Java memory allocation

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"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?

Posted 22-Sep-11 21:18pm

snp_shailesh

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Mehdi Gholam · Answer 1 · 2011-09-22T21:22:00

Solution 1

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.

Posted 22-Sep-11 21:22pm

Mehdi Gholam

Richard MacCutchan · Answer 2 · 2011-09-22T23:20:00

Solution 2

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:

Java

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

Posted 22-Sep-11 23:20pm

Richard MacCutchan

Comments

snp_shailesh 23-Sep-11 5:25am

hmmmm ..
got it now ...
thanks both of you for responding ...

Nagy Vilmos 23-Sep-11 6:44am

Good answer.