# Make a float using integers [closed]

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;
``````