That's not StringInput's job: the shell should have already parsed the string so that PHP/StringInput should be given the already resolved string. What you describe is not what StringInput::tokenize() is for.

That's not what StringInputTest suggests?

    public function testToString()
    {
        $input = new StringInput('-f foo');
        $this->assertEquals('-f foo', (string) $input);

        $input = new StringInput('-f --bar=foo "a b c d"');
        $this->assertEquals('-f --bar=foo '.escapeshellarg('a b c d'), (string) $input);

        $input = new StringInput('-f --bar=foo \'a b c d\' '."'A\nB\\'C'");
        $this->assertEquals('-f --bar=foo '.escapeshellarg('a b c d').' '.escapeshellarg("A\nB'C"), (string) $input);
    }

I've got a similar issue without using escapeshellarg.
echo new StringInput('--foo="bar"'); prints --for=bar instead of --foo="bar".
That causes a problem when you pass encoded JSON string as command line argument.

Hey, thanks for your report!
There has not been a lot of activity here for a while. Is this bug still relevant? Have you managed to find a workaround?

Hey, thanks for your report!
There has not been a lot of activity here for a while. Is this bug still relevant? Have you managed to find a workaround?

Yes, still relevant.

Hey, thanks for your report!
There has not been a lot of activity here for a while. Is this bug still relevant? Have you managed to find a workaround?

Yes, still relevant, stop trying to avoid issues.

@bohwaz are you up to fixing the issue?

@Chi-teck the issue is on your side: you need to escape the json content. Eg echo new StringInput('--foo=\'"bar"\'');

@bohwaz see #45088