Write accurate calculation code requires to know about the edge cases. Let’s explore what happens when casting a float into an int, in particular when the values overflow.

int intValue = (int) Float.NEGATIVE_INFINITY;
System.out.println(intValue);
// => -2147483648

The conversion to integers returns the closest value below lowest value of the primitive type. Narrowing from float to long works the same. There is nothing surprising about this, but what about smaller primitive types such as byte, char and short ?

short shortValue = (short) Float.NEGATIVE_INFINITY;
System.out.println(shortValue);
// => 0

Outch. The Java specification reveals that the float is converted to an int, and clamped to the negative value of largest magnitude: 0x80000000. Then the 32 bit binary representation is truncated to a 16 bit short: 0x0000.

It is possible compute the float range that is converted without causing overflow:

Float.intBitsToFloat(Float.floatToIntBits(Short.MAX_VALUE+1)-1);
// => 32767.998
Float.intBitsToFloat(Float.floatToIntBits(Short.MIX_VALUE-1)-1);
// => -32768.996

The trick consists in selecting the first integer that cause an overflow by explicitly going over the short range boundaries. This is what Short.MAX_VALUE + 1 achieves.

Then the integer value is converted to float. This is an exact conversion given that float has a 23 bits mantissa. This float value is the smallest one that causes an overflow.

Finally the next float value toward zero is selected. As a consequence the rounding toward zero will match the right boundary of short value interval.

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