Double to int

[enemy]

Hello!
I tried to convert double number m to integer.
Problem is, program is automatically converting p also to integer!
But I wrote just m to be converted to integer, not p!

Please, could someone help me get along with the mess?
Many thanks!

  double p3=d3*100/s;
double m3= p3*s/100;
int b3=(int)m3;

[Chervil]

Your question isn't clear. What are d3 and s

[enemy]

I am counting a procent.


(ie. we have d3 = 3 red apples, s= all the apples, p3= % red apples from all the apples)

d3 from the whole amount of s is a procent (p3)

problem is just that I converted to integer double m3 and not double p3.

But when I wrote cout << p3;

then p3 was alo rounded to a n integer.

But with the expression
int b3=(int)m3;

I thought, I converted jsut doube m3 to integer, not also p3.

it means, that , somehow automaticaly, p3 also "converted " from double to integer.

is that because they are both "connected" withe the sedond expression? I did not know that shoul dhave some influence, cause that is the row BEFORE.

many thanks!

[Annatar]

I believe since you're using both integers and doubles in the same equation, it is causing an issue.

Look for a warning message when you compile your program that indicates a conversion problem.

You might want to use all doubles and setprecision(0) for the numbers you want ?

Someone else should confirm what I'm saying, I'm still a noob but I had a similar problem just last night and solved it that way.

[a k n]

Setprecision() works for the output streams , doubles have as much precision as the mantissa can hold.
Try using all doubles, and writing 100.0 instead of just 100.

[Chervil]

I think the previous comments probably give the correct answer.

My question about d3 and s, I really meant, what type are they, double or int or something else.

But 100 is definitely of type int. In this case it would probably be better to use 100.0 instead, which is type double.