Floating point binary representation -

i have problem understading floating point representation(two's placement-sign mantissa exponent), check, doing right?

-1/7

-1*1/7*2^0=-1*4/7*2^1=-1*4/7*2^2=-1*8/7*2^3

so in binary itd like:

1 00000011 1.001 001 001 001 001 001 001

1/1357 1*1/1357*2^0=1*2048/1357*2^-11 0 | 11110101 | 1.100 000 100...  -205,34 1,60422*2^7 1| 0000011  1 | ... 

my main problem when know exponent negative, give me tips?

i assume you're talking float (i.e. ieee754 binary32)?

in binary, exact value is

-1/7 = -1.001001001001001001001001001…₂ × 2⁻³

firstly, exponent in range [-126,12], don't need worry underflow or overflow.

we round significand 24 bits:

-1.00100100100100100100101₂ × 2⁻³

(note last bit of significand rounded up)

we rewrite exponent in it's biased form (the exponent bias 127):

-1.00100100100100100100101₂ × 2^124−127

then can read off bit pattern directly:

1|01111100|00100100100100100100101 

where:

1. the sign negative, sign bit 1 (1 bit)
2. 124 in binary 01111100 (8 bits)
3. we drop implicit leading 1 significand (23 bits)