`ZipFile.testzip()` Skips Earlier Duplicate Members

ZipFile.testzip から始め、現在のエントリオープン動作を、issue にある重複メンバーの reproducer と比較します。reproducer と関連する zipfile テストを実行します。完了とは、すべての ZipInfo が個別にチェックされ、先にある破損した重複が duplicate.txt として報告されることを意味します。

Bug report

Summary

ZipFile.testzip() can incorrectly report that an archive is valid when an earlier member is corrupt and a later member has the same filename. The method iterates over all ZipInfo objects, but opens each one by filename instead of by its ZipInfo instance. Filename lookup selects the last duplicate, so earlier duplicates are checked against the wrong archive entry.

Minimal Reproducer

import io
import struct
import zipfile

buffer = io.BytesIO()
with zipfile.ZipFile(buffer, "w") as zf:
    zf.writestr("duplicate.txt", b"corrupt me")
    zf.writestr("duplicate.txt", b"valid data")

raw = bytearray(buffer.getvalue())

# Corrupt the payload of the first duplicate only.
with zipfile.ZipFile(io.BytesIO(raw)) as zf:
    first = zf.infolist()[0]
    name_length, extra_length = struct.unpack_from(
        "<HH", raw, first.header_offset + 26
    )
    data_offset = (
        first.header_offset + zipfile.sizeFileHeader +
        name_length + extra_length
    )

raw[data_offset] ^= 1

with zipfile.ZipFile(io.BytesIO(raw)) as zf:
    first = zf.infolist()[0]

    try:
        zf.read(first)
    except zipfile.BadZipFile:
        print("The first duplicate is corrupt")

    print(zf.testzip())

Observed output:

The first duplicate is corrupt
None

testzip() should return "duplicate.txt".

Expected Behavior

Every member in infolist() should be validated individually. For duplicate names, testzip() should preserve the identity of each ZipInfo object and open the entry with:

self.open(zinfo, "r")

Rather than:

self.open(zinfo.filename, "r")

CPython versions tested on:

CPython main branch

Operating systems tested on:

Linux

Linked PRs

• gh-156540