Find the single precision format of -118.375

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Answer 1

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Remember the sign, and consider the original [decimal] number as a positive valued mantissa.

Determine the value of the mantissa as a minimum [decimal] integer times an exponent of 2. If the mantissa has a non-0 fraction, then while it has a non-0 fraction, multiply it by 2, and decrement the exponent of 2. If the mantissa has no fraction, then while it is evenly divisible by 2, divide it by 2 and increment the exponent of 2.

The result is the sign (1 or -1), its power of 2 exponent (2^N), and [decimal] value of the mantissa.

-118.375 -> -1 * 2^+0 * 118.375

-> -1 * 2^-1 * 236.75

-> -1 * 2^-2 * 473.5

-> -1 * 2^-3 * 947.

Determine the binary value of the mantissa.

-1 * 2^-3 * 947. -> -1 * 2^-3 * 1 * 512 + 435

-> -1 * 2^-3 * 1 * 512 + 1 * 256 + 179

-> -1 * 2^-3 * 1 * 512 + 1 * 256 + 1 * 128 + 0 * 64 + 1 * 32 + 19

-> -1 * 2^-3 * 1 * 512 + 1 * 256 + 0 * 128 + 1 * 64 + 1 * 32 + 1 * 16 + 3

-> -1 * 2^-3 * 1 * 512 + 1 * 256 + 0 * 128 + 1 * 64 + 1 * 32 + 1 * 16 + 0 * 8 + 0 * 4 + 1 * 2 + 1

-> -1 * 2^-3 * 1 * 512 + 1 * 256 + 0 * 128 + 1 * 64 + 1 * 32 + 1 * 16 + 0 * 8 + 0 * 4 + 1 * 2 + 1 * 1

-> -1 * 2^-3 * 0b1101110011

Increment the exponent of 2, to move the [imaginary] decimal point of the mantissa left, until the value of the mantissa is 1 + a fraction.

2^-3 * 0b1101110011. -> 2^-2 * 0b110111001.1

-> 2^-1 * 0b11011100.11

-> 2^+0 * 0b1101110.011

-> 2^+1 * 0b110111.0011

-> 2^+2 * 0b11011.10011

-> 2^+3 * 0b1101.110011

-> 2^+4 * 0b110.1110011

-> 2^+5 * 0b11.01110011

-> 2^+6 * 0b1.101110011

If the original value was negative, the sign bit becomes 1, otherwise 0.

Sign bit: 1

Adjust the exponent by adding the bias (128) to form a positive value, and convert to binary.

Exponent: 2^+5 -> 5 + 128 (bias)

-> 133

-> 1 * 128 + 5

-> 1 * 128 + 0 * 64 + 0 * 32 + 0 * 16 + 0 * 8 + 1 * 4 + 0 * 2 + 1 * 1

-> 0b10000101

Drop the leading 1 bit of the mantissa, leaving only the fractional bits.

Mantissa: [0b1.]101110011

Put it all together:

-> 1 10000101 10111001100000000000000

-> 1100 0010 1101 1100 1100 0000 0000 0000

-> C 2 E C C 0 0 0