Why double-to-float Conversion Always Needs an Explicit Cast

A few days ago, I finally found the answer to the question that I posed in my previous post, thanks to Steve Luke's reply at JavaRanch. I created a post in the Beginning Java forum there, which led to a refresher for me on the subject of floating-point types and their ranges. An int literal may be implicitly narrowed on assignment to a byte variable if the literal value is within the range of a byte (The range of a floating-point type (double or float) is a little more complex. Apart from the possibility of overflow or underflow to positive or negative infinity, there is also the possibility of underflow to zero. This could occur if the absolute value of a number is too small for the type. So, while a double may be within the minimum and maximum float values, it may be too close to zero to fit into a float. Here is an illustration of underflow to zero from The Java Language Specification (Third Edition) (sec. 5.1.3, pg. 84). The output of:

System.out.println("(float)1e-50==" + (float)1e-50);

is:

(float)1e-50==0.0

So, the compiler refuses to implicitly convert any double value to a float. I guess that is simpler for the lazy compiler than to calculate whether the literal double value can be accurately represented as a float. Reading Section 4.2.3 Floating-Point Types, Formats, and Values of the Java Language Specification, I can see why this might not be a trivial deduction.

Primitive Assignments in Java

While studying to upgrade my Sun Certified Java Programmer status from version 1.4 to version 6, I came across what appears to be an anomaly in the Java compiler. In Java, a literal integer (like 8) is always implicitly an int, i.e. a 32-bit value. A byte is only 8 bits. However, the following narrowing assignment is legal. No explicit cast is required.

     byte b = 30;

The compiler automatically narrows the literal 30 to a byte. It can do this because 30 is a compile-time constant that is within the range of a byte. Now, recall that a floating point literal (like 43.4) is implicitly a double, i.e. a 64-bit value; but a float is only 32 bits. Consider the following code:

     float f = 43.4;

You might think that this is legal because 43.4 is a compile-time constant that is well within the range of a float. If the compiler obliges us by implicitly casting an int constant to a byte, why shouldn't it do the same for double to float assignments? However, the compiler does no such thing. The above code does not compile. The literal value has to be either explicitly cast to a float, or have an f (or F) appended to it to make it a float literal. Thus, the following assignments are all legal and compile successfully:

     float f = (float) 43.4;
     float f1 = 43.4f;
     float f2 = 43.4F;

Why is the compiler so finicky about floating points? I have not been able to think of an answer. Can you?