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float problem

 
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what's the difference between declaration of x & y

thanks in advance
 
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None.
 
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Refer to 5.1.2 "Widening Primitive Conversion" from "The Java Language Specification 2nd Edition"
- int to long, float or double
 
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They even look the same when disassembled with javap.
Both

and

result in the following output using javap -c Foo
 
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Originally posted by Ilja Preuss:
None.


is it? 0f is forced to a float while 0 is forced to a double which is then narrowed to a float to fit in the float.
In this particular instance the result is the same of course. In other circumstances different results may well arise.
Think of things like

d would (if memory serves) produce a float widened to a double while d2 would produce a double directly.
The precision of the result in d2 would be greater.
Note: this example is very contrived and might not be entirely correct but it serves the purpose.
 
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0f is forced to a float while 0 is forced to a double which is then narrowed to a float to fit in the float.


But all this is happening at compile time, so there really is no actual difference. (BTW, I'd think that 0 is an int literal. Is it really converted to a double before being narrowed to a float?)


d would (if memory serves) produce a float widened to a double while d2 would produce a double directly.


Actually, 1/3 produces an int of the value 0.
And if I remember correctly, 1/(3f) produces a double. I would be surprised if (1/3)f where a valid expression, but I didn't check.

Note: this example is very contrived and might not be entirely correct but it serves the purpose.


You are, of course, correct that you can find circumstances where using the f qualifier (is that the correct term?) makes a difference - else there wouldn't be a need for it. But in the specific example of the original poster, it doesn't.
 
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