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Why does 0x1p3 equal 8.0? Why does 0x1e3 equal 483, whereas 0x1e3d equals 7741? It is confusing since 1e3d equals 1000.0.
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As such, hexadecimal floating-point constants have exact representations in binary floating-point, unlike decimal floating-point constants, which in general do not.
Java supports hexadecimal floating-point constants much like C, with some slight difference in format. For example, it doesn’t print the ‘+’ sign for positive exponents, and it prints powers of two with a radix point followed by a zero (for example, 2 8 prints as 0x1.0p8).
This is specified as part of the Java Language Specification 4.2: The floating-point types are float, whose values include the 32-bit IEEE 754 floating-point numbers, and double, whose values include the 64-bit IEEE 754 floating-point numbers. a sign bit: 0 meaning it is positive.
A floating-point number is typically expressed in the scientific notation, with a fraction ( F ), and an exponent ( E) of a certain radix ( r ), in the form of F×r^E. Decimal numbers use radix of 10 ( F×10^E ); while binary numbers use radix of 2 ( F×2^E ). Representation of floating point number is not unique.
0x1e3 and 0x1e3d are hexadecimal integer literals. Note that e and d are hexadecimal digits, not the exponent indicator or double type indicator in this case.
1e3d is a decimal floating-point literal. The e is the exponent indicator, the d says that this is a double rather than a float.
The notation 0x1p3 is a way to express a floating-point literal in hexadecimal, as you can read in section 3.10.2 of the Java Language Specification. It means 1 times 2 to the power 3; the exponent is binary (so, it's 2-to-the-power instead of 10-to-the-power).
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0x1e3 is hex for 483, as is 0x1e3d hex for 7741. The e is being read as a hex digit with value 14.
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