C braced initializer for -non- array

Question

Please consider the following absurd but compiling code:

int main(int argc, char *argv[]) {
    void *a, *b, *c, *d;

    void *the_good_array[] = { a, b, c, d };
    void *the_bad_array    = { a, b, c, d };

    return 0;
}

I accidentally produced something similar and, to my astonishment, it fully compiles.

• Why does it compile?
• What is { a, b, c, d } supposed to mean as an initializer for a non array? Without the braces, I would understand it's an accidental use of the , operator resulting in the value of the last element (d). But, braces? What are they supposed to mean here?

If possible, please refer to the C standard when answering.

Edit 1: Compiles with both:

gcc -std=iso9899:1990 /tmp/bad_array.c     # gcc (Debian 12.2.0-14)`
clang -std=iso9899:1990 /tmp/bad_array.c   # Debian clang version 14.0.6

Edit 2: Apparently

int i = { 0 };

is perfectly valid C.

I would like to know why but don't have the standard to read it.

Answer 1

What is { a, b, c, d } supposed to mean as an initializer for a non array?

Having multiple intializer elements for single object is a constraint violation. N2176 (C17 standard draft) §6.7.9 Initialization:

Constraints

2. No initializer shall attempt to provide a value for an object not contained within the entity being initialized.

In this case it's depends on compiler what happens.

Why does it compile?

C compilers take backwards compatibility quite seriosly, and will accept a lot of code that would be considered invalid nowadays.

But, braces? What are they supposed to mean here?

Braces are allowed allowed even around single expression:

Syntax

1. • initializer:
     • assignment-expression
     • { initializer-list }
     • { initializer-list , }
   • initializer-list:
     • designation_opt initializer
     • initializer-list , designation_opt initializer
       ...

With rules above, initializer can after a few steps be expanded to { assignment-expression }.