Expose `resolvedModules` property of `SourceFile`

TypeScript Version: 3.6.4

Code

import * as ts from 'typescript'

function createCompilerHost(): ts.CompilerHost {
  function fileExists(fileName: string): boolean {
    return ts.sys.fileExists(fileName)
  }

  function readFile(fileName: string): string | undefined {
    return ts.sys.readFile(fileName)
  }

  function getSourceFile(
    fileName: string,
    languageVersion: ts.ScriptTarget,
  ) {
    const sourceText = ts.sys.readFile(fileName)
    return sourceText !== undefined
      ? ts.createSourceFile(fileName, sourceText, languageVersion)
      : undefined
  }

  function resolveModuleNames(moduleNames) {
    return moduleNames.map(() => undefined)
  }

  return {
    getSourceFile,
    getDefaultLibFileName: () => 'lib.d.ts',
    writeFile: (fileName, content) => ts.sys.writeFile(fileName, content),
    getCurrentDirectory: () => ts.sys.getCurrentDirectory(),
    getDirectories: path => ts.sys.getDirectories(path),
    getCanonicalFileName: fileName =>
      ts.sys.useCaseSensitiveFileNames ? fileName : fileName.toLowerCase(),
    getNewLine: () => ts.sys.newLine,
    useCaseSensitiveFileNames: () => ts.sys.useCaseSensitiveFileNames,
    fileExists,
    readFile,
    resolveModuleNames,
  }

}

const options = {
  module: ts.ModuleKind.ES2015,
  target: ts.ScriptTarget.ES5,
}
const host = createCompilerHost()
const sourceFiles = ts.createProgram(['./test.ts'], options, host).getSourceFiles()
const matchsFile = sourceFiles
  .filter(s => {
    return s.resolvedModules.has('typescript')
  })

Expected behavior:
Property 'resolvedModules' should exist on type 'SourceFile'

Actual behavior:

Commencez par localiser le type public SourceFile et le chemin createProgram().getSourceFiles() présenté dans la reproduction. Vérifiez comment resolvedModules est représenté en interne, puis vérifiez que la propriété est exposée sur SourceFile et que l’exemple TypeScript fourni peut y accéder.