C++: How to round a double to an int? [duplicate]

Answer 1

It is worth noting that what you're doing isn't rounding, it's casting. Casting using (int) x truncates the decimal value of x. As in your example, if x = 3.9995, the .9995 gets truncated and x = 3.

As proposed by many others, one solution is to add 0.5 to x, and then cast.

Answer 2

add 0.5 before casting (if x > 0) or subtract 0.5 (if x < 0), because the compiler will always truncate.

float x = 55; // stored as 54.999999...
x = x + 0.5 - (x<0); // x is now 55.499999...
int y = (int)x; // truncated to 55

C++11 also introduces std::round, which likely uses a similar logic of adding 0.5 to |x| under the hood (see the link if interested) but is obviously more robust.

A follow up question might be why the float isn't stored as exactly 55. For an explanation, see this stackoverflow answer.

Answer 3

Casting to an int truncates the value. Adding 0.5 causes it to do proper rounding.

int y = (int)(x + 0.5);

Answer 4

Casting is not a mathematical operation and doesn't behave as such. Try

int y = (int)round(x);

Answer 5

#include <iostream>
#include <cmath>
using namespace std;

int main()
{
    double x=54.999999999999943157;
    int y=ceil(x);//The ceil() function returns the smallest integer no less than x
    return 0;
}