Deno.run(...) in test

[tamusjroyce]

I have a deno test which executes version. Pretty simple. And deno even detects that this is leaking resources!!!

deno test --allow-run

deno.test.ts

Deno.test("deno runs", async () => {
    const p = Deno.run({
        cmd: ["deno", "--version"],
        stdout: "piped",
        stderr: "piped",
    });
    console.log(JSON.stringify(await p) + '');
    const output = await p.output();
    const outStr = new TextDecoder().decode(output);
    // const err = await p.error();
    // const errStr = new TextDecoder().decode(err);
    p.close();
});

Output:
deno runs
AssertionError: Test case is leaking resources.
"2": "stderr"
}
After: {
"0": "stdin",
"1": "stdout",
"2": "stderr",
"4": "childStderr"
}

vs

Deno.test("deno runs", async () => {
    const p = Deno.run({
        cmd: ["deno", "--version"],
        stdout: "piped",
        stderr: "piped",
    });
    console.log(JSON.stringify(await p) + '');
    // const output = await p.output();
    // const outStr = new TextDecoder().decode(output);
    // const err = await p.error();
    // const errStr = new TextDecoder().decode(err);
    p.close();
});

Before: {
"0": "stdin",
"1": "stdout",
"2": "stderr"
}
After: {
"0": "stdin",
"1": "stdout",
"2": "stderr",
"4": "childStderr"
}

Should awaiting on output() be required to clean up resources? How do I await on p.error()? error isn't a function

[lucacasonato]

await p.stderrOutput()

[tamusjroyce]

I should have dug harder. Now I can get my much improved unit test working. Thank you!