IEEE 754 to decimal

Question

I'm currently trying to convert the following IEEE 754 hex number 0x805c00f0 to its decimal equivalent which according to online converters is about -8.44920195816662938E-39. Working it out on paper step-by-step, I get the following:

805c00f0 = 1000 0000 0101 1100 0000 0000 1111 0000 Leftmost 1 means the number is negative. The next eight bits, 000 0000 0 means an exponent of -127 after subtracting the bias. I'm left with bits 101 1100 0000 0000 1111 0000, the mantissa.

After recalling the implicit 1, I have -1.101 1100 0000 0000 1111 0000 * 2^-127. Moving the decimal point to the left 127 places, I have -0.00(...)1101 1100 0000 0000 1111 0000. Summing up, I get -1(2^(-127)+2^(-128)+2^(-130)+2^(-131)+2^(-132)+2^(-143)+2^(-144)+2^(-145)+2^(-146)) = -1.01020727331947522E-38. This is not equal to what converters have given me and I cannot understand why. What am I getting wrong?

For what it's worth, I used binaryconvert.com/result_float.html?hexadecimal=805C00F0 to check my answer — user1800967
If you guys are stumped too, that's okay. I'd like to at least hear from you as you ponder this. — user1800967

Rob Starling Rob Starling · Accepted Answer · 2014-02-19T03:30:49

per wikipedia "The stored exponents 00 and FF are interpreted specially." and the formula to use is (−1)^(signbits)×2^(−126)×0.(significandbits)

805c00f0 =>
8 0 5 c 0 0 f 0 =>
1000 0000 0101 1100 0000 0000 1111 0000 =>
1 00000000 101 1100 0000 0000 1111 0000 =>
-1 x 2^(-126) x 0.10111000000000011110000 =>
-1 x 2^(-126) x (2^-1 + 2^-3 + 2^-4 + 2^-5 + 2^-16 + 2^-17 + 2^-18 + 2^-19) =>
-1 * (2^-127 + 2^-129 + 2^-130 + 2^-131 + 2^-142 + 2^-143 + 2^-144 + 2^-145) =>
-8.449202e-39

IEEE 754 to decimal

1 Answers