Let's say that we have declared the following variables

`float a = 1.2291;`

`float b = 3.99;`

`float`

variables have precision 6, which (if I understand correctly) means that the difference between the number that the computer actually stores and the actual number that you want will be less than `10^-6`

that means that both `a`

and `b`

have some error that is less than `10^-6`

so inside the computer `a`

could actually be `1.229100000012123`

and `b`

could be `3.9900000191919`

now let's say that you have the following code

```
float c = 0;
for(int i = 0; i < 1000; i++)
c += a + b;
```

my question is,

will `c`

's final result have a precision error that is less than `10^-6`

as well or not?

and if the answer is negative, how can we actually know this precision error and what exactly happens if you apply any kind of operations, as many times you wish and in any order?

float variables have precision 6, which (if I understand correctly) means that the difference between the number that the computer actually stores and the actual number that you want will be less than 10^-6

that means that both a and b have some error that is less than 10^-6

The 10^{-6} figure is a rough measure of the **relative** accuracy when representing arbitrary constants as floats. Not all numbers will be represented with an absolute error of 10^{-6}. The number 8765432.1, for instance, can be expected to be represented approximately to the unit. If you are at least a little bit lucky, you will get 8765432 when representing it as a `float`

. On the other hand, `1E-15f`

can be expected to be represented with an absolute error of at most about 10^{-21}.

so inside the computer a could actually be 1.229100000012123 and b could be 3.9900000191919

No, sorry, the way it works is not that you write the entire number and add six zeroes for the possible error. The error can be estimated by counting six zeroes from the **leading** digit, not from the last digit. Here, you could expect 1.22910012123 or 3.990000191919.

(Actually you would get exactly 1.2290999889373779296875 and 3.9900000095367431640625. Don't forget that representation error can be negative as well as positive, as it is for the first number.)

now let's say that you have the following code […]

my question is,

will

`c`

's final result have a precision error that is less than 10^-6 as well or not?

No. The total absolute error will be the sum of all the representation errors for `a`

and `b`

for each of the thousand times you used them, plus the errors of the 2000 additions you did. That's 4000 different sources of error! Many of them will be identical, some of them will happen to compensate each other, but the end result will probably not be to 10^{-6} relative accuracy, more like 10^{-5} relative accuracy (suggestion done without counting).