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Is primitive assigned a memory address?

Tags:

java

primitive

I am trying to understand the process of declaration and assignment of a primitive type at the back stage.

  1. int i;
  2. i = 3;

For 1), on the memory stack, it assigns a space for storing an int type value named i For 2), it assigns the value 3 to the space preserved above

Is there a memory address there? From my impression, memory address is always associated with the objects on the heap?

Update:

Regarding the replies:

So, for every variable on the stack, they are all assigned a memory address as well just like the objects on the heap. Am I correct?

But for Java, this is not the case?

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user36064 Avatar asked Nov 25 '08 02:11

user36064


2 Answers

There are not always addresses involved. The compiler can put variables into registers if it finds that their address is never taken by the programmer. So you wouldn't need any access to the main memory. For example in your code above, what the compiler could generate could be as simple as

add $2, $0, 3

to put value 3 into register 2. As soon as you create a pointer and make it point to that variable, then you have an address, actually. And then the variable cannot be in a register only anymore.

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Johannes Schaub - litb Avatar answered Sep 21 '22 23:09

Johannes Schaub - litb


Assuming you're talking about C or C++ (I can't tell), yes. You can access the address like so:

int i = 3;

int *k = &i; // k now is a pointer to i

*k = 4; // assigns the value k points to (i) to 4, i is now 4
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Kyle Cronin Avatar answered Sep 18 '22 23:09

Kyle Cronin