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java rule round up #113

 
Greenhorn
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Can somebody explain to me why the following code compiles and runs:
double x; x=24.0/0;
The answer says it will compile and run because floating point nummers don't produce a divide-by-zero ArithmeticException. They will give a result which is Not a Number value. But for me it looks like 24.0 is a double and not a floating point number because it has no f attached to it.
 
Author & Gold Digger
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A double primitive is also a floating point number. By default, a floating point value is a double. If you append a 'f' after the number, it becomes a float value.
 
Ranch Hand
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Hi Jeroen,
Welcome to Javaranch .
Both float(32 bits long) and double(64 bits long) are floating point values (that is both are decimal values) with different ranges. Default floating point value is 'double'.
For example
float f = 12.0, this will give a compile time as error as '12.0' denotes a double value.to make this statement compile we need to modify it as float f = 12.0F.
So the above answer mentioned is right.
reshma
 
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