• 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 Pie Elite all forums
this forum made possible by our volunteer staff, including ...
Marshals:
  • Campbell Ritchie
  • Paul Clapham
  • Ron McLeod
  • Jeanne Boyarsky
  • Tim Cooke
Sheriffs:
  • Liutauras Vilda
  • paul wheaton
  • Henry Wong
Saloon Keepers:
  • Tim Moores
  • Tim Holloway
  • Stephan van Hulst
  • Carey Brown
  • Frits Walraven
Bartenders:
  • Piet Souris
  • Himai Minh

casting

 
Greenhorn
Posts: 8
  • Mark post as helpful
  • send pies
    Number of slices to send:
    Optional 'thank-you' note:
  • Quote
  • Report post to moderator
by default every integer literal is an int
so how does this works witjout typecasting
byte b=6;
 
Ranch Hand
Posts: 439
  • Mark post as helpful
  • send pies
    Number of slices to send:
    Optional 'thank-you' note:
  • Quote
  • Report post to moderator
it looks at a value of 6 and sees that it's in a range of byte so no casting is necessary. But if you do this
byte b = 6;
b = b + 1; // Error because binary operation causes an error you try to add 1 which is an int and the result is not in the range of byte so you must cast
operations like this b++; will not caouse an error because implicit cast is being performed.
 
Greenhorn
Posts: 14
  • Mark post as helpful
  • send pies
    Number of slices to send:
    Optional 'thank-you' note:
  • Quote
  • Report post to moderator
Hi!!
constants which r in range wont throw any error.like byte b=10;though 10 is in range ,still the compiler knows that it is a constant whose value is not going to change and it is within limits so it is okie
another thing which will work is
final int j=30;
byte b=j;
this will also work fine.
 
Greenhorn
Posts: 5
  • Mark post as helpful
  • send pies
    Number of slices to send:
    Optional 'thank-you' note:
  • Quote
  • Report post to moderator
So why doesn't this work with floats/doubles, etc?
e.g., why can't I do:
float f = 3.14; //clearly within range & precision of f.

Originally posted by puneet pruthi:
Hi!!
constants which r in range wont throw any error.like byte b=10;though 10 is in range ,still the compiler knows that it is a constant whose value is not going to change and it is within limits so it is okie
another thing which will work is
final int j=30;
byte b=j;
this will also work fine.


 
Ranch Hand
Posts: 115
  • Mark post as helpful
  • send pies
    Number of slices to send:
    Optional 'thank-you' note:
  • Quote
  • Report post to moderator
Rules apply on char,short and byte. If the integer literal is in range of the given primitive, compiler allow narrowing cast.
float f=3.14; here 3.14 is double and you cannot assign it without explicit cast. If expression is float f=30; compiles fine.
 
The moth suit and wings road is much more exciting than taxes. Or this tiny ad:
Free, earth friendly heat - from the CodeRanch trailboss
https://www.kickstarter.com/projects/paulwheaton/free-heat
reply
    Bookmark Topic Watch Topic
  • New Topic