124 lines
4.8 KiB
Python
124 lines
4.8 KiB
Python
"""探索清单形状校验纯函数：顺序、身份、类型与来源绑定的确定性拒绝；离线不连库。"""
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from dataclasses import replace
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import pytest
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from muse.上下文.接口 import 上下文错误, 探索条目, 校验清单, 记录探索
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from muse.任务运行.接口 import 工具来源, 工具结果
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范围 = {
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"work_id": "work-a",
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"as_of": 2,
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"source_ids": ["doc-1", "plan-1", "fact-lin"],
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"documents": [
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{
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"chapter_id": "ch-1",
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"branch_id": "main",
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"revision": 1,
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"document_id": "doc-1",
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"document_hash": "hash-doc-1",
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"position": 1,
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}
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],
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"directory_revision": 3,
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"authorized_by": "author-a",
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"run_purpose": "production",
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"content_purpose": "generation",
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"target_plan": {
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"chapter_id": "ch-3",
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"plan_id": "plan-1",
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"revision": 1,
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"content_hash": "hash-plan-1",
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},
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"facts": {"system_revision": 5},
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}
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读取结果 = 工具结果(
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内容={"chapter_id": "ch-1", "text": "林深停在码头"},
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来源=(
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工具来源("doc-1", "1", "hash-doc-1", "visible-text-codepoints/v1:production:generation"),
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),
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)
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def 条目(调用ID: str = "call-1", 工具名: str = "read_history_document", 参数: dict | None = None):
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"""用真实读取结果形状构建顺序为 1 的清单条目；覆盖字段由用例自行 replace。"""
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return 记录探索(
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(),
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调用ID,
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工具名,
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参数 if 参数 is not None else {"chapter_id": "ch-1", "revision": 1},
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读取结果,
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)[0]
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def 顺序清单(*调用: tuple[str, str, dict]) -> tuple[探索条目, ...]:
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"""按调用次序经记录探索构建顺序连续的合法清单。"""
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清单: tuple[探索条目, ...] = ()
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for 调用ID, 工具名, 参数 in 调用:
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清单 = 记录探索(清单, 调用ID, 工具名, 参数, 读取结果)
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return 清单
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def test_顺序断裂拒绝__b7e0b1() -> None:
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跳号 = (条目("call-1"), replace(条目("call-2"), 顺序=3))
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with pytest.raises(上下文错误) as 异常:
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校验清单(跳号, 范围)
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assert 异常.value.错误码 == "CONTEXT_INPUT_INVALID"
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assert "顺序" in 异常.value.说明
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倒序 = (replace(条目("call-1"), 顺序=2), replace(条目("call-2"), 顺序=1))
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with pytest.raises(上下文错误) as 异常:
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校验清单(倒序, 范围)
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assert "顺序" in 异常.value.说明
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def test_重复调用ID拒绝__b7e0b2() -> None:
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重复 = (条目("call-1"), replace(条目("call-1"), 顺序=2))
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with pytest.raises(上下文错误) as 异常:
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校验清单(重复, 范围)
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assert 异常.value.错误码 == "CONTEXT_INPUT_INVALID"
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assert "call-1" in 异常.value.说明 and "重复" in 异常.value.说明
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def test_布尔与类型混淆拒绝__b7e0b3() -> None:
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# 顺序是 bool 而不是 int：True 不能冒充 1
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with pytest.raises(上下文错误) as 异常:
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校验清单((replace(条目("call-1"), 顺序=True),), 范围)
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assert 异常.value.错误码 == "CONTEXT_INPUT_INVALID"
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# 参数里的版本是 bool：JSON Schema 的 integer 不接受布尔
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with pytest.raises(上下文错误) as 异常:
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校验清单((条目("call-1", 参数={"chapter_id": "ch-1", "revision": True}),), 范围)
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assert 异常.value.错误码 == "CONTEXT_INPUT_INVALID"
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# 条目本身不是探索条目对象：字典形状不能冒充
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with pytest.raises(上下文错误) as 异常:
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校验清单(({"call_id": "call-1", "tool": "read_history_document"},), 范围)
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assert 异常.value.错误码 == "CONTEXT_INPUT_INVALID"
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def test_缺来源绑定与非法消费哈希拒绝__b7e0b4() -> None:
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缺绑定 = (replace(条目("call-1"), 来源引用=()),)
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with pytest.raises(上下文错误) as 异常:
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校验清单(缺绑定, 范围)
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assert 异常.value.错误码 == "CONTEXT_INPUT_INVALID"
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assert "来源绑定" in 异常.value.说明
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非法哈希 = (replace(条目("call-1"), 消费哈希="不是十六进制"),)
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with pytest.raises(上下文错误) as 异常:
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校验清单(非法哈希, 范围)
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assert 异常.value.错误码 == "CONTEXT_INPUT_INVALID"
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assert "消费哈希" in 异常.value.说明
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def test_合法清单按序通过且空清单拒绝__b7e0b5() -> None:
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清单 = 顺序清单(
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("call-1", "query_work_plan", {}),
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("call-2", "query_facts_as_of", {}),
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("call-3", "read_history_document", {"chapter_id": "ch-1", "revision": 1}),
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("call-4", "read_history_document", {"chapter_id": "ch-1", "revision": 1, "end": 6}),
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)
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assert [项.顺序 for 项 in 清单] == [1, 2, 3, 4]
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校验清单(清单, 范围) # 不抛异常即通过
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with pytest.raises(上下文错误) as 异常:
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校验清单((), 范围)
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assert 异常.value.错误码 == "CONTEXT_INPUT_INVALID"
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