将float转换为int

Question

I come from Python background and taking a dive into Java world. I am trying to convert Float to int in Java. Something we do like this in Python int_var = int(float_var)

public class p1 {
    public static void main(String args[]) {
        Integer a = new Integer(5);
        Float b;
        b = new Float(3.14);
        a = (int)b;
        System.out.println(a);
    }
}

Yields the following error -

 p1.java:7: error: inconvertible types a = (int)b; ^ required: int found: Float 1 error 

Answer 1

你可以做到

a = b.intValue()

Answer 2

That's one of the annoying things in Java. Fix your problem with this:

a = (int)(float)b;

Unboxing will require that you cast from Float to float and then to int

Answer 3

Use primitives types, and you will be fine:

int a = 5;
float b = 3.14f;  // 3.14 is by default double. 3.14f is float.

a = (int)b;

The reason it didn't work with Wrappers is those types are non-covariant, and thus are incompatible.

---

And if you are using Integer types (Which you really don't need here), then you should not create Wrapper type objects using new . Just make use of auto-boxing feature:

Integer a = 5;  // instead of `new Integer(5);`

the above assignment works after Java 1.5, performing auto-boxing from int primitive to Integer wrapper. It also enables JVM to use cached Integer literals if available, thus preventing creation of unnecessary objects.

Answer 4

Since you're using boxed primitive, a = b.intValue(); should suit your needs.

Answer 5

 float a =10.0f;
 int k=(int)a;

Answer 6

使用此代码
a = b.intValue();

Answer 7

使用Math.round方法：

int a = Math.round(b);

Answer 8

for the best always use:

Math.floor(d)
Math.round(d)
Math.abs(d)

These are meant for conversions.

Answer 9

您可以简单地将Use Math.round（）用作

   int a = Math.round(14.32f)