Java内存分配 [英] Java memory allocation

问题描述

java中的"Short"在内存中占据2个字节，数值范围为-32768到32767.
让我们将该类型的变量声明为，
short var;
var = 5;

现在，该值5也在字节"数据类型的范围内(从-128到127)，但由于它是短"变量，因此仍在内存中占据2个字节.

为什么在处理此类情况以有效利用内存时Java不允许隐式类型转换?

"Short" in java occupies 2 bytes in memory with numeric range as -32768 to 32767 .
Let''s declare a variable of that type as,
short var ;
var=5;

Now, this value 5 also comes under the range of "Byte" datatype(from -128 to 127) ,but still it occupies 2 bytes in memory as it is a "short" variable.

Why did java not allowed implicit typecasting while dealing with such scenarios to make efficient use of memory?

推荐答案

在将来的某个时间点，您将1000分配给并将其存储在byte中?

您会破坏内存！

这些选择都是经过深思熟虑后做出的，除非您已经检查了所有内容并且提供了更好的东西，否则请不要质疑它们.

What in a future point in time you assign 1000 to your var and it was stored in a byte?

You would corrupt the memory!

These choices were made with a lot of thinking behind them, please do not question them unless you have checked everything, and are offering something better.

类型转换不会产生任何区别，因为var声明为short类型，因此在程序运行时无法进行更改.考虑类似的东西:

Typecasting would not make any difference as var is declared as a short type so it cannot be changed as the program is running. Consider something like:

short var;
var = 5;
// ...
// do some processing
// ...
var = 300;
// what happens now if var had been dynamically changed to byte?

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