is there any way???

[dolphin1]

i want using some number floating point
but in the end of the my routine i want convert this(floating) code to integer but i cant find any way please write for me a few line code
thanx to every body...


const int R=384;
int Rx //Rx is variable reading from port
float z,t
main(void){
read by y from RA1;
z=y/x;
if(z<1){t=1-z;x=x+x*t;} //add(or subtruct) tolerance to R(esistance)
if(z>1){t=z-1;x=x-x*t;}

[Ttelmah]

In C, this is semi automatic, but can also be forced with a 'cast'.
If you have:

float val;
int ival;

Then you can say:
ival=val*10.213;

and ival, will receive the integer part of the float sum 'val' times 10.213.

Hence as I say 'semi automatic'.

However you can also 'force' the conversion at particular points. So for instance:

ival = ((int)(val*10.213))/5;

Will take the float number 'val', use floating point arithmetic to multiply this by 10.213, then convert the result to integer (This is the '(int)' statement), then divide the integer by 5, using integer arithmetic.
This bracketted numeric type, is called a 'cast', and is particularly useful to override the defaults, where otherwise a number might overflow. If (for instance), you calculate:

int8 ival2,ival;
ival=(ival*128)/ival2;

Then the multiplication will be done using 8bit arithmetic since all the numbers concerned are 8bit values. However:

ival=(ival*128L)/ival2;

Will force the arithmetic to use 16bit, since '128' is now declared as a 'long'. Similarly:

ival=((int16)ival*128)/ival2;

Will do the same, since ival, will now be converted to a 16bit type, before the sum.

Best Wishes