Java中的数字是否有默认类型 [英] Is there a default type for numbers in Java

问题描述

如果我写这样的东西

System.out.println(18);

哪种类型有'18'？
是 int 还是 byte ？
或者它还没有类型？

Which type has the '18'? Is it int or byte? Or doesn't it have a type yet?

它不能是int，因为这样的东西是正确的：

It can't be int, because something like this is correct:

byte b = 3;

这是不正确的：

int i = 3;
byte bb = i; //error!

编辑：
我认为我在作业转换：

常量的编译时缩小意味着代码如：

The compile-time narrowing of constants means that code such as:

byte theAnswer = 42;

byte theAnswer = 42;

是允许的。如果没有缩小，整数文字42具有int类型的事实意味着需要转换为字节：

is allowed. Without the narrowing, the fact that the integer literal 42 has type int would mean that a cast to byte would be required:

byte theAnswer =（byte）42; //允许施放但不是必需的

byte theAnswer = (byte) 42; // cast is permitted but not required

推荐答案

此

18

被称为整数字面值。有各种各样的文字，浮点，字符串，字符等。

is known as an integer literal. There are all sorts of literals, floating point, String, character, etc.

在下面，

byte b = 3;

文字 3 是一个整数文字。它也是一个不变的表达。由于Java可以判断 3 适合字节，因此它可以安全地应用缩小基元转换并将结果存储在 byte 变量。

the literal 3 is an integer literal. It's also a constant expression. And since Java can tell that 3 fits in a byte, it can safely apply a narrowing primitive conversion and store the result in a byte variable.

在此

int i = 3;
byte bb = i; //error!

文字 3 是一个常量表达式，但变量 i 不是。编译器只是决定，因为 i 不是一个常量表达式，因此不会用它来计算它的值，转换为 byte 可能会丢失信息（如何将 12345 转换为字节？）因此不被允许。您可以通过使 i 一个常量变量来重写此行为

the literal 3 is a constant expression, but the variable i is not. The compiler simply decides that since i is not a constant expression, and therefore doesn't go out of its way to figure out its value, a conversion to byte may lose information (how to convert 12345 to a byte?) and should therefore not be allowed. You can override this behavior by making i a constant variable

final int i = 3;
byte bb = i; // no error!

或指定一个明确的演员

int i = 3;
byte bb = (byte) i; // no error!

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