Vrftjkry

I want to make a floating point number that has only one decimal point.
I have separate integers for both side.
Ex:

int n1 = 8;

int n2 = 2;

I want to make 8.2 as float value using separate integers.
Please give me a solution.

Seems too simple:

float x = n1 + n2 * 0.1;

Is there a trick?

Edit: The method proposed by Michel Keijzers, namely

float x = n1 + n2 / 10.0;

(I removed the redundant casts) can be slightly more accurate, but takes
longer to compute, because division is significantly slower than
multiplication on the Uno. Computing n2/10.0 always yields the
correctly rounded result, namely the float that best approximates the
exact mathematical result. On the other hand, n2*0.1 involves two
rounding operations: one at compile time, in the representation of 0.1
(which is not an exact float), another at run time, which rounds the
result of the multiplication. If n2 is between 0 and 8, you end up
getting the correctly rounded result anyway, just as with n2/10.0.
However, if n2 is 9, then

• 
  n2*0.1 yields 0.900000035762786865234375 (error ≈ 3.6e-8)


• 
  n2/10.0 yields 0.89999997615814208984375 (error ≈ -2.4e-8)

The former carries a rounding error 1.5 larger than the latter.

float x = (float) n1 + (float) (n2 / 10.0);

This works only if the value of n2 has 1 digit. Otherwise the following make it smaller until n2 gets below 1.0:

float x = (float) n1;



float f2 = (float) n2;

while (f2 >= 1.0)

{

   f2 /= 10.0;

}

x = n1 + f2;