• Post Reply Bookmark Topic Watch Topic
  • New Topic
programming forums Java Mobile Certification Databases Caching Books Engineering Micro Controllers OS Languages Paradigms IDEs Build Tools Frameworks Application Servers Open Source This Site Careers Other all forums
this forum made possible by our volunteer staff, including ...
Marshals:
  • Jeanne Boyarsky
  • Liutauras Vilda
  • Campbell Ritchie
  • Tim Cooke
  • Bear Bibeault
Sheriffs:
  • Paul Clapham
  • Junilu Lacar
  • Knute Snortum
Saloon Keepers:
  • Ron McLeod
  • Ganesh Patekar
  • Tim Moores
  • Pete Letkeman
  • Stephan van Hulst
Bartenders:
  • Carey Brown
  • Tim Holloway
  • Joe Ess

Errata: OCP Oracle Certified Professional Java SE 8 Programmer II - Chapter 7 page 373  RSS feed

 
Greenhorn
Posts: 8
1
  • Mark post as helpful
  • send pies
  • Quote
  • Report post to moderator
First sentence on page 373 says that "This code consistently outputs the first value in the serial stream, 1." where the code, the author is referring to is How is it possible for findAny to return predictable results?

on the same page too i.e. 373 the same mistake was made:



The result is that the output could be 4, 1, or really any value in the stream. You can see that with parallel streams, the results of findAny() are no longer predictable.

Somehow it feels like author is talking about findFirst();
 
author & internet detective
Marshal
Posts: 38506
653
Eclipse IDE Java VI Editor
  • Mark post as helpful
  • send pies
  • Quote
  • Report post to moderator
Desire,
That's correct in the book. That section is explaining that findAny() is not guaranteed to return a certain element on a parallel stream. It just happens to on a serial stream.

Whereas findFirst() loses some of the benefit of parallelism because it forces the stream to be ordered.

Does that make sense now?
 
It is sorta covered in the JavaRanch Style Guide.
  • Post Reply Bookmark Topic Watch Topic
  • New Topic
Boost this thread!