Conversion from double to integer [duplicate]

Question

I am stuck in problem where the double number is not getting properly converted to integer.

In this case->

int x=1000; double cuberoot=pow(x,(1/(double)3)); int a=cuberoot; cout<<"cuberoot="<<cuberoot<<endl; cout<<"a="<<a<<endl;

Output:

cuberoot=10 a=9

Why here a=9 and not 10?

Any solution to this problem??

Also I don't want to round the value..if a=3.67 then it should be converted to 3 only and not 4.

You can add a small positive to number and then calculate the root. Like x=1000+0.05. This would give a=10. — user2734982
– user2734982, Commented Jun 14, 2014 at 18:03

yizzlez · Accepted Answer · 2014-06-14 18:11:12Z

1

Because the cuberoot is very close to 10 but not quite 10. std::cout truncates and rounds the number to 10, but a double to integer conversion will strip the decimal which is why a = 9. To solve this problem, you can use std::round():

int a=round(cuberoot);

edited Jun 14, 2014 at 18:11

answered Jun 14, 2014 at 17:59

yizzlez

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7 Comments

user2826957 Over a year ago

So any solution to this so that a=10 here?

OldProgrammer Over a year ago

use a round() function.

user2826957 Over a year ago

Actually problem is I don't want to round exactly.

user2826957 Over a year ago

I want that if a=3.99 then it should be converted to 3 only....any solution to this?

yizzlez Over a year ago

Right now cube = 9.99... and it is converting that to a=9

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druckermanly · Accepted Answer · 2014-06-14 18:03:09Z

Try this and see why!

#include <iostream> #include <iomanip> #include <cmath> using namespace std; int main(){ int x = 1000; double cube = pow(x, 1.0/3.0); int a = cube; cout<<"cube="<< fixed << setprecision(16) << cube<<endl; cout<<"a="<<a<<endl; }

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