(take the half of the previous in each step) We introduce weights for each digit: bits in M: 0 1 0 0 1. are the ones that are actually coded in m. M must be interpreted the following way: It is interpreted as a number beginning with a 1 followed by a point (. Note that you must subtract 127 from E to obtain the original exponent e = E - 127 = 2 s=+1Į is to be interpreted as unsigned number, thus representing 129. S=0 denotes that it is a positive number, i.e. Then, you can write any number x as s * m * 2^e where ^ denotes exponentiation.ĥ.2 should be represented as follows: 0 10000001 01001100110011001100110 mantissa m (essentially the digits of your number - 24 bits.sign s (denoting whether it's positive or negative) - 1 bit.I think the diagram is not one hundret percent correct. This is not exactly how it works, but that is more or less what happens when a number like 5.2 is converted to binary. Now you only have to decide little or big endian. Now you only have to add the bias of 127 (i.e. make sure the top bit is just before the decimal point) and adjust the exponent accordingly: 1.010011001100110011001100110011.
That one missing bit doesn't matter when dividing by 5, so 1.0000. → F / 5 = 3, so it is easy to see that 0.FFFF. (This can be followed more easily if you consider one bit less: 0.FFFF. As was said, 5.2 is represented as a sign bit, an exponent and a mantissa.
0 kommentar(er)
