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:

Inizia individuando il tipo pubblico SourceFile e il percorso createProgram().getSourceFiles() mostrato nella riproduzione. Controlla come viene rappresentato internamente resolvedModules, quindi verifica che la proprietà sia esposta su SourceFile e che l’esempio TypeScript fornito possa accedervi.