I have a no-op Transform stream test that outputs the list of all files passed through it. Why does the following code outputs the list of files only once? How do I compose two streams properly?
var composed = test().pipe(test());
gulp.src('source/*').pipe(composed);
To compose streams, you pipe() the first to the second, but you also have to modify pipe() and unpipe() so they apply to the second stream. This way additional calls to pipe() on the composed stream will use the output of the second stream instead of the first.
function compose(s1, s2) {
s1.pipe(s2);
s1.pipe = function (dest) {
return s2.pipe(dest);
}
s1.unpipe = function (dest) {
return s2.unpipe(dest);
}
return s1;
}
var composed = compose(test(), test());
gulp.src('source/*')
.pipe( composed ) // src -> s1 -> s2
.pipe( ... ); // s2 -> ...
compose operation is inconsistent: it has pipe of second stream, but events and properties of first. The cleaner approach would be to create a new stream class from scratch, but there's a wide variety of stream types, and the resulting stream could be Readable, Writable of Transform depending on types of passed streams. I haven't found a way to check for stream types to make this possible.
on() method to link to s2 for event, but that still doesn't give properties. The complete solution would be a to implement a full new stream.
pipe() returns the stream being piped to, not from (to allow chaining). So the second line is basically trying to pipe to the second instance of test(). What you need to do instead is:
var first = test();
first.pipe(test());
gulp.src('source/*').pipe(first);
Or:
gulp.src('source/*')
.pipe(test())
.pipe(test());
