printf("%a"): how the format and parameters of hexadecimal floating-point constant are selected?

Consider this simple code (t0.c):

#include <stdio.h>
#include <float.h>

#if DBL_HAS_SUBNORM == 1
double d = 0x23d804860c09bp-1119;

int main(void)
{
    printf("%a\n", d);
    return 0;
}
#endif

Invocation and output:

# host: CPU: Intel, OS: Windows 10
$ gcc t0.c -std=c11 && ./a.exe
0x1.2p-1070

# host: CPU: Intel, OS: Windows 10
$ clang t0.c -std=c11 && ./a.exe
0x1.2p-1070

# host: CPU: Intel, OS: Linux
$ gcc t0.c -std=c11 && ./a.out
0x0.0000000000012p-1022

# host: CPU: Intel, OS: Linux
$ clang t0.c -std=c11 && ./a.out
0x0.0000000000012p-1022

Question: For conversion specifier %a, how:

the exact format of hexadecimal floating-point constant is selected, and
the parameters of this hexadecimal floating-point constant are selected?

For example, why 0x1.2p-1070 and not 0x0.0000000000012p-1022 (or other variations) (and vise versa)?

Solution

C allows some latitude in the details

A double argument representing a floating-point number is converted in the style [-]0xh.hhhhp±d, where there is one hexadecimal digit (which is nonzero if the argument is a normalized floating-point number and is otherwise unspecified) before the decimal-point character and the number of hexadecimal digits after it is equal to the precision; if the precision is missing and FLT_RADIX is a power of 2, then the precision is sufficient for an exact representation of the value ... (more concerning base 10 encodings, Inf and NaN)
C2xdr § 7.21.6.1 8

Notable variations I have seen is if the first h digit is '0'-'F' or limited to '0'-'1'.

why 0x1.2p-1070 and not 0x0.0000000000012p-1022

Leading digit specified as non-zero for normal values. Yet as OP's value looks like a sub-normal one, either would have been acceptable. It is unspecified.