Is there a simple way to turn an Iterator[Byte] into an inputstream lazily?-CodePudding

Given an Iterator[Byte] can I produce an InputStream without materializing the entire Iterator into memory?

CodePudding user response：

It is not too difficult. Just construct an InputStream and implement its read() method in terms of the Iterator. In Java it may look like this:

public InputStream streamFromIterator(Iterator<Byte> iterator) {
    return new InputStream() {

        @Override
        public int read() {
            if (iterator.hasNext())
                return Byte.toUnsignedInt(iterator.next());
            else
                return -1;  // end of stream
        }
    };
}

CodePudding user response：

If you are using Scala, there is a one line solution:

scala> import java.io.InputStream
                                                                                                                                                                           
scala> val itr = Iterator.range(0, 10).map(_.toByte)
val itr: Iterator[Byte] = <iterator>

scala> val ins: InputStream = () => if(itr.hasNext) itr.next & 255 else -1  // Here it is, or
//     val ins: InputStream = () => itr.nextOption.fold(-1)(_ & 255)
val ins: java.io.InputStream = anon$1@5dc218e2

scala> ins.readAllBytes()
val res0: Array[Byte] = Array(0, 1, 2, 3, 4, 5, 6, 7, 8, 9)