For value-type fields, this storage holds the actual value. The values of each of these types would require 16 bytes of memory assuming a bit word size. The field I in each case takes 4 bytes to store its value, the field S takes 4 bytes to store its reference, and the field L takes 8 bytes to store its value. So the memory for the value of both RefType and ValType looks like this:.
Now if you had three local variables in a function, of types RefType , ValType , and int[] , like this:. Memory at address 0x4AB68 value of refType would be something like:. Memory at address 0x4AA4C value of intArray would be something like:. Now, if you passed intArray to another function, the value pushed onto the stack would be 0x4AA4C , the address of the array, not a copy of the array. Arrays in Java store one of two things: either primitive values int , char , So, new Integer[10] creates space for 10 Integer references only.
It does not create 10 Integer objects or even free space for 10 Integer objects. So long as you know what the semantics are, the only consequences of stack vs heap are in terms of making sure you don't overflow the stack, and being aware that there's a cost associated with garbage collecting the heap. For instance, the JIT could notice that a newly created object was never used outside the current method the reference could never escape elsewhere and allocate it on the stack.
It doesn't do that at the moment, but it would be a legal thing to do. Likewise the C compiler could decide to allocate all local variables on the heap - the stack would just contain a reference to an instance of MyMethodLocalVariables and all variable access would be implemented via that.
In fact, variables captured by delegates or iterator blocks already have this sort of behaviour. But it decays to a pointer to an array when used in appropriate context. How is the array stored in memory? Asked 5 Months ago Answers: 5 Viewed 31 times.
Your array is allocated on the heap, and the ints are not boxed. So the memory for the value of both RefType and ValType looks like this: 0????????????????????? S"; refType. S"; valType. Memory at address 0x4AB68 value of refType would be something like: 0?????????????????????
If you only want to know if the memory is on the stack or heap, read the other answers, they are much more informed than me.
A statically-created array will be on the stack or in the. A dynamically-created array with new or malloc will be allocated on the heap. First, you pointer has to be of type int. An array is just a group of integer, saved in the memory as single integer, but in one row.
A integer has 4-Byte in the memory, so you can access each value of your array by increasing your pointer by 4. Stack Overflow for Teams — Collaborate and share knowledge with a private group. Create a free Team What is Teams?
Collectives on Stack Overflow. Learn more. How is the array stored in memory? Ask Question. Asked 9 years, 5 months ago. Active 1 month ago. Viewed 56k times. Improve this question. Maybe I've missed something, but why do you need to know whether it is stored in the stack or on the heap?
Have you tried p[0]? Perhaps you can ask a question by saying what problem you are having so that we have a starting point As Ray says, depending on what you're doing it may well "not matter" Either way, those aren't the only two classes of storage Still, this may be useful: gotw. The question seems to be about byte order.
Add a comment. Active Oldest Votes. Improve this answer. NAND 7 7 silver badges 22 22 bronze badges. Alok Save Alok Save k 44 44 gold badges silver badges bronze badges. Since I can't add comments just yet, here's my two cents in an answer: If you only want to know if the memory is on the stack or heap, read the other answers, they are much more informed than me.
NickO NickO 1 1 gold badge 6 6 silver badges 19 19 bronze badges. Still, one should always remember that compilers are different. So just because you observe a pattern in your experiments doesn't mean it's a "truth".
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