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What is the difference between a double and a float?

 
Matt Kidd
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I keep encountering an error in my programs whenever I have float variables. In debugging the error usually says something along the lines of found a double required a float. I thought if I had a floating point result and I used floating point variables I would get a float as out put i.e float = float * float.
 
Anthony Villanueva
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Can you post a simple test case?
 
Matt Kidd
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Here is the code:

Actual error is Found: double Required: float

(Marilyn reformated long line in code)
[ August 09, 2002: Message edited by: Marilyn de Queiroz ]
 
Anthony Villanueva
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Hi Matt,
After running the program I discovered 2 things:
1. the literal 1.5 is a double. Use 1.5f instead.
2. I need a raise.
 
Matt Kidd
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What makes the compiler believe 1.5 is a double and not a float?
 
Anthony Villanueva
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Because of the JLS:

A floating-point literal is of type float if it is suffixed with an ASCII letter F or f; otherwise its type is double and it can optionally be suffixed with an ASCII letter D or d.
 
Khurram Shahood(SCJP2)
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hi
with respect to capacity.
 
Marilyn de Queiroz
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Originally posted by Matt Kidd:
What makes the compiler believe 1.5 is a double and not a float?


In java the default decimal is a double, not a float. You have to specifically state when you want a float (22.5F)
 
Wilfried LAURENT
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Hi Matt,
To be convinced of the difference between float and double ,and the danger in type casting, you can run this simple test program:

The output ought to be
f = 0.95
d = 0.95
d==f? false
(double)f= 0.949999988079071
 
Thomas Markl
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C:\Java\EigeneJavaProgramme>javac GrossPay1.java
GrossPay1.java:4: possible loss of precision
found : double
required: float
float hoursWorked = 3.056;
^
GrossPay1.java:5: possible loss of precision
found : double
required: float
float workRate = 4.0125;
^
GrossPay1.java:9: possible loss of precision
found : double
required: float
yourPay = ((hoursWorked - 40) * (workRate * 1.5)) + (40 * workRate);
^
3 errors
Why is double found? The input of the yourPay calculation is float so the output to yourpay should
Be float, too and not double.
Thank you very much for your answers.
Thomas
 
Peter Kristensson
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Note the "1.5", that's treated as a double, so the whole result returns a double. Use 1.5f instead.
 
Anthony Villanueva
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Hi Thomas. The primitive double is like the Borg among other primitive types -- any primtive type that interacts with a double will return a double. For example,
double * float
double * long
double * int
double * short
double * byte'
will return a double result.
 
Marilyn de Queiroz
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Put 'F' after each literal and it will work. If you have an expression which includes a double and a float, the float will be automatically promoted to a double.
 
Matt Kidd
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Thanks everyone...
 
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