缩小类型转换：为什么允许将int赋值给声明中的字节？ [英] Narrowing Type Conversion: Why is assignment of int to a byte in a declaration allowed?

问题描述

对于中级Java程序员来说，这听起来太微不足道了。但在我审查Java基础知识的过程中，发现了一个问题：

This may sound too trivial for an intermediate Java programmer. But during my process of reviewing Java fundamentals, found a question:

为什么缩小转换范围如下：

Why is narrowing conversion like:

byte b = 13;

将被允许而

int i = 13;
byte b = i;

会被编译器投诉吗？

推荐答案

因为字节b = 13; 是一个常量的赋值。它的值在编译时是已知的，因此编译器可以/应该/将会抱怨如果赋值的值会导致溢出（尝试 byte b = 123456789; ，看看是什么发生。）

Because byte b = 13 ; is assignment of a constant. Its value is known at compile time, so the compiler can/should/will whine if assignment of the constant's value would result in overflow (try byte b = 123456789 ; and see what happens.)

将它分配给变量后，您将分配表达式的值，尽管它可能是不变，编译器不知道。表达式可能导致溢出，因此编译器发出呜呜声。

Once you assign it to a variable, you're assigning the value of an expression, which, while it may well be invariant, the compiler doesn't know that. That expression might result in overflow and so the compiler whines.

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