What is the meaning of casting an integer number wrapped by curly braces in C language?

Question

Happened to see a usage like this:

uint8_t a = 0x1a;
...
foo(&(uint32_t){ a })
...

So I tried below code snippet:

#include <stdio.h>
#include <stdlib.h>

int main(int argc, const char * argv[])
{
    uint8_t a = 0x1a;
    
    printf("0x%x\n", a);
    printf("%p\n", &a);
    printf("%p\n", &a + 1);
    printf("===============\n");
    printf("0x%x\n", (uint32_t){a});
    printf("%p\n", &(uint32_t){a});
    printf("%p\n", (&(uint32_t){a} + 1));
    printf("===============\n");
    printf("%d\n", (&(uint32_t){a} + 1) == (&(uint32_t){a}));
    printf("%ld\n", (&(uint32_t){a} + 1) - (&(uint32_t){a}));
    
    return 0;
}

On M1 MacBook, compiled successfully and ran:

$ gcc try.c -o try.bin
$
$ ./try.bin
0x1a
0x16bb773ef
0x16bb773f0
===============
0x1a
0x16bb773e4
0x16bb773e4
===============
0
2

I got confused with below questions based on the printing:

1. Is (uint32_t){a} just a casting for a(like (uint32_t)a) in C language ?
   • If no, the values are the same
   • If yes, the addresses are different
2. Is &(uint32_t){a} and &(uint32_t){a} + 1 same in this case ?
   • If no, the addresses are printed same
   • If yes, == and - proved they are different