How to reinterpret the bits of a float as an int

Question

What is the Java equivalent of following C++ code?

 float f=12.5f;
 int& i = reinterpret_cast<int&>(f);

Did you check if the code runs successfully?? it should be `int& i = r_c(f)` — UltraInstinct, Jan 01 '10 at 17:06
It is not really translatable into any language. It is interpreting the bits of the float as an int!!! What does that mean? Because the representation of floating point is undefined by the standard it only has meaning if you understand what floating point representation is used by your compiler/hardware here and know that this maps to the Java floating point representation. — Martin York, Jan 01 '10 at 17:31
In addition to what Martin York said, it is technically performing an *implementation-defined mapping* from float reference to int reference. Strictly speaking, there's no guarantee that the resulting int reference will alias the same object. — jalf, Jan 01 '10 at 17:52
See also http://stackoverflow.com/questions/4805058/is-there-cast-in-java-similar-to-reinterpret-cast-in-c — Raedwald, Mar 24 '16 at 09:36

score 26 · Accepted Answer · answered Jan 01 '10 at 16:29

26

float f = 12.5f;
int i = Float.floatToIntBits(f);

answered Jan 01 '10 at 16:29

missingfaktor

1

Out of curiosity, what is the difference between what you posted and a int i = (int)f;? – Alberto Zaccagni Jan 01 '10 at 16:35
4

@Montecristo : (int) f in Java is like static_cast(f) in C++. – missingfaktor Jan 01 '10 at 16:39
Ok... so I think I'll have to look for those two c++ methods to get the whole picture, thanks :) – Alberto Zaccagni Jan 01 '10 at 16:46
@Montecristo : They are cast operators, not methods. – missingfaktor Jan 01 '10 at 16:51
13

Note the C++ version uses int reference. Thus modifying the float will cause the int to change (in some undefined way). The Java version does not provide that functionality. Every change to 'f' must be followed by a call to floatToIntBits(). – Martin York Jan 01 '10 at 17:35

1 Answers1