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Harakim · Jun 6th, 2006, 2:12 AM

This is a horribly complicated subject for me to explain, but I'll try.

If I understand you correctly, you want to pass the array and modify it and not have to return it. In Java, you can do this the same way as C++.

The references/pointers you pass in Java/C++ pass by value. They are numbers.

Just for sake of an example:

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char *temp = new char[x];
foo( temp );

void foo( char *ptr )
{
	ptr = new char[y];
}

`temp` still points at "new char[x]". Why?
I'll just show you a simplified example with sample addresses. It would probably help if you kept track on paper.

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char *temp;

temp is created at location 0x00

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temp = new char[x];

a new array is allocated starting at location 0x10
the value of the array's address (0x10) is stored in 0x00 (temp)

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foo( temp );

`ptr` is created at location 0x04

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ptr = new char[y];

a new array is allocated starting at location 0x20
the value of the array's address (0x20) is stored in 0x04 (ptr)

The chart would look like this:
location value
0x00 0x10
0x04 0x20
0x10 the value of temp[0]
0x20 the value of ptr[0]

Put another way, it is like saying this:
int temp = 1;
int ptr = temp;
ptr = 100;

Temp won't change, just ptr. char * is just an int that holds a memory address. (on 32-bit machines)

Jun 6th, 2006, 2:12 AM	#18
Harakim Hobbyist Programmer Join Date: May 2006 Location: West Jordan, Utah, United States Posts: 176 Rep Power: 3	This is a horribly complicated subject for me to explain, but I'll try. If I understand you correctly, you want to pass the array and modify it and not have to return it. In Java, you can do this the same way as C++. The references/pointers you pass in Java/C++ pass by value. They are numbers. Just for sake of an example: (Toggle Plain Text) char temp = new char[x]; foo( temp ); void foo( char ptr ) { ptr = new char[y]; } char temp = new char[x]; foo( temp ); void foo( char ptr ) { ptr = new char[y]; } `temp` still points at "new char[x]". Why? I'll just show you a simplified example with sample addresses. It would probably help if you kept track on paper. (Toggle Plain Text) char temp; char temp; temp is created at location 0x00 (Toggle Plain Text) temp = new char[x]; temp = new char[x]; a new array is allocated starting at location 0x10 the value of the array's address (0x10) is stored in 0x00 (temp) (Toggle Plain Text) foo( temp ); foo( temp ); `ptr` is created at location 0x04 (Toggle Plain Text) ptr = new char[y]; ptr = new char[y]; a new array is allocated starting at location 0x20 the value of the array's address (0x20) is stored in 0x04 (ptr) The chart would look like this: location value 0x00 0x10 0x04 0x20 0x10 the value of temp[0] 0x20 the value of ptr[0] Put another way, it is like saying this: int temp = 1; int ptr = temp; ptr = 100; Temp won't change, just ptr. char * is just an int that holds a memory address. (on 32-bit machines)