Implicit narrowing conversions

[Jimmy Blakely]

Am I correct in concluding that integer literals are the only primitive literals that the compiler performs implicit narrowing conversions on at compile time?
For example:
short s = 12; // okay...literal 12 implicitly narrowed
char c = 15; // okay...literal 15 implicitly narrowed
byte b = 8.0; // not okay....no implicit conversion for double literal
int i = 12.0; // not okay....same as above
float f = 14.0; // believe it or not...this isn't allowed either

[Ragu Sivaraman]

Originally posted by Jimmy Blakely:
Am I correct in concluding that integer literals are the only primitive literals that the compiler performs implicit narrowing conversions on at compile time?
For example:
short s = 12; // okay...literal 12 implicitly narrowed
char c = 15; // okay...literal 15 implicitly narrowed
byte b = 8.0; // not okay....no implicit conversion for double literal
int i = 12.0; // not okay....same as above
float f = 14.0; // believe it or not...this isn't allowed either

Yup, BSC <----I
Implicit narrow only for integers
Even within BSC we need explicit casting
oh btw, BSC is Byte, Short , Char
Note: The value of int MUST be with the range
of the destination type
HTH
[This message has been edited by Ragu Sivaraman (edited August 13, 2001).]

[Paul Anilprem]

This should help you: http://enthuware.com/jqplus/FAJQuestions.html
-Paul.
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[leena rane]

>Am I correct in concluding that integer literals are the only >primitive literals that the compiler performs implicit narrowing >conversions on at compile time?
Yes,that's right .
For example:
short s = 12; // okay...literal 12 implicitly narrowed
char c = 15; // okay...literal 15 implicitly narrowed
byte b = 8.0; // not okay....no implicit conversion for double literal
int i = 12.0; // not okay....same as above
float f = 14.0; // believe it or not...this isn't allowed
An integer literal can be assigned to a byte,short,char if value is within bounds.(this answers 1&2)
Definitely they can't take double values. (this answers 3)
An integer cannot take long,float or double values(this answers4)
float cannot take a double value. 14.0 is a double literal.
try this:
float f=14.0f;
it should work,
(this answers 5)

hope this helps ,
Best regards,
leena.

[Guoqiao Sun]

And remember that is is always possible to assign an integral type value to a float type value:
For example:

float  f1 = 100;  
float  f2 = 100L; // A
double d1 = 100;
double d2 = 100L;

Case // A is feasible although some precision might be lost.
Guoqiao
[This message has been edited by Guoqiao Sun (edited August 13, 2001).]