Why converting from float to double changes the value?

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

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.

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.

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.