As Rob mentioned, you probably just want to print the float to 1 decimal place. In this case, you can do something like the following:
#include <stdio.h>
#include <stdlib.h>
int main()
{
float conver = 45.592346543;
printf("conver is %0.1f\n",conver);
return 0;
}
If you want to actually round the stored value, that's a little more complicated. For one, your one-decimal-place representation will rarely have an exact analog in floating-point. If you just want to get as close as possible, something like this might do the trick:
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
int main()
{
float conver = 45.592346543;
printf("conver is %0.1f\n",conver);
conver = conver*10.0f;
conver = (conver > (floor(conver)+0.5f)) ? ceil(conver) : floor(conver);
conver = conver/10.0f;
//If you're using C99 or better, rather than ANSI C/C89/C90, the following will also work.
//conver = roundf(conver*10.0f)/10.0f;
printf("conver is now %f\n",conver);
return 0;
}
I doubt this second example is what you're looking for, but I included it for completeness. If you do require representing your numbers in this way internally, and not just on output, consider using a fixed-point representation instead.