Long is 64 bytes and both int and float are 32 bytes. Why can a long be assigned to a float without an explicit cast but an int can not? If a long is declared as follows: long l = 123456L; This doesn't implicitly convert int i = l; // Gives a compile error as expected but doing this safely compiles and run. float f = l; Can someone please explain? Thanks

The way floats are stored internally (in the form of exponents) gives it a larger range and hence makes it safe to cast a long to a float. Long to int of course might lead to possible loss of info.

Good question. A floating point number uses essentially the binary version of scientific notation... so it can express the very large possible values of long variables, but with a loss of *precision*. That is, I might convert (this example by analogy is actually using decimal precisions) 23472166003027 into (2.34722 * 10^13) as a floating point number. Whether this should be considered a widening conversion (and thus not need a cast) is a debatable point, but the Java design team decided that it should. On the pro side, you do get a reasonable value, even if precision is lost (compare with, for example, converting long to int, where a value out of range creates a practically random numeric result). On the con side, it *is* a loss of precision, so perhaps the user should be warned about it. I'm always in favor of compiler checks, so I disagree with Java's choice on this one. If this would help, this is the data from IEEE 754 Floating Point which Java complies : A float is 4 bytes, 32 bits, With 6 to 7 significant digits of accuracy. covers a range from �1.40129846432481707e-45 to �3.40282346638528860e+38, is formed of 3 fields: 1-bit sign 8-bit base 2 exponent biased+127 23-bit fraction, lead 1 implied e.g. 3. = 0x404000 -3. = 0xC04000

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