89
votes

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.

10
Can't you cast both of them to a double (and add 0.00)? Not sure if this is possible in C, just a suggestion.Kevin
Look in the source code, it states right there that doubleVar is a double. No need(and not possible either) to check it at runtime.Habalusa
In response to Edit #1: have you considered using function pointers (like a vtable) to solve your issue?Michael Foukarakis
If you like the sizeof methode, read that article about the tgmath implementation of gcc.quinmars
@Michael Foukarakis Would you provide an example?con-f-use

10 Answers

57
votes

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)
21
votes

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.

14
votes

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.

7
votes

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;
7
votes

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!");
}
6
votes

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).

3
votes

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

2
votes

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.

0
votes

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.

0
votes

For that purpose I have written a simple C program for that... It is in github...GitHub Link

Here how it works... First convert your double into a char string named s..

char s[50];
sprintf(s,"%.2f", yo);

Then use my dtype function to determine the type... My function will return a single character...You can use it like this...

char type=dtype(s);
//Return types are :
//i for integer
//f for float or decimals
//c for character...

Then you can use comparison to check it... That's it...