Why converting from float to double changes the value?

问题

I\'ve been trying to find out the reason, but I couldn\'t. Can anybody help me?

Look at the following example.

float f = 125.32f;
System.out.println(\"value of f = \" + f);
double d = (double) 125.32f; 
System.out.println(\"value of d = \" + d);

This is the output:

value of f = 125.32
value of d = 125.31999969482422

回答1:

The value of a float does not change when converted to a double. There is a difference in the displayed numerals because more digits are required to distinguish a double value from its neighbors, which is required by the Java documentation. That is the documentation for toString, which is referred (through several links) from the documentation for println.

The exact value for 125.32f is 125.31999969482421875. The two neighboring float values are 125.3199920654296875 and 125.32000732421875. Observe that 125.32 is closer to 125.31999969482421875 than to either of the neighbors. Therefore, by displaying “125.32”, Java has displayed enough digits so that conversion back from the decimal numeral to float reproduces the value of the float passed to println.

The two neighboring double values of 125.31999969482421875 are 125.3199996948242045391452847979962825775146484375 and 125.3199996948242329608547152020037174224853515625.
Observe that 125.32 is closer to the latter neighbor than to the original value (125.31999969482421875). Therefore, printing “125.32” does not contain enough digits to distinguish the original value. Java must print more digits in order to ensure that a conversion from the displayed numeral back to double reproduces the value of the double passed to println.

回答2:

1. When you convert a float into a double, there is no loss of information. Every float can be represented exactly as a double.
2. On the other hand, neither decimal representation printed by System.out.println is the exact value for the number. An exact decimal representation could require up to about 760 decimal digits. Instead, System.out.println prints exactly the number of decimal digits that allow to parse the decimal representation back into the original float or double. There are more doubles, so when printing one, System.out.println needs to print more digits before the representation becomes unambiguous.

回答3:

The conversion from float to double is a widening conversion, as specified by the JLS. A widening conversion is defined as an injective mapping of a smaller set into its superset. Therefore the number being represented does not change after a conversion from float to double.

More information regarding your updated question

In your update you added an example which is supposed to demonstrate that the number has changed. However, it only shows that the string representation of the number has changed, which indeed it has due to the additional precision acquired through the conversion to double. Note that your first output is just a rounding of the second output. As specified by Double.toString,

There must be at least one digit to represent the fractional part, and beyond that as many, but only as many, more digits as are needed to uniquely distinguish the argument value from adjacent values of type double.

Since the adjacent values in the type double are much closer than in float, more digits are needed to comply with that ruling.

回答4:

The 32bit IEEE-754 floating point number closest to 125.32 is in fact 125.31999969482421875. Pretty close, but not quite there (that's because 0.32 is repeating in binary).

When you cast that to a double, it's the value 125.31999969482421875 that will be made into a double (125.32 is nowhere to be found at this point, the information that it should really end in .32 is completely lost) and of course can be represented exactly by a double. When you print that double, the print routine thinks it has more significant digits than it really has (but of course it can't know that), so it prints to 125.31999969482422, which is the shortest decimal that rounds to that exact double (and of all decimals of that length, it is the closest).

回答5:

The issue of the precision of floating-point numbers is really language-agnostic, so I'll be using MATLAB in my explanation.

The reason you see a difference is that certain numbers are not exactly representable in fixed number of bits. Take 0.1 for example:

>> format hex

>> double(0.1)
ans =
   3fb999999999999a

>> double(single(0.1))
ans =
   3fb99999a0000000

So the error in the approximation of 0.1 in single-precision gets bigger when you cast it as double-precision floating-point number. The result is different from its approximation if you started directly in double-precision.

>> double(single(0.1)) - double(0.1)
ans =
     1.490116113833651e-09

回答6:

As already explained, all floats can be exactly represented as a double and the reason for your issue is that System.out.println performs some rounding when displaying the value of a float or double but the rounding methodology is not the same in both cases.

To see the exact value of the float, you can use a BigDecimal:

float f = 125.32f;
System.out.println("value of f = " + new BigDecimal(f));
double d = (double) 125.32f;
System.out.println("value of d = " + new BigDecimal(d));

which outputs:

value of f = 125.31999969482421875
value of d = 125.31999969482421875

回答7:

it won`t work in java because in java by default it will take real values as double and if we declare a float value without float representation like 123.45f by default it will take it as double and it will cause an error as loss of precision

回答8:

The representation of the values changes due to contracts of the methods that convert numerical values to a String, correspondingly java.lang.Float#toString(float) and java.lang.Double#toString(double), while the actual value remains the same. There is a common part in Javadoc of both aforementioned methods that elaborates requirements to values' String representation:

There must be at least one digit to represent the fractional part, and beyond that as many, but only as many, more digits as are needed to uniquely distinguish the argument value from adjacent values

To illustrate the similarity of significant parts for values of both types, the following snippet can be run:

package com.my.sandbox.numbers;

public class FloatToDoubleConversion {

    public static void main(String[] args) {
        float f = 125.32f;
        floatToBits(f);
        double d = (double) f;
        doubleToBits(d);
    }

    private static void floatToBits(float floatValue) {
        System.out.println();
        System.out.println("Float.");
        System.out.println("String representation of float: " + floatValue);
        int bits = Float.floatToIntBits(floatValue);
        int sign = bits >>> 31;
        int exponent = (bits >>> 23 & ((1 << 8) - 1)) - ((1 << 7) - 1);
        int mantissa = bits & ((1 << 23) - 1);
        System.out.println("Bytes: " + Long.toBinaryString(Float.floatToIntBits(floatValue)));
        System.out.println("Sign: " + Long.toBinaryString(sign));
        System.out.println("Exponent: " + Long.toBinaryString(exponent));
        System.out.println("Mantissa: " + Long.toBinaryString(mantissa));
        System.out.println("Back from parts: " + Float.intBitsToFloat((sign << 31) | (exponent + ((1 << 7) - 1)) << 23 | mantissa));
        System.out.println(10D);
    }

    private static void doubleToBits(double doubleValue) {
        System.out.println();
        System.out.println("Double.");
        System.out.println("String representation of double: " + doubleValue);
        long bits = Double.doubleToLongBits(doubleValue);
        long sign = bits >>> 63;
        long exponent = (bits >>> 52 & ((1 << 11) - 1)) - ((1 << 10) - 1);
        long mantissa = bits & ((1L << 52) - 1);
        System.out.println("Bytes: " + Long.toBinaryString(Double.doubleToLongBits(doubleValue)));
        System.out.println("Sign: " + Long.toBinaryString(sign));
        System.out.println("Exponent: " + Long.toBinaryString(exponent));
        System.out.println("Mantissa: " + Long.toBinaryString(mantissa));
        System.out.println("Back from parts: " + Double.longBitsToDouble((sign << 63) | (exponent + ((1 << 10) - 1)) << 52 | mantissa));
    }
}

In my environment, the output is:

Float.
String representation of float: 125.32
Bytes: 1000010111110101010001111010111
Sign: 0
Exponent: 110
Mantissa: 11110101010001111010111
Back from parts: 125.32

Double.
String representation of double: 125.31999969482422
Bytes: 100000001011111010101000111101011100000000000000000000000000000
Sign: 0
Exponent: 110
Mantissa: 1111010101000111101011100000000000000000000000000000
Back from parts: 125.31999969482422

This way, you can see that values' sign, exponent are the same, while its mantissa was extended retained its significant part (11110101010001111010111) exactly the same.

The used extraction logic of floating point number parts: 1 and 2.

回答9:

Both are what Microsoft refers to as "approximate number data types."

There's a reason. A float has a precision of 7 digits, and a double 15. But I have seen it happen many times that 8.0 - 1.0 - 6.999999999. This is because they are not guaranteed to represent a decimal number fraction exactly.

If you need absolute, invariable precision, go with a decimal, or integral type.

来源：https://stackoverflow.com/questions/17504833/why-converting-from-float-to-double-changes-the-value