Is it not possible to use brackets when using C _Generic?

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I'm using the online compiler and debugger for C. This webpage uses GCC 5.4.1 C99. When I test this code,

https://www.onlinegdb.com/online_c_compiler

#include <stdio.h>
#include <stdint.h>
#define SIMPLE(T) _Generic( (T), char: 'x', int: 2, long: 3, default: 0)
#define BRAC(T) _Generic( (T), uint8_t*: {1, 1}, uint16_t*: {2, 2}, default: {0xFA, 0xFA})

int main(){
    int num = SIMPLE(777); // this works
    uint8_t arr[] = BRAC("DEFAULT"); // error: expected expression before ‘{’ token
    printf("num = %d\n", num); // prints 2
    printf("0x%x\n", arr[1]);
    return 0;
}

The error message is

main.c: In function ‘main’:
main.c:4:42: error: expected expression before ‘{’ token
 #define BRAC(T) _Generic( (T), uint8_t*: {1, 1}, uint16_t*: {2, 2}, default: {0xFA, 0xFA})
                                          ^
main.c:8:21: note: in expansion of macro ‘BRAC’
     uint8_t arr[] = BRAC("DEFAULT"); // error: expected expression before ‘{’ token
                 ^~~~

This compiler supports _Generic but I have this problem when using {} brackets.

How can I solve this when I want to initialize the array using the _Generic keyword?

c generics
1 Answers

_Generic does not work this way. It is not a "classic" macro. It has to understand the types and it is "expanded" to the value or code by the compiler.

#include <stdio.h>
#include <stdint.h>
#include <string.h>

#define SIMPLE(T) _Generic( (T), char: 'x', int: 2, long: 3, default: 0)
#define BRAC(T) _Generic( (T), uint8_t*: (uint8_t []){1, 1}, uint16_t*: (uint8_t []){2, 2}, default: (uint8_t []){0xFA, 0xFB})

int main(){
    int num = SIMPLE(777); // this works
    uint8_t *arr = BRAC("DEFAULT"); 
    uint8_t arr1[2];
    memcpy(arr1, BRAC("DEFAULT"), sizeof(arr1));
    printf("num = %d\n", num); // prints 2
    printf("0x%x\n", arr[1]);
    printf("0x%x\n", arr1[0]);
    return 0;
}
user_61100940
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