Memory allocation in C vs Java

Question

I am trying to find if a number is even or odd without using Modulus operation. According to my understanding, when I define an int in C or in Java, that number is stored as 64 bytes of memory. I can access Zeroth(0th) bit of that number using square bracket[].

However, I am not able to do it in Java.

So, my question is how memory allocation and syntax is different in Java?

Answer 1

This really is nothing to do with "memory" or "memory allocation". You're talking about a simple property of a number.

In binary, the least significant bit of a number is 1 if the number is odd, or 0 if the number is even. OK, so how do we isolate the bottom bit? The answer is the 'and' operation, expressed in both C and Java as '&'.

  boolean isEven = (number & 1) == 0;  
  boolean isOdd = (number & 1) != 0;  

By the way, in C

• An integer certainly does not occupy 64 bytes. Maybe 64 bits, depending on computer, C implementation, etc.

• Given a single integer, brackets don't give you bit-level access. Brackets are for array indexing: ie, picking out the N'th number in an array of numbers.

Answer 2

You cannot access memory in Java like you can in C

You can, however, use the bit wise & operator to pick out one or more bits of the integer value. If you do it for the bit corresponding to even/odd and see if the result is zero or not, you will have your answer without using modulo

Answer 3

You can't access individual bits in an integer type in C or Java using the [] operator. The operand of the [] operator must be an array type (C and Java) or a pointer type (C).

Integer types in C and Java behave similarly, although integer types in Java have a fixed size (C integer types have a minimum size). In both languages, integer types are allocated the same way.

As for testing odd/even integers without using bitwise operations (which may have endianness issues) or the modulus operator, you can take advantage of integer arithmetic.

In both C and Java, dividing an integer by an integer gives an integer result. 3/2 == 1 , not 1.5 . 100/26 == 3 , etc. So, if you divide the original value by 2, then multiply that result by 2, you'll get the original value back only if the original value is even:

3 / 2 == 1, 1 * 2 == 2, 2 != 3 
4 / 2 == 2, 2 * 2 == 4, 4 == 4
5 / 2 == 2, 2 * 2 == 4, 4 != 5

You get the idea. So,

int tmp = x / 2;
if ( x == tmp * 2 )
  // x is even
else
  // x is odd