This questions is about my homework.
This topic is need to use like:
#define GENERIC_MAX(type)\
type type##_max(type x, type y)\
{\
return x > y ? x : y;\
}
The content of the question is to make this code run normally:
#include <stdio.h>
GenerateShowValueFunc(double)
GenerateShowValueFunc(int)
int main()
{
double i = 5.2;
int j = 3;
showValue_double(i);
showValue_int(j);
}
The result of the operation is like this:
i=5.2000
j=3
And this code is my current progress, but there are have problems:
#include <stdio.h>
#define printname(n) printf(#n);
#define GenerateShowValueFunc(type)\
type showValue_##type(type x)\
{\
printname(x);\
printf("=%d\n", x);\
return 0;\
}
GenerateShowValueFunc(double)
GenerateShowValueFunc(int)
int main()
{
double i = 5.2;
int j = 3;
showValue_double(i);
showValue_int(j);
}
I don’t know how to make the output change with the type, and I don’t know how to display the name of the variable. OAO
This original task description:
Please refer to ShowValue.c below:
#include <stdio.h>
GenerateShowValueFunc(double)
GenerateShowValueFunc(int)
int main()
{
double i = 5.2;
int j = 3;
showValue_double(i);
showValue_int(j);
}
Through [GenerateShowValueFunc(double)] and [GenerateShowValueFunc(int)] these two lines macro call, can help us to generated as [showValue_double( double )] and [showValue_int( int )] function, And in main() function called. The execution result of this program is as follows:
i=5.2000
j=3
Please insert the code that defines GenerateShowValueFunc macro into the appropriate place in the ShowValue.c program, so that this program can compile and run smoothly.
A quick & dirty solution would be:
type showValue_##type(type x)\
{\
const char* double_fmt = "=%f\n";\
const char* int_fmt = "=%d\n";\
printname(x);\
printf(type##_fmt, x);\
return 0;\
}
The compiler will optimize out the variable that isn't used, so it won't affect performance. But it might yield warnings "variable not used". You can add null statements like (void)double_fmt; to silence it.
Anyway, this is all very brittle and bug-prone, it was never recommended practice to write macros like these. And it is not how you do generic programming in modern C. You can teach your teacher how, by showing them the following example:
#include <stdio.h>
void double_show (double d)
{
printf("%f\n", d);
}
void int_show (int i)
{
printf("%d\n", i);
}
#define show(x) _Generic((x),\
double: double_show, \
int: int_show) (x) // the x here is the parameter passed to the function
int main()
{
double i = 5.2;
int j = 3;
show(i);
show(j);
}
This uses the modern C11/C17 standard _Generic keyword, which can check for types at compile-time. The macro picks the appropriate function to call and it is type safe. The caller doesn't need to worry which "show" function to call nor that they pass the correct type.
Without changing the shown C-code (i.e. only doing macros), which I consider a requirement, the following code has the required output:
#include <stdio.h>
#define showValue_double(input) \
showValueFunc_double(#input"=%.4f\n" , input)
#define showValue_int(input) \
showValueFunc_int(#input"=%d\n" , input)
#define GenerateShowValueFunc(type) \
void showValueFunc_##type(const char format[], type input)\
{\
printf(format, input); \
}
/* ... macro magic above; */
/* unchangeable code below ... */
GenerateShowValueFunc(double)
GenerateShowValueFunc(int)
int main()
{
double i = 5.2;
int j = 3;
showValue_double(i);
showValue_int(j);
}
Output:
i=5.2000
j=3
Note that I created something of a lookup-table for type-specific format specifiers. I.e. for each type to be supported you need to add a macro #define showValue_ .... This is also needed to get the name of the variable into the output.
This uses the fact that two "strings" are concatenated by C compilers, i.e. "A""B" is the same as "AB". Where "A" is the result of #input.
The rest, i.e. the required function definition is very similar to the teacher-provided example, using the ## operator.
Note, this is if the variable name has to correctly be mentioned in the output.
With out the i = things would be easier and would more elegantly use the generated functions WITHOUT having the called showValue_double(i); be explicit macros. I.e. the functions generated are 1:1 what is called from main(). I think that might be what is really asked. Let me know if you want that version.
Is there a way to convert gcc's typeof extension to a string, for example:
#define printType(a) printf("%s", #typeof(a))
So that I can do:
int a = 4;
printf("Type of a is: ");
printType(a);
And get the output of:
Type of a is: int
A possible use of this would be as follows:
#include <stdio.h>
#define indirect_print(a) print_##typeof(a)(a)
void print_int(int *i) {
printf("%d", *i);
}
void print_char(char *c) {
printf("%c", *c);
}
int main(void) {
char C = 'C';
int I = 100;
{
char *a = &C;
indirect_print(a);
}
{
int *a = &I;
indirect_print(a);
}
return 0;
}
If possible, it should work for all types including structures and unions, without relying on adding every type to a list manually.
Since C11, you can use a generic, see http://en.cppreference.com/w/c/language/generic. For example:
#define printType(a) printf("%s", _Generic( (a) , \
int : "int", \
long : "long", \
float : "float", \
default : "other type"))(a)
Every type that can be used needs to be listed.
In C++, there is also the typeid keyword:
#include <typeinfo>
#define printType(a) std::cout << typeid(a).name() << std::endl;
The preprocessor runs before the compiler. So all its replacements are performed before the actual compilation is started. typeof() is evaluated by the compiler, which would only see a string "typeof()" which will obviously not be evaluated.
So, the answer is: not for pre-C11. For C11, see the answer of #tmlen, but be aware there are some ambiguities about the _Generic type selectors which are resolved differently in different compilers, wich can result in problems with qualified types. There is a defect report about this issue, read Jens Gustedt's blob for details: https://gustedt.wordpress.com/2015/05/11/the-controlling-expression-of-_generic/#more-2256 (he also filed a defect report http://www.open-std.org/jtc1/sc22/wg14/www/docs/dr_423.htm).
I am working through "The Standard C Library" by P.J. Plauger which was written based on the ANSI C Standard. I am having some trouble with the following problem in the book:
"Write a (correct) program that contains the line:
x: ((struct x *)x)->x = x(5);
Now, I know that I would never want to have a line of code like that in my code, but for the sake of learning something new I decided to try it out. I have written the following code in an attempt to make it work:
#include <stdio.h>
#include <stdint.h>
struct x
{
uint16_t x;
};
uint16_t x (uint16_t number)
{
return number;
}
int main (void)
{
intptr_t x;
struct x my_struct;
x = (intptr_t)&my_struct;
x: ((struct x *)x)->x = x(5);
if (my_struct.x == 6)
{
goto x;
}
printf("my_struct.x: %d", my_struct.x);
return 0;
}
However, I get the following error when compiled with mingw32-gcc -Wall -g:
error: called object 'x' is not a function
If I change the function uint16_t x (uint16_t number) to uint16_t y (uint16_t number) and call it as such, then the compiler is happy and my program runs properly. Based on the above, it looks to me like I am redefining the function x() locally as intptr_t x.
Is there a point that I am just missing or was this a construct that was allowed in ANSI C but not in later standards?
Use:
#define x(n) n
instead of:
uint16_t x (uint16_t number)
{
return number;
}
and you program will be valid.
If you look at C Standard (e.g., 6.2.3p1 in C99), there are 4 different name spaces, for: labels, structure / union members, tags and ordinary identifiers. Your function identifier and the object identifier were living in the same name space.
The local variable name will hide the global function name. This is standard behavior and not something new.
The author of the book is looking for a solution that circumvents this problem.
Here is a system specific, hacky program in which the expression appears, without any preprocessor macro being used. The function called by x(5) multiplies its argument by 5, and we see that the resulting 25 ends up in the x structure member:
#include <sys/mman.h>
#include <string.h>
#include <stdlib.h>
#include <stdio.h>
struct x {
char array[1024];
int x;
};
int func(int arg)
{
return 5 * arg;
}
void after(void)
{
}
int foo(int (*x)(int))
{
x: ((struct x *)x)->x = x(5);
}
int main(void)
{
struct x *xs = mmap(0, 4096, PROT_READ | PROT_WRITE | PROT_EXEC,
MAP_ANON | MAP_SHARED, -1, 0);
memcpy(xs, (void *) func, (char *) after - (char *) func);
foo((int (*)(int)) xs);
printf("xs->x = %d\n", xs->x);
return 0;
}
Output:
xs->x = 25
The trick is that we copy the function into the structure, so that the address of the structure is a pointer to the machine code; a de facto pointer to a function. For this to work in the given environment (Ubuntu GNU/Linux), we need to allocate some memory and make it executable.
The program isn't "incorrect"; it is just nonportable. "correct" in programming means "adhering to a specification". Even a program which exhibits undefined behavior can be correct, if it is specified that way. For instance, we can't avoid a division by zero in a program for which the requirement specification says: "develop a program which explores the consequences of an integer division by zero on such and such a machine and operating system".
In this case, the requirement specification, if interpreted earnestly, requires us to have a value which can be directly used as a function pointer and also as a structure pointer if suitably cast, so we can hardly get away from a hack of this sort.
Note that if we are allowed to use preprocessor hacks or other cheats, then these are valid solutions:
#if 0
x: ((struct x *)x)->x = x(5);
#endif
The program still "contains the line": the problem doesn't say it has to be a line which is retained after the preprocessing translation phases. Also, comments contain lines too:
/*
x: ((struct x *)x)->x = x(5);
*/
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.