C convert array to a arguments for a function

Question

I have a list of arguments of different types:

value_info args_array[10];

value_info is defined as follows:

typedef int int_type;
typedef float float_type;
typedef double double_type;
typedef bool bool_type;
typedef char *string_type;

typedef enum {
    INT_TYPE,
    FLOAT_TYPE,
    DOUBLE_TYPE,
    BOOL_TYPE,
    STRING_TYPE,
} data_type;

typedef struct {
    data_type type;
    union {
        int_type val_int;
        float_type val_float;
        double_type val_double;
        bool_type val_bool;
        string_type val_string;
        function_info val_function;
    } value;
} value_info;

I need to call functions from the following table:

#define MAX_ARGS 10

typedef struct {
    union {
        int (*f_int)();
        float (*f_float)();
        double (*f_double)();
        bool (*f_bool)();
        char *(*f_string)();
    } function;
    int argc;
    data_type argv[MAX_ARGS];
    data_type ret_type;
} function_info;

struct init {
    char *fname;
    function_info fnct;
};

struct init arith_fncts[] = {
    {"sin", {.function.f_double = sin, 1, {DOUBLE_TYPE}, DOUBLE_TYPE}},
    {"cos", {.function.f_double = cos, 1, {DOUBLE_TYPE}, DOUBLE_TYPE}},
    {"tan", {.function.f_double = tan, 1, {DOUBLE_TYPE}, DOUBLE_TYPE}}};

I woud like to call the function by string from the arith_fncts passing the correct number of arguments from the value_info array to the function.

I have tried using the _Generic feature of C and writing a macro like this:

#define run(func, ...)                                                 \
    _Generic(&(__VA_ARGS__),                                           \
        int(*)[3]: _Generic((func), void (*)(int, int, int): (func)))( \
        __VA_ARGS__[0], __VA_ARGS__[1], __VA_ARGS__[2])

But this only works for array with 3 arguments, i would like it to be generic and aplicable to as many arguments as I want. If that is not possible, at least for 10 arguments would be sufficient. How can I achieve this in C?