# whats the difference between float & double

posted 5 years ago

- 0

Dear all

I have got some question about float & double , my answer is due to precision because float is 32 bit integer and double is 64 bit integer, However I am not sure...please help me

different

same

In first case why it is not equal and in case the value is same,why is it so?

I have got some question about float & double , my answer is due to precision because float is 32 bit integer and double is 64 bit integer, However I am not sure...please help me

**Output :**different

same

In first case why it is not equal and in case the value is same,why is it so?

Ian Lubelsky

Ranch Hand

Posts: 49

posted 5 years ago

- 0

I think you are getting "different" for the 3.2 because a computer's floating point unit works with base 2 binary and 0.2 can't be represented precisely in binary, it is called a repeater fraction. In base-2 only numbers with denominators that are powers of 2 are terminating, which I think is only .25, .50, and .75, which is why 6.5 shows up as "same".

Hope this helps, but there may be better minded people out there that can give you a better explanation perhaps.

Hope this helps, but there may be better minded people out there that can give you a better explanation perhaps.

Jim Hoglund

Ranch Hand

Posts: 525

posted 5 years ago

- 0

Float and double values are not integers, but rather floating point numbers. The Java

integer types are: byte, short, int, long (and char - 16 bits unsigned). In both cases

your code is comparing a float value to a double, as the compiler sees 3.2f as a float

and 3.2 (without the f) as a double. I suspect the precision of the compare logic has

something to do with the different results. Someone else may comment further.

Jim ... ...

integer types are: byte, short, int, long (and char - 16 bits unsigned). In both cases

your code is comparing a float value to a double, as the compiler sees 3.2f as a float

and 3.2 (without the f) as a double. I suspect the precision of the compare logic has

something to do with the different results. Someone else may comment further.

Jim ... ...

BEE MBA PMP SCJP-6

posted 5 years ago

- 0

According to JLS - 3.10.2 Floating-Point Literals...

So in line 6, f1 is assigned the value 3.2f, which is a

But in line 9, this value is compared to the value 3.2, which is a

These are not equal because the value 3.2 cannot be precisely stored (in binary) within the range of a float.

A floating-point literal is of type float if it is suffixed with an ASCII letter F or f; otherwise its type is double...

So in line 6, f1 is assigned the value 3.2f, which is a

**float.**But in line 9, this value is compared to the value 3.2, which is a

**double.**These are not equal because the value 3.2 cannot be precisely stored (in binary) within the range of a float.

"We're kind of on the level of crossword puzzle writers... And no one ever goes to them and gives them an award." *~Joe Strummer*

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Sebastian Hatt

Greenhorn

Posts: 1

posted 5 years ago

- 0

As Ian pointed out, 6.5 can be represented

To quickly refresh how binary numbers work:

100 -> 4

10 -> 2

1 -> 1

0.1 -> 0.5 (or 1/2)

0.01 -> 0.25 (or 1/4)

etc.

6.5 in binary: 110.1 (exact result, the rest of the digits are just zeroes)

3.2 in binary: 11.001100110011001100110011001100110011001100110011001101... (here precision matters!)

A float only has 24 bits precision (the rest is used for sign and exponent), so:

3.2f in binary: 11.0011001100110011001100 (not equal to the double precision approximation)

There's a handy calculator for things like this at http://www.digitconvert.com/

Basically it's the same as when you're writing 1/5 and 1/7 in decimal numbers:

1/5 = 0,2

1,7 = 0,14285714285714285714285714285714...

**exactly**in binary, whereas 3.2 can't. That's why the difference in precision doesn't matter for 6.5, so 6.5 == 6.5f.To quickly refresh how binary numbers work:

100 -> 4

10 -> 2

1 -> 1

0.1 -> 0.5 (or 1/2)

0.01 -> 0.25 (or 1/4)

etc.

6.5 in binary: 110.1 (exact result, the rest of the digits are just zeroes)

3.2 in binary: 11.001100110011001100110011001100110011001100110011001101... (here precision matters!)

A float only has 24 bits precision (the rest is used for sign and exponent), so:

3.2f in binary: 11.0011001100110011001100 (not equal to the double precision approximation)

There's a handy calculator for things like this at http://www.digitconvert.com/

Basically it's the same as when you're writing 1/5 and 1/7 in decimal numbers:

1/5 = 0,2

1,7 = 0,14285714285714285714285714285714...

posted 5 years ago

- 0

A

A

`float`is a 32-bit, single-precision IEEE 754 floating point number.A

`double`is a 64-bit, double-precision IEEE 754 floating point number.Consider Paul's rocket mass heater. |