//! Disc-wide test of one bit: does `kind & 1` mean "this element has a parent"?
//!
//! The port is blocked on what `0x3003` is, having only `0x3002` in its rule.
//! `0x3002` and `0x3003` differ in bit 0 alone, and the struct doc claims bit 0
//! is "has a parent". That is a falsifiable claim over every element on the
//! disc, so it is tested here rather than argued from two screens.
//!
//!     cargo run -p sylpheed-formats --example kind_bit0_census
use std::collections::BTreeMap;
use std::path::PathBuf;
use sylpheed_formats::{pak::PakArchive, ui_layout};

fn main() {
    let root = PathBuf::from(std::env::var("SYLPHEED_DISC").expect("SYLPHEED_DISC"));
    let mut agree = 0usize;
    let mut disagree = 0usize;
    let mut kinds: BTreeMap<u32, usize> = BTreeMap::new();
    let mut kind_parent: BTreeMap<(u32, bool), usize> = BTreeMap::new();
    let mut examples: Vec<String> = Vec::new();

    let mut paks: Vec<_> = std::fs::read_dir(root.join("dat"))
        .expect("dat")
        .filter_map(|e| e.ok())
        .map(|e| e.path())
        .filter(|p| p.extension().map(|x| x == "pak").unwrap_or(false))
        .collect();
    paks.sort();
    for p in &paks {
        let Ok(ar) = PakArchive::open(p) else {
            continue;
        };
        for (i, ent) in ar.entries().iter().enumerate() {
            let Ok(by) = ar.read(ent) else { continue };
            let Some(b) = ui_layout::parse_build(&by) else {
                continue;
            };
            for el in &b.elements {
                let has_parent = el.parent.is_some();
                let bit0 = el.kind & 1 == 1;
                *kinds.entry(el.kind).or_default() += 1;
                *kind_parent.entry((el.kind, has_parent)).or_default() += 1;
                if bit0 == has_parent {
                    agree += 1
                } else {
                    disagree += 1;
                    if examples.len() < 10 {
                        examples.push(format!(
                            "{} entry {i} {} kind={:#x} parent={:?}",
                            p.file_name().unwrap().to_string_lossy(),
                            el.name,
                            el.kind,
                            el.parent
                        ));
                    }
                }
            }
        }
    }
    println!("kind&1 == has_parent :  agree {agree}   DISAGREE {disagree}");
    for e in &examples {
        println!("   counterexample: {e}");
    }
    println!("\nkind histogram (count, and how many of each have a parent):");
    for (k, n) in &kinds {
        let wp = kind_parent.get(&(*k, true)).copied().unwrap_or(0);
        println!(
            "  {k:#06x}  n={n:6}   with parent {wp:6}   without {:6}",
            n - wp
        );
    }
}