Floating point allocation of bits........

[flukey2005]

I apologise if this is in the wrong forum.
Ok guys,

Right

i have converted 2659 into floating point format which is:

Sign ¦ Exponent ¦ Mantissa
0 ¦ 1100 ¦ 0.101001100011


0.5 =1
0.25 =0
0.125 =1
0.0625 =0
0.03125 =0
0.015625 =1
0.0078125 =1
0.00390625 =0
0.001953125 =0
0.000976563 =0
0.000488281 =1
0.000244141 =1

so all that added up works out to and then multipled by 2^12 works out to be 2659 which is all dandy.

However, for the question we're using a 16 bit processor.

12bits mantissa, 4bits exponent and 1bit for sign = 17bits which is useless for a 16 bit processor.

Can i have 11 bits for the mantissa instead so i can allocate it for a 16 bit processor?
But if i allocate 11 bits for the mantissa wouldn't this make the calculation wrong?

Cheers

[BobTheBuilder.]

Can you show how you converted this please, because that may help. I don't know about anyone else, but I've never converted a number like this so anything I post would be a guess at best!

[flukey2005]

Can you show how you converted this please, because that may help. I don't know about anyone else, but I've never converted a number like this so anything I post would be a guess at best!

Yeah sure.

2659 converted into binary is 101001100011.

Moved the decimal point 12 places so i get.

0.101001100011

Exponent is 12.

so that mantissa multiplied by 10^12 = 2659. Make sense?

Whats a better way to work it out?

Cheers.

[BobTheBuilder.]

Yeah sure.

2659 converted into binary is 101001100011.

Moved the decimal point 12 places so i get.

0.101001100011

Exponent is 12.

so that mantissa multiplied by 10^12 = 2659. Make sense?

Whats a better way to work it out?

Cheers.

Eh yeah...still not following. You are saying 0.101001100011 * 10^12 = 2659?

Because i get 0.101001100011 * 10^12 = 101001100011.

[flukey2005]

Eh yeah...still not following. You are saying 0.101001100011 * 10^12 = 2659?

Because i get 0.101001100011 * 10^12 = 101001100011.

equals 2659 because i converted back into decimal. Sorry i forgot to say

[BobTheBuilder.]

Ok sorry for the mass questions but...what's the point then? You converted, shifted the decimal, shifted it back (during the multiplication), the converted back.

[flukey2005]

Ok sorry for the mass questions but...what's the point then? You converted, shifted the decimal, shifted it back (during the multiplication), the converted back.

Just making surei its correct.

[zaza]

Hi,

Can i have 11 bits for the mantissa instead so i can allocate it for a 16 bit processor?
But if i allocate 11 bits for the mantissa wouldn't this make the calculation wrong?

You just can't do this calculation. To represent a number > 2047 in binary you need 12 digits. Then you need another 4 to represent the xponent in binary and 1 for the sign, which is 17. If you try to use 11 digits in the mantissa, then you can only represent numbers up to 11111111111 = 2047. You still need 4 for the exponent and one for the sign, which is the 16.

zaza

[kedaman]

Usually, there is an excess by 2^(the amount of bits for exponent) - 1 on the exponent. Depends on your cpu. The half precision format should have 5 digits for exponent and 11 for mantissa, but its not standard. If your cpu can't do it, then there are always ways around it although they may waste a lot of cycles.