Re: HLASM Floating Point Field - Remainder?

Hansedl wrote:
(snip)
uses 32 bits of data from the registers. The field contains
x'43E88000'. This means the x'43' denotes the sign bit and the
characteristic or
exponent. The x'E88000' is the mantissa or fraction (binary '1110
1000 1000'). The exponent is binary '100 0011' (67 decimal). In
order to support
negative characteristics, the assembler increases the exponent by 64
before converting it to binary. The bias is 64 so 67 - 64 = 3 (the
exponent).
So the value is '111.010001000' because we move the radix point three
places to the right. This field contains the Elapsed time, in seconds
which is

The exponent is base 16, so you move three hex digits, or 12 bits,
such that the value us X'E88' or decimal 3720.

(snip)

Another number I have is x'423c0000'. So x'42' is binary '1000010'
or decimal 66. And 66 - 64 = 2. So x'3C' is binary '111100'
adjusted two

Again two hex digits to the right for X'42' or 66.

If this is homework, please reference the newsgroup.

-- glen

.

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