5 Doomed Romance Leonardo DiCaprio Movi ...
Wes Anderson's 5 Best Commercials
Can 'World War Z' Break Even?
Steve Soderbergh On Cinema, Studios, Mor ...
Recap: 'The King Of Comedy' 30th Anniversary ...
Excl: Lake Bell Joins 'Million Dollar Ar ...
10 Essential Cinematic AntiheroesThe MSN report also delivers some new plot details, like how the titular character (voiced by the always-wonderful John C. Reilly) is a "9-foot-tall, 643-pound video game villain" who apparently destroys buildings in a "Rampage"-style game. Our hero, Fix-It Felix (Jack McBrayer) is the guy that cleans up after his path of destruction. The movie begins when Ralph decides he doesn't want to be the villain anymore, and embarks on an odyssey through classic videogame archetypes (including games that pay homage to "Halo" and "Mario Kart") to find a game where he can be the virtuous leading man.
What we've suspected for some time, and what the MSN report confirms, is that actual classic videogame characters will intermingle with the new characters, in a way that seems reminiscent of Robert Zemeckis' groundbreaking "Who Framed Roger Rabbit." The report cites Bowser from "Super Mario Bros" and a ghost from "Pac-Man" making appearances. (Apparently other villains like Doctor Robotnik from "Sonic the Hedgehog," Kano from "Mortal Kombat," and Coily the Snake from "Q*bert" also pop up.) This is very exciting indeed and evocative of the irreverent tone being set by the film's director, Rich Moore, who directed classic episodes of "The Simpsons" (including the Sideshow Bob-centric "Cape Feare" and the Conan O'Brien-penned "Marge vs. the Monorail") as well as the Emmy-winning "Futurama" time travel episode "Roswell That Ends Well."
We can't wait to stick a couple of quarters in when "Wreck-It Ralph" appears on November 2nd. Expect a teaser trailer in front of "Brave" (which has a beautiful, brand new trailer attached to "Chimpanzee" this weekend) in June.
1 Comment
Jeff | April 17, 2012 5:46 PM
Is it just me or does Wreck-It Ralph look just like Heat Miser?