i=m.i*1.00 isn't integer so it'll be stored as float.

>This is official and must trust me, or must i control it ?
>
>Then:
>

>i=1-2^31     && i value is stored  into 4 bytes
>i=m.i-1      && i value is convert to a 8 bytes
>i=m.i+1      && i value is convert to a 4 bytes
>i=m.i*1.00   && i value is 4 or 8 bytes ?
>

>