Thread: Differnce between C and Java returning an Object reference/pointer

Moreover the array declaration

Code:

char a[]="Sumit";

Here a also points to the string literal.
What is the difference?

Originally Posted by AndrewHunter

In Salem's example you are returning a pointer, which since C is pass by value, gets copied. That pointer points to a string literal which stays present for the life of the program.

Be careful. It does not point to a string literal.

The "Sumit" here is just one way of writing:

Code:

 char a[] = {'S','u','m','i','t',0};

Saving ones fingers does not change the semantics of variable storage. The array (meaning its contents) are stored locally in the scope of the nearest block.

Originally Posted by sumit180288

Moreover the array declaration

Code:

char a[]="Sumit";

Here a also points to the string literal.
What is the difference?

Again, in that case, you are copying the character literal into a local array of bytes.

@whiteflags and @all dear members

Can u tell me the difference in the mechanisms of memory storage in case of a string literal and the one mentioned by @whiteflags i.e.

Code:

char a[] = {'S','u','m','i','t',0};

Is String literal not stored as a[] is stored?

Originally Posted by whiteflags

Be careful. It does not point to a string literal.

The "Sumit" here is just one way of writing:

Code:

 char a[] = {'S','u','m','i','t',0};

Saving ones fingers does not change the semantics of variable storage. The array (meaning its contents) are stored locally in the scope of the nearest block.

@laserlight.
Nice thoughts there. Could you please tell me a link or a reference where i can get these nice facts about the scope of a literal.

Originally Posted by laserlight

What do you mean?

EDIT:
Oh, I think I understand. Let me just say this: the string literal has static storage duration. Its lifetime is for the entire program. A local array's lifetime, on the other hand, is only until its scope ends, unless it has say, static storage duration due to being declared static.

Originally Posted by sumit180288

@Salem. Sorry to say i didn't see the pointer version. I thought u are asking about the array version.
I am also not knowing why this verison works?
Anyone plz clear this doubt.

Regards

Ok... all the way back to your first message...

The Java version works because Java has a native string type that is "reference counted". That is the memory occupied by the string is not released if there is anything referring to it, like in your first example.

However, C does not have a string type (despite some who argue to the contrary). C programs use character arrays to store groups of characters with a terminating null. We call it a string but it isn't really a string (in the Pascal/Basic/Java sense) at all. It is nothing but an array, that gets destroyed in the stack cleanup when the function exits, leaving you with a pointer to garbage.

Why does the one in message 15 work? Because the literal string "Sumit" is not created or stored in the program stack area used by the function, therefore it's address remains valid after the function exits... that is, it doesn't get invalidated by the stack cleanup.


This is one of the pitfalls of comparing languages... as they say "This ain't That"... and it's a mistake to try to understand one by knowing the other.

@Tater
Correctly said... But i will one more argument in support.
Every thing is an object in Java (including arrays) i.e. Memory for these objects are allocated dynamically out of heap.
Hence as u mentioned It remains there until all references to that memory location dies out. (i.e. reference count becomes zero).
This is internally performed internally using the GC (Garbage collector) which is a very nice functionality provided by Java and we dont have to worry about
freeing up this dynamic heap memory. It is automatically managed by JDK.


Memory allocated out of heap space will remain for a lifetime which is till program ends or if memory is freed forcefully.

1. The memory which is allocated using malloc() etc function
2. String literal. (Dynamic allocation not done but literals have static scope)
etc.

@ALL PLZ ADD ANYTHING ELSE WHERE MEMORY IS DYNAMICALLY ALLOCATED.

Originally Posted by CommonTater

Ok... all the way back to your first message...

The Java version works because Java has a native string type that is "reference counted". That is the memory occupied by the string is not released if there is anything referring to it, like in your first example.

However, C does not have a string type (despite some who argue to the contrary). C programs use character arrays to store groups of characters with a terminating null. We call it a string but it isn't really a string (in the Pascal/Basic/Java sense) at all. It is nothing but an array, that gets destroyed in the stack cleanup when the function exits, leaving you with a pointer to garbage.

Why does the one in message 15 work? Because the literal string "Sumit" is not created or stored in the program stack area used by the function, therefore it's address remains valid after the function exits... that is, it doesn't get invalidated by the stack cleanup.


This is one of the pitfalls of comparing languages... as they say "This ain't That"... and it's a mistake to try to understand one by knowing the other.