How do I convert the float value of 12345.12346f to a 开发者_开发百科String of binary values, i.e. "0011010101010101", and vice-versa?
I'm not sure it is what you want, but here's a solution to have the binary representation of the IEEE 754 floating-point "double format" bit layout for a float (it is basically the memory representation of a float) :
int intBits = Float.floatToIntBits(yourFloat);
String binary = Integer.toBinaryString(intBits);
For the reverse procedure :
int intBits = Integer.parseInt(myString, 2);
float myFloat = Float.intBitsToFloat(intBits);
Working sample:
class F {
public static void main( String ... args ) {
System.out.println(
Integer.toBinaryString(
Float.floatToIntBits(12345.12346f)
)
);
}
}
Output:
1000110010000001110010001111110
For signed Floats, use Long or BigInteger to parse the string. Casting by int causes the digit at first of 32 bits be intepreted as sign digit. procedure :
int intBits = Float.floatToIntBits(yourFloat);
String binary = Integer.toBinaryString(intBits);
reverse procedure :
int intBits = new BigInteger(myString, 2).intValue();
// int intBits = (int) Long.parseLong(myString, 2);
float myFloat = Float.intBitsToFloat(intBits);
You can get the bit representation of a Java Float value using
Float.floatToIntBits(float f);
Once you have done that you can test each bit of the resulting int in turn, starting with the highest, and find out if it set, writing a '1' if it is set and a '0' if not.
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