java float 向double 隐式转换精度会有丢失
float f = 8.69f; int a = Float.floatToIntBits(f); String floatStr = Integer.toBinaryString(a); double d1 = f; long b = Double.doubleToLongBits(d1); String convertStr = Long.toBinaryString(b); double d2 = 8.69d; long c = Double.doubleToLongBits(d2); String doubleStr = Long.toBinaryString(c); System.out.println(floatStr); System.out.println(convertStr); System.out.println(doubleStr);输出为
1000001000010110000101000111101 100000000100001011000010100011110100000000000000000000000000000 100000000100001011000010100011110101110000101000111101011100001
想问下，为什么在隐式转换的过程中，jvm只是单纯的拷贝float中的尾数部分，然后补0，而不是精确的计算尾数部分的值？
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There is no loss of precision when adding 0 to the mantissa, this is how floating point numbers are represented
float occupies 4 bytes, while double occupies 8 bytes.
During the memory copy process, only 4 bytes are copied in, and the remaining bytes default to 0. (This is not necessarily the case in release mode and debug mode.)