OJ2/apps/api/src/judge/ast.ts

import {
  AST_NODE_TARGETS_BY_LANGUAGE,
  AST_OPERATOR_TARGETS_BY_LANGUAGE,
  astNodeLabel,
  astOperatorLabel,
  astRuleIsMeaningful,
  astTargetNodeType,
  astRuleSchema,
  AST_SUPPORTED_LANGUAGES,
  type AstRequirement,
  type AstRequirements,
  type AstRule,
  type AstRules,
} from "@oj2/contract"
import { Language, Parser, type Node } from "web-tree-sitter"
// 语法 wasm 内嵌成资源。原来是 `Bun.resolveSync(pkg + "/" + name, import.meta.dir)`，
// 编译成单二进制后 import.meta.dir 是 /$bunfs/root，解析不到 node_modules。见 vendor/jieba.ts
import cWasmPath from "tree-sitter-c/tree-sitter-c.wasm" with { type: "file" }
import cppWasmPath from "tree-sitter-cpp/tree-sitter-cpp.wasm" with { type: "file" }
import pythonWasmPath from "tree-sitter-python/tree-sitter-python.wasm" with { type: "file" }
// web-tree-sitter 自己的运行时 wasm，Parser.init() 要用
import treeSitterWasmPath from "web-tree-sitter/web-tree-sitter.wasm" with { type: "file" }

// AstRule 的形状在契约里（astRuleSchema）—— 原来这份和前端两份各写各的。
// 这里 re-export，判题机的调用方不用再去 import 契约。
export type { AstRule } from "@oj2/contract"

export interface AstResult {
  description: string
  passed: boolean
  /** count_* 引擎实际数到的次数。失败时前端拿它补一句「当前 N 次」 */
  actual?: number
  /** 没过的「不能用 X」「X 太多」：X 在第几行（1 起），挑战没做到时编辑器里标出来 */
  lines?: number[]
}

/** 节点所在的行，去重排序。submission.code 只是学生写的那段，不含题目模板，行号不用再扣 */
function linesOf(nodes: Node[]) {
  return [...new Set(nodes.map((node) => node.startPosition.row + 1))].sort((a, b) => a - b)
}

/** 没过才带行号：过了的规则不用标 */
function withLines(result: AstResult, nodes: Node[]): AstResult {
  return result.passed || nodes.length === 0 ? result : { ...result, lines: linesOf(nodes) }
}

let initPromise: Promise<void> | undefined
const languages = new Map<string, Language>()

async function loadLanguage(language: string) {
  if (!AST_SUPPORTED_LANGUAGES.includes(language)) return null
  // locateFile 指到内嵌的 tree-sitter.wasm：emscripten 默认按脚本所在目录找，
  // 单二进制里那个目录是 /$bunfs/root，它自己找不着
  if (!initPromise) initPromise = Parser.init({ locateFile: () => treeSitterWasmPath })
  await initPromise

  const cached = languages.get(language)
  if (cached) return cached

  const wasmPath = language === "C" ? cWasmPath : language === "C++" ? cppWasmPath : pythonWasmPath
  const loaded = await Language.load(wasmPath)
  languages.set(language, loaded)
  return loaded
}

function collectNodes(root: Node, type: string, result: Node[] = []) {
  if (root.type === type) result.push(root)
  for (const child of root.children) collectNodes(child, type, result)
  return result
}

function hasNode(root: Node, type: string): boolean {
  if (root.type === type) return true
  return root.children.some((child) => hasNode(child, type))
}

function targetName(rule: AstRule, language?: string) {
  const target = rule.target ?? ""
  return rule.label || astNodeLabel(target, language) || "指定语法"
}

function countPhrase(verb: string, rule: AstRule) {
  if (rule.exact !== undefined) return `${verb} ${rule.exact} 次`
  if (rule.min !== undefined && rule.max !== undefined) return `${verb} ${rule.min}～${rule.max} 次`
  if (rule.min !== undefined) return `至少${verb} ${rule.min} 次`
  if (rule.max !== undefined) return `至多${verb} ${rule.max} 次`
  return ""
}

/**
 * 一条规则的中文描述。判题结果（statistic_info.ast_results）和题目页的「要求」
 * 用的是同一份 —— 原来前端 ProblemContent.vue 里另有一份几乎一样的实现，
 * 只有 min/max 同时给出时的措辞不一样（生产库里没有这种规则）。
 */
export function describeAstRule(rule: AstRule, language?: string): string {
  if (rule.message) return rule.message
  const name = targetName(rule, language)
  const target = rule.target ?? ""
  switch (rule.engine) {
    case "must_exist_node":
      return `必须使用 ${name}`
    case "must_not_exist_node":
      return `不能使用 ${name}`
    case "count_node":
      return `${name} ${countPhrase("出现", rule)}`.trim()
    case "must_call_function":
      return `必须调用 ${target}()`
    case "must_not_call_function":
      return `不能调用 ${target}()`
    case "count_function_call":
      return `${target}() ${countPhrase("调用", rule)}`.trim()
    case "must_call_method":
      return `必须调用 .${target}()`
    case "must_not_call_method":
      return `不能调用 .${target}()`
    case "must_use_operator":
      return `必须使用 ${astOperatorLabel(target, language)} 运算符`
    case "must_have_nesting": {
      // 这两个走 astNodeLabel 而不是裸值 —— 少了这一步文案就是
      // 「必须使用 for_loop 嵌套」，旧栈 ast_checker/engines/nesting.py 是翻的
      const outer = astNodeLabel(rule.outer ?? "", language)
      const inner = astNodeLabel(rule.inner ?? "", language)
      return outer === inner ? `必须使用 ${outer} 嵌套` : `必须在 ${outer} 中嵌套使用 ${inner}`
    }
  }
}

/** 标签配色用的粗分类，见契约 astRequirementSchema */
function requirementKind(engine: AstRule["engine"]): AstRequirement["kind"] {
  if (engine.startsWith("must_not")) return "forbid"
  if (engine.startsWith("count")) return "count"
  return "require"
}

/**
 * 把规则原文投影成下发给学生的「代码要求」。规则里的 engine / target 不出现在
 * 响应里 —— 阶段 3 泄露评审收掉 ast_rules 时要的就是这个，见契约的注释。
 *
 * 只投影判题机真检查得了的语言。原来这里不看语言，给 C++ 题配的规则照样渲染成
 * 「必须使用 for 循环」挂在题目页上，而 loadLanguage 对 C++ 返回 null、
 * checkAst 直接放行 —— 学生看得见要求，判题从不检查。
 */
export function astRequirements(value: unknown): AstRequirements | null {
  const grouped =
    value && typeof value === "object" && !Array.isArray(value)
      ? (value as Record<string, unknown>)
      : null
  if (!grouped) return null
  const out: AstRequirements = {}
  for (const [language, rules] of Object.entries(grouped)) {
    if (!Array.isArray(rules)) continue
    if (!AST_SUPPORTED_LANGUAGES.includes(language)) continue
    const items = rules.flatMap((rule) => {
      const parsed = astRuleSchema.safeParse(rule)
      if (!parsed.success) return []
      if (!astRuleIsMeaningful(parsed.data)) return []
      return [
        {
          description: describeAstRule(parsed.data, language),
          kind: requirementKind(parsed.data.engine),
        },
      ]
    })
    if (items.length > 0) out[language] = items
  }
  return Object.keys(out).length > 0 ? out : null
}

/**
 * AST 规则的语义校验。zod 只管形状（engine 在枚举里、min 是整数），管不了
 * 「给 C 题选了只有 Python 才有的 list_comprehension」这类组合 —— 那种规则存得进去，
 * 判题时 astTargetNodeType() 找不到就回落成裸名去比节点类型，永远失败或永远通过，
 * 两头都不报错，只有学生受着。
 *
 * 放这儿而不是 astRulesSchema 的 refine 上：那个 schema 同时用于**读**后台题目详情，
 * 在读路径上抛错会让历史脏数据直接把题目详情打不开。
 */
export function astRulesError(astRules: AstRules | null): string | null {
  if (!astRules) return null
  for (const [language, rules] of Object.entries(astRules)) {
    if (rules.length === 0) continue
    if (!AST_SUPPORTED_LANGUAGES.includes(language)) {
      return `代码规则暂不支持 ${language}，判题机只检查 ${AST_SUPPORTED_LANGUAGES.join(" / ")}`
    }
    const nodes = AST_NODE_TARGETS_BY_LANGUAGE[language] ?? {}
    const operators = AST_OPERATOR_TARGETS_BY_LANGUAGE[language] ?? {}
    for (const [index, rule] of rules.entries()) {
      const at = `代码规则 ${language} 第 ${index + 1} 条`
      const target = rule.target ?? ""
      if (rule.engine.endsWith("_node")) {
        if (!(target in nodes)) return `${at}：${language} 没有「${target}」这种语法`
      } else if (rule.engine === "must_use_operator") {
        if (!(target in operators)) return `${at}：${language} 没有「${target}」运算符`
      } else if (rule.engine === "must_have_nesting") {
        for (const value of [rule.outer ?? "", rule.inner ?? ""]) {
          if (!(value in nodes)) return `${at}：${language} 没有「${value}」这种语法`
        }
      } else if (!target.trim()) {
        return `${at}：要检查的函数名/方法名不能为空`
      }
      if (!astRuleIsMeaningful(rule)) return `${at}：次数检查至少要填一个数字`
    }
  }
  return null
}

/**
 * 保存前清掉够不着的规则分组：不在题目 languages 里的（老师改过语言列表），
 * 以及判题机检查不了的（C++ / Java / …）。两者编辑器都不给开 tab，留着就是死数据。
 *
 * 必须先剔除再校验，否则历史脏数据会把老师锁死：一道 languages 含 C++ 的题，
 * 早年配过 C++ 规则，如今 tab 里看不到那组规则，保存却被「暂不支持 C++」拦下，
 * 老师在界面上无从修改。
 */
export function pickAstRules(astRules: AstRules | null, languages: string[]): AstRules | null {
  if (!astRules) return null
  const out: AstRules = {}
  for (const [language, rules] of Object.entries(astRules)) {
    if (!languages.includes(language)) continue
    if (!AST_SUPPORTED_LANGUAGES.includes(language)) continue
    if (rules.length > 0) out[language] = rules
  }
  return Object.keys(out).length > 0 ? out : null
}

function rangePassed(count: number, rule: AstRule) {
  if (rule.exact !== undefined && count !== rule.exact) return false
  if (rule.min !== undefined && count < rule.min) return false
  if (rule.max !== undefined && count > rule.max) return false
  return true
}

const CALL_NODE_TYPES: Record<string, string> = {
  C: "call_expression",
  "C++": "call_expression",
  Python: "call",
}

function functionCalls(root: Node, target: string, language: string) {
  const callType = CALL_NODE_TYPES[language] ?? "call"
  return collectNodes(root, callType).filter((call) => {
    const fn = call.childForFieldName("function")
    if (!fn) return false
    if (fn.type === "identifier") return fn.text === target
    // `std::sort(...)` 是 qualified_identifier。学生写 sort 还是 std::sort 取决于
    // 有没有 using namespace std，两种都得认，所以末段也比一次
    if (language === "C++" && fn.type === "qualified_identifier") {
      return fn.text === target || fn.text.split("::").pop() === target
    }
    return false
  })
}

function methodCalls(root: Node, target: string, language: string) {
  // C++ 的 `a.push_back()` / `p->push_back()` 都是 call_expression + field_expression，
  // 和 Python 的 attribute 不是一回事 —— 少了这个分支，C++ 的「必须调用 .push_back()」
  // 会静默地永远失败
  if (language === "C++") {
    return collectNodes(root, "call_expression").filter((call) => {
      const fn = call.childForFieldName("function")
      return fn?.type === "field_expression" && fn.childForFieldName("field")?.text === target
    })
  }
  if (language !== "Python") return []
  return collectNodes(root, "call").filter((call) => {
    const fn = call.childForFieldName("function")
    return fn?.type === "attribute" && fn.childForFieldName("attribute")?.text === target
  })
}

function evaluateRule(root: Node, rule: AstRule, language: string): AstResult | null {
  const target = rule.target ?? ""
  const nodeType = astTargetNodeType(target, language)

  switch (rule.engine) {
    case "must_exist_node":
      return {
        description: describeAstRule(rule, language),
        passed: hasNode(root, nodeType),
      }
    case "must_not_exist_node": {
      const nodes = collectNodes(root, nodeType)
      return withLines(
        { description: describeAstRule(rule, language), passed: nodes.length === 0 },
        nodes,
      )
    }
    case "count_node": {
      const nodes = collectNodes(root, nodeType)
      return withLines(
        {
          description: describeAstRule(rule, language),
          passed: rangePassed(nodes.length, rule),
          actual: nodes.length,
        },
        // 次数太少时没有「哪几行」可标
        rule.max !== undefined || rule.exact !== undefined ? nodes : [],
      )
    }
    case "must_call_function":
      return {
        description: describeAstRule(rule, language),
        passed: functionCalls(root, target, language).length > 0,
      }
    case "must_not_call_function": {
      const calls = functionCalls(root, target, language)
      return withLines(
        { description: describeAstRule(rule, language), passed: calls.length === 0 },
        calls,
      )
    }
    case "count_function_call": {
      const calls = functionCalls(root, target, language)
      return withLines(
        {
          description: describeAstRule(rule, language),
          passed: rangePassed(calls.length, rule),
          actual: calls.length,
        },
        rule.max !== undefined || rule.exact !== undefined ? calls : [],
      )
    }
    case "must_call_method":
      return {
        description: describeAstRule(rule, language),
        passed: methodCalls(root, target, language).length > 0,
      }
    case "must_not_call_method": {
      const calls = methodCalls(root, target, language)
      return withLines(
        { description: describeAstRule(rule, language), passed: calls.length === 0 },
        calls,
      )
    }
    case "must_use_operator":
      return {
        description: describeAstRule(rule, language),
        passed: hasNode(root, nodeType),
      }
    case "must_have_nesting": {
      const outer = rule.outer ?? ""
      const inner = rule.inner ?? ""
      const outerType = astTargetNodeType(outer, language)
      const innerType = astTargetNodeType(inner, language)
      const passed = collectNodes(root, outerType).some((node) =>
        node.children.some((child) => hasNode(child, innerType)),
      )
      return { description: describeAstRule(rule, language), passed }
    }
    default:
      return null
  }
}

export async function checkAst(
  code: string,
  language: string,
  rules: AstRule[],
): Promise<{ passed: boolean; results: AstResult[] }> {
  if (rules.length === 0) return { passed: true, results: [] }

  const treeSitterLanguage = await loadLanguage(language)
  if (!treeSitterLanguage) return { passed: true, results: [] }

  const parser = new Parser()
  try {
    parser.setLanguage(treeSitterLanguage)
    const tree = parser.parse(code)
    if (!tree) return { passed: true, results: [] }
    try {
      const results = rules
        .filter(astRuleIsMeaningful)
        .map((rule) => evaluateRule(tree.rootNode, rule, language))
        .filter((result): result is AstResult => result !== null)
      return { passed: results.every((result) => result.passed), results }
    } finally {
      tree.delete()
    }
  } catch {
    return { passed: true, results: [] }
  } finally {
    parser.delete()
  }
}

/**
 * 一份代码对着几组规则各判一遍，只解析一次。挑战用：一道题几条挑战各是一组，补历史记录时
 * 一份代码要判全部挑战，每组 checkAst 一次就是把同一份代码解析好几遍。
 *
 * 和 checkAst 不同，**判不了时返回 null 而不是「通过」**：checkAst 是题目的硬要求，
 * 判不了就放行、别冤枉学生；挑战判不了就是没做到，放行等于白送。
 */
export async function checkAstGroups(
  code: string,
  language: string,
  groups: AstRule[][],
): Promise<{ passed: boolean; results: AstResult[] }[] | null> {
  const treeSitterLanguage = await loadLanguage(language)
  if (!treeSitterLanguage) return null
  const parser = new Parser()
  try {
    parser.setLanguage(treeSitterLanguage)
    const tree = parser.parse(code)
    if (!tree) return null
    try {
      return groups.map((rules) => {
        const results = rules
          .filter(astRuleIsMeaningful)
          .map((rule) => evaluateRule(tree.rootNode, rule, language))
          .filter((result): result is AstResult => result !== null)
        return { passed: results.length > 0 && results.every((r) => r.passed), results }
      })
    } finally {
      tree.delete()
    }
  } catch {
    return null
  } finally {
    parser.delete()
  }
}