Why does the following code require the (int)?

Question

The Java method Math.round can be used to round numbers. Which of the following code fragments converts a floating-point number to the nearest integer?

The right answer is:

double f = 4.65          
int n = (int) Math.round(f);

Why is it not the following:

double f = 4.65;      
int n = Math.round(f);

Answer 1

Math.round(double)返回long ，因此缩小范围。

Answer 2

Math has two round methods.

static long round(double a) 
//Returns the closest long to the argument.

static int round(float a) 
//Returns the closest int to the argument.

You are using the first one, which returns a long value, which can store larger integers than int, and cannot be implicitly cast to int.

Answer 3

If you pass a double to Math.Round then you get a long as result.
only if you pass a float you get a int as result.

From the Java Docs:

round( double a)
Returns the closest long to the argument.

round( float a)
Returns the closest int to the argument.

Answer 4

根据Java文档， Math.round(double)将返回long并且因为long不是int ，因此必须使用(int)进行(int) 。