带时间的浮点精度-乘除法

Question

Will these statements execute with exactly the same results across different platforms?

    currentTime = System . nanoTime ( ) * 0.000000001f ; // 01

    currentTime = System . nanoTime ( ) / 1000000000f ; // 02

The concern is loss of precision using a small floating point value.

Answer 1

Just leave it as a long. No loss of precision and perfect Determinism across the board.