNED_CHORT_INT,
TYPENAME_INT,
/* ... */
TYPENAME_POINTER_TO_INT,
TYPENAME_OTHER
};
And then use _Generic to match types to this enum:
#define typename(x) _Generic((x), \
_Bool: TYPENAME_BOOL, unsigned char: TYPENAME_UNSIGNED_CHAR, \
char: TYPENAME_CHAR, signed char: TYPENAME_SIGNED_CHAR, \
short int: TYPENAME_SHORT_INT, unsigned short int: TYPENAME_UNSIGNED_SHORT_INT, \
int: TYPENAME_INT, \
/* ... */ \
int *: TYPENAME_POINTER_TO_INT, \
default: TYPENAME_OTHER)
C does not support this form of type introspection. What you are asking is not possible in C (at least without compiler-specific extensions; it would be possible in C++, however).
In general, with C you're expected to know the types of your variable. Since every function has concrete types for its parameters (except for varargs, I suppose), you don't need to check in the function body. The only remaining case I can see is in a macro body, and, well, C macros aren't really all that powerful.
Further, note that C does not retain any type information into runtime. This means that, even if, hypothetically, there was a type comparison extension, it would only work properly when the types are known at compile time (ie, it wouldn't work to test whether two void * point to the same type of data).
As for typeof: First, typeof is a GCC extension. It is not a standard part of C. It's typically used to write macros that only evaluate their arguments once, eg (from the GCC manual):
#define max(a,b) \
({ typeof (a) _a = (a); \
typeof (b) _b = (b); \
_a > _b ? _a : _b; })
The typeof keyword lets the macro define a local temporary to save the values of its arguments, allowing them to be evaluated only once.
In short, C does not support overloading; you'll just have to make a func_a(struct a *) and func_b(struct b *), and call the correct one. Alternately, you could make your own introspection system:
struct my_header {
int type;
};
#define TYPE_A 0
#define TYPE_B 1
struct a {
struct my_header header;
/* ... */
};
struct b {
struct my_header header;
/* ... */
};
void func_a(struct a *p);
void func_b(struct b *p);
void func_switch(struct my_header *head);
#define func(p) func_switch( &(p)->header )
void func_switch(struct my_header *head) {
switch (head->type) {
case TYPE_A: func_a((struct a *)head); break;
case TYPE_B: func_b((struct b *)head); break;
default: assert( ("UNREACHABLE", 0) );
}
}
You must, of course, remember to initialize the header properly when creating these objects.
As other people have already said this isn't supported in the C language. You could however check the size of a variable using the sizeof() function. This may help you determine if two variables can store the same type of data.
Before you do that, read the comments below.
Gnu GCC has a builtin function for comparing types __builtin_types_compatible_p.
https://gcc.gnu.org/onlinedocs/gcc-3.4.5/gcc/Other-Builtins.html
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.
Used in your example:
double doubleVar;
if(__builtin_types_compatible_p(typeof(doubleVar), double)) {
printf("doubleVar is of type double!");
}
As others have mentioned, you can't extract the type of a variable at runtime. However, you could construct your own "object" and store the type along with it. Then you would be able to check it at runtime:
typedef struct {
int type; // or this could be an enumeration
union {
double d;
int i;
} u;
} CheesyObject;
Then set the type as needed in the code:
CheesyObject o;
o.type = 1; // or better as some define, enum value...
o.u.d = 3.14159;
As another answer mentioned, you can now do this in C11 with _Generic.
For example, here's a macro that will check if some input is compatible with another type:
#include <stdbool.h>
#define isCompatible(x, type) _Generic(x, type: true, default: false)
You can use the macro like so:
double doubleVar;
if (isCompatible(doubleVar, double)) {
printf("doubleVar is of type double!\n"); // prints
}
int intVar;
if (isCompatible(intVar, double)) {
printf("intVar is compatible with double too!\n"); // doesn't print
}
This can also be used on other types, including structs. E.g.
struct A {
int x;
int y;
};
struct B {
double a;
double b;
};
int main(void)
{
struct A AVar = {4, 2};
struct B BVar = {4.2, 5.6};
if (isCompatible(AVar, struct A)) {
printf("Works on user-defined types!\n"); // prints
}
if (isCompatible(BVar, struct A)) {
printf("And can differentiate between them too!\n"); // doesn't print
}
return 0;
}
And on typedefs.
typedef char* string;
string greeting = "Hello world!";
if (isCompatible(greeting, string)) {
printf("Can check typedefs.\n");
}
However, it doesn't always give you the answer you expect. For instance, it can't distinguish between an array and a pointer.
int intArray[] = {4, -9, 42, 3};
if (isCompatible(intArray, int*)) {
printf("Treats arrays like pointers.\n");
}
// The code below doesn't print, even though you'd think it would
if (isCompatible(intArray, int[4])) {
printf("But at least this works.\n");
}
Answer borrowed from here: http://www.robertgamble.net/2012/01/c11-generic-selections.html
From linux/typecheck.h:
/*
* Check at compile time that something is of a particular type.
* Always evaluates to 1 so you may use it easily in comparisons.
*/
#define typecheck(type,x) \
({ type __dummy; \
typeof(x) __dummy2; \
(void)(&__dummy == &__dummy2); \
1; \
})
Here you can find explanation which statements from standard and which GNU extensions above code uses.
(Maybe a bit not in scope of the question, since question is not about failure on type mismatch, but anyway, leaving it here).
This is crazily stupid, but if you use the code:
fprintf("%x", variable)
and you use the -Wall flag while compiling, then gcc will kick out a warning of that it expects an argument of 'unsigned int' while the argument is of type '____'. (If this warning doesn't appear, then your variable is of type 'unsigned int'.)
Best of luck!
Edit: As was brought up below, this only applies to compile time. Very helpful when trying to figure out why your pointers aren't behaving, but not very useful if needed during run time.
As of C2x, typeof is now a part of the language's standard. This allows the creation of a macro that compares the types of two values:
#include <stdlib.h>
#include <stdio.h>
#include <stdbool.h>
#define sametypeof(A,B) _Generic(A, typeof(B): true, default: false)
int main() {
if (sametypeof(1, 2)) {
printf("1 and 2 have the same type.\n");
} else {
printf("1 and 2 don't have the same type.\n");
}
}
(This compiles with the latest experimental version of GCC 13, using the -std=c2x flag)
If you want to compare between two types, you can use the following workaround:
#include <stdlib.h>
#include <stdio.h>
#include <stdbool.h>
#define sametype(A,B) _Generic(*((A*)0), B: true, default: false)
int main() {
if (sametype(void*, nullptr_t)) {
printf("void* and nullptr_t are the same type.\n");
} else {
printf("void* and nullptr_t are not the same type.\n");
}
}
Although *((A*)0) is not valid code at runtime, the compiler will still be able to deduce its type as A, so it will work in _Generic, as the code itself will not run and will be discarded. (as far as I remember, this trick has worked in every C11 compliant compiler I've used, including Clang and the Tiny C Compiler)
(you also cannot just do (A)0 because 0 cannot be cast to a struct)
C is statically typed language. You can't declare a function which operate on type A or type B, and you can't declare variable which hold type A or type B. Every variable has an explicitly declared and unchangeable type, and you supposed to use this knowledge.
And when you want to know if void * points to memory representation of float or integer - you have to store this information somewhere else. The language is specifically designed not to care if char * points to something stored as int or char.
One possible way is to have your variables names prepend your variable definitions with the type information.
Ex:
All integers will have i_
All floats will have f_
etc..
The variable name can be got out by the #<variable_name>,
This
There is a built-in function in GCC.
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.
i've searched a solution to solve the issue of controlling data type for while , and i thought that maybe my founding could add up well with the initial demand #con-f-use, even if it's no exactly the same issue.
An other way around to control the datatype could be done using an union with predefined type. In my case, i had a defined structure in which i was originally using a void* to allow divers data type to be passed :
originally:
//[main]:
uint32_t vtest3= 100000;
int32_t vtest2= 100000;
struct entity list[] = {
{ TYPE_INT32, s_int32_t, .label="tension", &vtest3},
{ TYPE_INT32, s_int32_t, .label="tension", &vtest3}
};
//[file.h]:
struct entity {
enum entity_type type;
uint32_t dimension;
char* label;
void* ptr_data;
uint32_t offset;
};
enum entity_type {
TYPE_NONE = 0,
TYPE_INT8 = 1,
TYPE_INT16 = 2,
TYPE_INT32 = 3,
TYPE_INT64 = 4,
TYPE_UINT8 = 5,
TYPE_UINT16 = 6,
TYPE_UINT32 = 7,
TYPE_UINT64 = 8,
TYPE_FLOAT32 = 9
};
The issue with this method is that it accept all type of variable in an uncontrolled way. There is no easy method to control the data type referenced by the void* pointer, Excepted maybe thought the use of a macro and _Generic as described before in this thread.
If the programmer decided to pass a type different from the list of type accepted ,there while be no error thrown at compile time.
. They other way around is by replacing the void* by an union , this way the structure while only accept specific data type defined inside the union list . If the programmer decide to pass a pointer with an type which is not already defined inside the ptr_data union{...} , it will throw an error.
//[file.h]:
enum entity_type {
TYPE_NONE = 0,
TYPE_INT8 = 1,
TYPE_INT16 = 2,
TYPE_INT32 = 3,
TYPE_INT64 = 4,
TYPE_UINT8 = 5,
TYPE_UINT16 = 6,
TYPE_UINT32 = 7,
TYPE_UINT64 = 8,
TYPE_FLOAT32 = 9
};
struct entity {
enum entity_type type;
uint32_t dimension;
char* label;
union {
uint8_t *uint8;
uint16_t *uint16;
uint32_t *uint32;
uint32_t *uint;
int16_t *int16;
int32_t *int32;
int64_t *int64;
float *f;
} ptr_data;
uint32_t offset;
};
[main:]
uint32_t vtest3= 100000;
int32_t vtest2= 100000;
struct entity list[] = {
{ TYPE_INT32, s_int32_t, .label="a", .ptr_data = {.uint16=&vtest1}
},
{ TYPE_INT32, s_int32_t, .label="b", .ptr_data = {.int32=&vtest2}
};
This method make use of the union to control implicitly the data type of the variable inserted by the programmer in the structure. If not correct the compiler while throw an error at compile time.
Obviously this code example is far from perfect and cannot be used directly but i tried to explain in a way as clear as possible the logic and the the idea that i proposed ;)
Is it possible to typecheck arguments to a #define macro? For example:
typedef enum
{
REG16_A,
REG16_B,
REG16_C
}REG16;
#define read_16(reg16) read_register_16u(reg16); \
assert(typeof(reg16)==typeof(REG16));
The above code doesn't seem to work. What am I doing wrong?
BTW, I am using gcc, and I can guarantee that I will always be using gcc in this project. The code does not need to be portable.
gcc supports typeof
e.g. a typesafe min macro taken from the linux kernel
#define min(x,y) ({ \
typeof(x) _x = (x); \
typeof(y) _y = (y); \
(void) (&_x == &_y); \
_x < _y ? _x : _y; })
but it doesn't allow you to compare two types. Note though the pointer comparison which Will generate a warning - you can do a typecheck like this (also from the linux kernel)
#define typecheck(type,x) \
({ type __dummy; \
typeof(x) __dummy2; \
(void)(&__dummy == &__dummy2); \
1; \
})
Presumably you could do something similar - i.e. compare pointers to the arguments.
The typechecking in C is a bit loose for integer-related types; but you can trick the compiler by using the fact that most pointer types are incompatible.
So
#define CHECK_TYPE(var,type) { __typeof(var) *__tmp; __tmp = (type *)NULL; }
This will give a warning, "assignment from incompatible pointer type" if the types aren't the same. For example
typedef enum { A1,B1,C1 } my_enum_t;
int main (int argc, char *argv) {
my_enum_t x;
int y;
CHECK_TYPE(x,my_enum_t); // passes silently
CHECK_TYPE(y,my_enum_t); // assignment from incompatible pointer type
}
I'm sure that there's some way to get a compiler error for this.
This is an old question, But I believe I have a general answer that according to Compiler Explorer apears to work on MSVC, gcc and clang.
#define CHECK_TYPE(type,var) { typedef void (*type_t)(type); type_t tmp = (type_t)0; if(0) tmp(var);}
In each case the compiler generates a useful error message if the type is incompatible. This is because it imposes the same type checking rules used for function parameters.
It can even be used multiple times within the same scope without issue. This part surprises me somewhat. (I thought I would have to utilize "__LINE__" to get this behavior)
Below is the complete test I ran, commented out lines all generate errors.
#include <stdio.h>
#define CHECK_TYPE(type,var) { typedef void (*type_t)(type); type_t tmp = (type_t)0; if(0) tmp(var);}
typedef struct test_struct
{
char data;
} test_t;
typedef struct test2_struct
{
char data;
} test2_t;
typedef enum states
{
STATE0,
STATE1
} states_t;
int main(int argc, char ** argv)
{
test_t * var = NULL;
int i;
states_t s;
float f;
CHECK_TYPE(void *, var); //will pass for any pointer type
CHECK_TYPE(test_t *, var);
//CHECK_TYPE(int, var);
//CHECK_TYPE(int *, var);
//CHECK_TYPE(test2_t, var);
//CHECK_TYPE(test2_t *, var);
//CHECK_TYPE(states_t, var);
CHECK_TYPE(int, i);
//CHECK_TYPE(void *, i);
CHECK_TYPE(int, s); //int can be implicitly used instead of enum
//CHECK_TYPE(void *, s);
CHECK_TYPE(float, s); //MSVC warning only, gcc and clang allow promotion
//CHECK_TYPE(float *, s);
CHECK_TYPE(float, f);
//CHECK_TYPE(states_t, f);
printf("hello world\r\n");
}
In each case the compiler with -O1 and above did remove all traces of the macro in the resulting code.
With -O0 MSVC left the call to the function at zero in place, but it was rapped in an unconditional jump which means this shouldn't be a concern. gcc and clang with -O0 both remove everything except for the stack initialization of the tmp variable to zero.
No, macros can't provide you any typechecking. But, after all, why macro? You can write a static inline function which (probably) will be inlined by the compiler - and here you will have type checking.
static inline void read_16(REG16 reg16) {
read_register_16u(reg16);
}
Building upon Zachary Vander Klippe's answer, we might even go a step further (in a portable way, even though that wasn't a requirement) and additionally make sure that the size of the passed-in type matches the size of the passed-in variable using the "negative array length" trick that was commonly used for implementing static assertions in C (prior to C11, of course, which does provide the new _Static_assert keyword).
As an added benefit, let's throw in some const compatibility.
#define CHECK_TYPE(type,var) \
do {\
typedef void (*type_t) (const type);\
type_t tmp = (type_t)(NULL);\
typedef char sizes[((sizeof (type) == sizeof (var)) * 2) - 1];\
if (0) {\
const sizes tmp2;\
(void) tmp2;\
tmp (var);\
}\
} while (0)
Referencing the new typedef as a variable named tmp2 (and, additionally, referencing this variable, too) is just a method to make sure that we don't generate more warnings than necessary, c.f., -Wunused-local-typedefs and the like. We could have used __attribute__ ((unused)) instead, but that is non-portable.
This will work around the integer promotion "issue" in the original example.
Example in the same spirit, failing statements are commented out:
#include <stdio.h>
#include <stdlib.h>
#define CHECK_TYPE(type,var) \
do {\
typedef void (*type_t) (const type);\
type_t tmp = (type_t)(NULL);\
typedef char sizes[((sizeof (type) == sizeof (var)) * 2) - 1];\
if (0) {\
const sizes tmp2;\
(void) tmp2;\
tmp (var);\
}\
} while (0)
int main (int argc, char **argv) {
long long int ll;
char c;
//CHECK_TYPE(char, ll);
//CHECK_TYPE(long long int, c);
printf("hello world\n");
return EXIT_SUCCESS);
}
Naturally, even that approach isn't able to catch all issues. For instance, checking signedness is difficult and often relies on tricks assuming that a specific complement variant (e.g., two's complement) is being used, so cannot be done generically. Even less so if the type can be a structure.
To continue the idea of ulidtko, take an inline function and have it return something:
inline
bool isREG16(REG16 x) {
return true;
}
With such as thing you can do compile time assertions:
typedef char testit[sizeof(isREG16(yourVariable))];
No. Macros in C are inherently type-unsafe and trying to check for types in C is fraught with problems.
First, macros are expanded by textual substitution in a phase of compilation where no type information is available. For that reason, it is utterly impossible for the compiler to check the type of the arguments when it does macro expansion.
Secondly, when you try to perform the check in the expanded code, like the assert in the question, your check is deferred to runtime and will also trigger on seemingly harmless constructs like
a = read_16(REG16_A);
because the enumerators (REG16_A, REG16_B and REG16_C) are of type int and not of type REG16.
If you want type safety, your best bet is to use a function. If your compiler supports it, you can declare the function inline, so the compiler knows you want to avoid the function-call overhead wherever possible.