Java : double to float type conversion is giving 'infinity' for larger values

Question

Suppose if i have a variable with some random larger values of type double:

double d = 4786777867867868654674678346734763478673478654478967.77;

now if i try to convert this into float at some point in my program, the output shows 'infinity' (in eclipse IDE):

float f = (float)d;    // inifinty
byte b = (byte)d;      // some valid value
short s = (short)d;    // some valid value
int i = (int)d;        // some valid value

Can someone give me any valid answers, how it is not converting for only for float datatype ?

Answer 1

Let's take a look at the result of casting this large double to each of the other numeric primitive types:

    double d = 4786777867867868654674678346734763478673478654478967.77;
    System.out.printf("float  %f\n", (float)d);
    System.out.printf("long   %d\n", (long)d);
    System.out.printf("int    %d\n", (int)d);
    System.out.printf("short  %d\n", (short)d);
    System.out.printf("byte   %d\n", (byte)d);

Output:

float Infinity
long  9223372036854775807
int   2147483647
short -1
byte  -1

Float

From the JLS :

A narrowing primitive conversion from double to float is governed by the IEEE 754 rounding rules (§4.2.4). This conversion can lose precision, but also lose range, resulting in a float zero from a nonzero double and a float infinity from a finite double .

Basically, IEEE 754 mandates this behaviour. IEEE 754 sets aside a specific bit pattern to represent infinity, and all floating-point operations involving this value are defined.

Long & Int

The JLS states that in the case of a narrowing primitive conversion from double to long or int , where the value is too large to fit in the range (64 or 32 bit signed integer), the largest representable value should be used, Long.MAX_VALUE and Integer.MAX_VALUE ;

Short & Byte

In casting a double to a short or byte , the number is first cast to an int , using the rule above. It is then cast from int to short or byte by discarding all but the n lowest bits (16 for short , 8 for byte ). Since all but the high bit of Integer.MAX_VALUE are set to 1, the short and byte values have all bits set, which corresponds to -1 in signed two's complement.

Answer 2

Float range is from 1.40129846432481707e-45 to 3.40282346638528860e+38.

The double value cannot fit in it. Java specs also say:

This conversion can lose precision, but also lose range, resulting in a float zero from a nonzero double and a float infinity from a finite double.

Answer 3

Double is 64 bits, float is 32 bits. Not all doubles will fit in a float.