Gulp: How to access filename inside the gulp-replace callback function

Question

I need to inject css in html. The project structure is as follows:

/
   html/
        my-app.html
        app-header.html
        ...
   css/
       my-app.css
       app-header.css
       ...

So, each of the html files contains:

<style>
    <!-- inject:css -->
</style>

And I want to inject the css into the html. So I found the following gulp task:

gulp.task('replace', function() {
    gulp.src('html/*.html')
        .pipe(replace(/<!-- inject:css -->/, function(s) {
        var style = fs.readFileSync('style.css', 'utf8');
        return style;
    }))
    .pipe(gulp.dest('./dist/tmp'));
});

As you can see, the input css filename is incorrect and should match the name of the html file (but html replaced with css of course!)

How can I access the filename inside that function? or is there a better way to get this job done ?