将十进制数转换为浮点表示法和 IEEE 754 格式

Question

I'm doing an assignment for one of my classes and I'm stuck on those two questions:

1. Express the decimal -412.8 using binary floating point notation using 11 fraction bits for the significand and 3 digits for the exponent without bias

I think I managed to solve it, but my exponent has 4 bits not 3. I don't really understand how you can convert -412.8 to floating point notation using only 3 bit exponents. Here is how I tried to solve it:

First of all, the floating point notation has three parts. The sign part, 0 for positive numbers and 1 for negative numbers, the exponent part and finally the mantissa. The mantissa in this case includes the leading 1. Since the number is negative, the sign bit is going to be 1. For the mantissa, I first converted 412.8 to binary, which gave me 110011100.11 and then I shifted the decimal point to the left 8 times, which gives me 1.1001110011. The mantissa is therefore 1100 1110 011 (11 bits as the teacher asked). Finally, the exponent is going to be 2^8, since I shifted the decimal 8 times to the right. 8 is 1000 in binary. So am I correct to assume that my floating point notation should be 1 1000 11001110011?

2. Represent the decimal number 16.1875×2-134 in single-precision IEEE 754 format. I'm completely stuck on this one. I don't know how to convert that number. When I enter it in wolfram, the decimal number is way beyond the limit of the single precision format. I do know that the sign bit is going to be 0 since the number is positive. I don't know what the mantissa is though, nor how to find it. I also don't know how to find the exponent. Can someone guide me through this problem? Thanks.

Answer 1

For 1, you appear to be correct -- there's no way to reperesent the exponent unbiased in 3 bits. Of course, the problem says "3 digits" and doesn't define a base for the digits...

2 is relatively straight-forward -- convert the value to binary gives you 10000.0011 then normalize, giving 1.00000011×2 ^-130 . Now -130 is too small for a single-precision exponent (minimum is -126), so we have to denormalize (continue shifting the point to get an exponent of -126), which gives us 0.000100000011×2 ^-126 . That's then our mantissa (with the 0 dropped) and an exponent field of 0: 0|00000000|00010000001100000000000 (vertical bars separating the sign/exponent/mantissa fields) or 0x00081800