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# What is the difference between a double and a float?

Matt Kidd
Ranch Hand
Posts: 267
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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Posts: 1055
Can you post a simple test case?

Matt Kidd
Ranch Hand
Posts: 267
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
Ranch Hand
Posts: 1055
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
Ranch Hand
Posts: 267
What makes the compiler believe 1.5 is a double and not a float?

Anthony Villanueva
Ranch Hand
Posts: 1055
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)
Greenhorn
Posts: 18
hi
with respect to capacity.

Marilyn de Queiroz
Sheriff
Posts: 9059
12
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
Ranch Hand
Posts: 269
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
Ranch Hand
Posts: 192

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
Ranch Hand
Posts: 118

Note the "1.5", that's treated as a double, so the whole result returns a double. Use 1.5f instead.

Anthony Villanueva
Ranch Hand
Posts: 1055
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
Sheriff
Posts: 9059
12

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
Ranch Hand
Posts: 267
Thanks everyone...