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test: add a test to make sure the modules can be required independently #24402

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test: add a test to make sure the modules can be required independently #24402

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152a8f6
stream: do not use crypto.DEFAULT_ENCODING in lazy_transform.js
joyeecheung Nov 16, 2018
987fe87
test: add a test to make sure the modules can be required independently
joyeecheung Nov 16, 2018
7925ff0
fixup! test: add a test to make sure the modules can be required inde…
joyeecheung Nov 22, 2018
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7 changes: 5 additions & 2 deletions lib/internal/streams/lazy_transform.js
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Original file line number Diff line number Diff line change
Expand Up @@ -5,7 +5,10 @@

const stream = require('stream');
const util = require('util');
const crypto = require('crypto');

const {
getDefaultEncoding
} = require('internal/crypto/util');

module.exports = LazyTransform;

Expand All @@ -22,7 +25,7 @@ function makeGetter(name) {
this._writableState.decodeStrings = false;

if (!this._options || !this._options.defaultEncoding) {
this._writableState.defaultEncoding = crypto.DEFAULT_ENCODING;
this._writableState.defaultEncoding = getDefaultEncoding();
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}

return this[name];
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43 changes: 43 additions & 0 deletions test/sequential/test-native-module-deps.js
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Original file line number Diff line number Diff line change
@@ -0,0 +1,43 @@
'use strict';

// This tests that all the native modules can be loaded independently
// Flags: --expose-internals

if (process.argv[3]) {
require(process.argv[3]);
return;
}

require('../common');
const {
cachableBuiltins
} = require('internal/bootstrap/cache');
const { fork } = require('child_process');
const assert = require('assert');

for (const key of cachableBuiltins) {
run(key);
}

function run(key) {
const child = fork(__filename,
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Maybe use:

Suggested change
const child = fork(__filename,
const child = execFile(process.execPath, ['-e', `require('${key}')])`

this way you don't need the self-reference, and eliminate L6-L9

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@joyeecheung joyeecheung Nov 23, 2018 •
edited
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@refack -e introduces noise in the dependency graph because that option also leads to additional module loads. Same goes to -p. It somewhat weakens the test.

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Then maybe use a fixture?
It seems to me like in this case the self referenced part makes the file look awkward. And the usual benefit of having all the test code in one place, is not that beneficial since the child code is just one expression.
But it's just a style nit, and I defer to your decision.

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New idea, a variation on (either with shell or by using child.stdin):

const child = exec(`echo "require('${key}')" | `${process.execPath}` --`, {shell: true});

[ '--expose-internals', key ],
{ silent: true }
);

let stdout = '';
let stderr = '';
child.stdout.on('data', (data) => (stdout += data.toString()));
child.stderr.on('data', (data) => (stderr += data.toString()));
child.on('close', (code) => {
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add common.mustCall

if (code === 0) {
return;
}
console.log(`Failed to require ${key}`)
console.log('----- stderr ----')
console.log(stderr);
console.log('----- stdout ----')
console.log(stdout);
assert.strictEqual(code, 0);
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@refack refack Nov 23, 2018 •
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IIUC this doesn't match L33
Ahh you want this to fail, so assert.fail(`exit code: ${code}`).

Another thought, this will lead to the test failing for the first bad module masking, any other possible fails. Maybe replace with

    common.mustCall(() => `exit code: ${code} for module: ${key}`);

The returned function will never get called, but will make the test fail when the process exits.

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@refack In general I think it's fine to just fail when we encounter the first module without a clean dependency graph, and fix them one by one?

});
}