### Problem with decimals

Hy,

I'm having a prob, with a program that's supposed to print values of y=2x+5, for x from -10 to 10.
I know it's really simple code, but it gets me a decimal value of i...when it should just subtract 0.2, so it ends up like -4.60001, or 7.79999...http://prntscr.com/2rxep2

Here is the code:

 ``123456789`` `````` #include using namespace std; int main() { for(float i(-10); i<=10; i+=0.2) cout<<"Za i="<
The output of your code is normal, I'm not sure what more you are looking for. If you don't want decimal numbers then use integers or use output stream manipulators:
http://en.cppreference.com/w/cpp/io/manip/setprecision

`std::cout << std::fixed << std::setprecision(0) << "Za i = " << i << " : " << 2 * i + 5 << std::endl;`
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You've tested this code and got the normal output ?
Because look at this: http://prntscr.com/2rxep2

And decimals don't bother me..they need to be cuz there is a step of 0.2.....I'm sorry if I have expressed myself wrong in previous post.

What bothers me is those 0's and then 1 in output eg. -4.60001.

Do not use float as iterating variable. 0.2 is about 1/5, so stepping from -50 to 50 would give you how many steps? More or less than your loop?

Wait, "0.2 is about 1/5"? Isn't it exactly so? No. Floating point numbers are not continuous; they are discrete, with limited precision. You do see that imprecision in your results. I guess that demostrating this feature is a purpose of your homework(?).

You can use tools from <iomanip> to format the output.
Floating-point numbers are inexact.
Not every real number can be represented perfectly as a `float` or `double`, so we have to stick with rough approximations, which is what you're seeing.
For instance, at least on my machine, setting `float f = 0.02;` will actually make f closer to 0.2000000029802322387695312500.

In your case, you can either reduce the precision of cout:
`cout.precision(3); // Display 3 digits `

or you can switch to `double` (which has more precision, but it's still inexact) instead of `float`.

In either case, your i = 0 is still going to be messed up, because you start at i = -10 and then you keep adding approximations of 0.2, so you'll actually end up "missing" 0 by a very small margin (I end up with -5.03957e-006).
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Great, thank you for your explanation long double main
I'll just set the digits to display then.

Thanks to all of you
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