For loop to increment by decimal

Question

How can one increment in a for loop such that:

for (int j = 0; j < 1.0; j+=0.01) {}

Immediately I know that having a decimal makes the use of int odd.

EDIT:

Before anybody gets cheeky, I already know int is incorrect...the question is how to perform the method since int is not a valid manner to do so.

Why are you using `int` when you need to increment by non-integer amounts? — ShadowRanger, Apr 27 '17 at 18:19
You can't do it using an `int`, unless the value indicates hundredths. — Fred Larson, Apr 27 '17 at 18:19
It seems you need to learn about the [fundamental types](http://en.cppreference.com/w/cpp/language/types) of C++ — UnholySheep, Apr 27 '17 at 18:21
Please tell me you know the difference between integers and rational numbers. It is elementary set theory — Ed Heal, Apr 27 '17 at 18:21
So all the comments are exactly what my question is about. I've only ever incremented with `int` (haven't been doing C++ long) so what manner should one increment if a decimal increment is required? — , Apr 27 '17 at 18:21
Use double / float instead of int if you want to work with non-integer amounts — Math Newbie, Apr 27 '17 at 18:22

score 8 · Accepted Answer · edited May 23 '17 at 12:02

8

I would advise not doing this and use an integer instead. Floating point math is weird and can give you undesired results. Using your range for example we could use

for (int i = 0; i < 100; ++j)
{
    double step = static_cast<double>(i) / 100;
}

This guarantees we get 100 iterations but still allows you to get a decimal "step" for each iteration.

edited May 23 '17 at 12:02

Community

1
1

answered Apr 27 '17 at 18:22

NathanOliver

171,901
28
288
402

As so I can use the `step` variable in my loop then for say what I would have used `x = x1 + j(x2-x1)` I just replace `j` with `step`. – Apr 27 '17 at 18:24
@pasta_sauce Yes. – NathanOliver Apr 27 '17 at 18:24
Great explanation! Thanks! – Apr 27 '17 at 18:30
@pasta_sauce No problem. Glad to help. – NathanOliver Apr 27 '17 at 18:36

score 3 · Answer 2 · answered Apr 27 '17 at 18:22

Either use double and accept the accumulating errors, or use int in fixed point form, so you can compute a more accurate value on each loop.

With double:

for (double j = 0; j < 1.0; j+=0.01) {}

With fixed point to get maximum accuracy:

for (int i = 0; i < 100; ++i) {
    double j = i / 100.0;
}

score 1 · Answer 3 · answered Apr 27 '17 at 18:21

1

Use a floating point type instead of an integral type.

for (float j = 0; j < 1.0; j+=0.01) {}

or

for (double j = 0; j < 1.0; j+=0.01) {}

answered Apr 27 '17 at 18:21

R Sahu

204,454
14
159
270

score 1 · Answer 4 · answered Apr 27 '17 at 18:21

Since the type of j is int, the result of any arithmetic you do on it will be truncated to an integer value. So those increments will have no effect. If you change the type of j to double thing will work more like you expect. Keep in mind, though, that 0.01 can't be represented exactly as an IEEE floating-point value (which is what your hardware almost certainly uses).

For loop to increment by decimal

4 Answers4