Problem of automatic widening conversion

Question

Hello,
base one link below:
http://docstore.mik.ua/orelly/java-ent/jnut/ch02_04.htm
Finally, the notation Y* means that the conversion is an automatic widening conversion, but that some of the least significant digits of the value may be lost by the conversion. This can happen when converting an int or long to a float or double. The floating-point types have a larger range than the integer types, so any int or long can be represented by a float or double. However, the floating-point types are approximations of numbers and cannot always hold as many significant digits as the integer types.
int = 32 bits
float = 32 bits
why from int to float, will have problem? can show example or diagram of bits of problems?
long = 64 bits
double = 64 bits
why from long to float, will have problem? can show example or diagram of bits of problems?
Thanks.

Answer 1

812322 wrote:
... the notation Y* means that the conversion is an automatic widening conversion, but that some of the least significant digits of the value may be lost by the conversion. This can happen when converting an int or long to a float or double. The floating-point types have a larger range than the integer types, so any int or long can be represented by a float or double. However, the floating-point types are approximations of numbers and cannot always hold as many significant digits as the integer types.
int = 32 bits
float = 32 bits
why from int to float, will have problem? can show example or diagram of bits of problems?
long = 64 bits
double = 64 bits
why from long to float, will have problem? can show example or diagram of bits of problems?Considering just positive double (ignoring 32-bit float).
In a Java double (IEEE754 floating point)
one bit is used for the sign,
11 bits are used for the magnitude and
52 bits are used for the mantissa (the digits)
A Java double can store an int because an int only has 32 bits.
A Java double can store a long that uses only 52 bits.
When converting long values in the range [ 0x0010000000000000 ; 0x7FFFFFFFFFFFFFFF ]
because 11 bits are used for the magnitude
there are not enough bits in the mantissa to represent every long value.
Exercise:
Write a Java application that converts randomly selected long values
in the range [ 0x0010000000000000L ; 0x7FFFFFFFFFFFFFFFL ] to double and
then back to long - compute the divergence (if any) - print (in hexadecimal)
the long value and the divergence.