Java双初始化 [英] Java double initialization

问题描述

这些陈述在哪些方面有所不同？

In what way are these statements different?

1. double dummy = 0;


2. double dummy = 0.0;


3. double dummy = 0.0d;


4. double dummy = 0.0D;

1. double dummy = 0;
2. double dummy = 0.0;
3. double dummy = 0.0d;
4. double dummy = 0.0D;

推荐答案

尝试了一个简单的程序（同时使用0和100，以显示特殊常量和常规常量之间的区别），Sun Java 6编译器将为1和2输出相同的字节码（就编译器而言，情况3和4与2相同）。

Having tried a simple program (using both 0 and 100, to show the difference between "special" constants and general ones) the Sun Java 6 compiler will output the same bytecode for both 1 and 2 (cases 3 and 4 are identical to 2 as far as the compiler is concerned).

例如：

double x = 100;
double y = 100.0;

编译为：

0:  ldc2_w  #2; //double 100.0d
3:  dstore_1
4:  ldc2_w  #2; //double 100.0d
7:  dstore_3

然而，我看不到任何东西Java语言规范保证这种常量表达式的编译时扩展。对于以下情况，编译时缩小：

However, I can't see anything in the Java Language Specification guaranteeing this compile-time widening of constant expressions. There's compile-time narrowing for cases like:

byte b = 100;

在第5.2节，但这并不完全相同。

as specified in section 5.2, but that's not quite the same thing.

也许眼睛比我更敏锐的人可以在某处找到保证...

Maybe someone with sharper eyes than me can find a guarantee there somewhere...

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