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volatile · Jun 2nd, 2006, 2:24 PM

I see I'm not the only one confused. Why do you want them to be adjacent, why do you care. Thing is, they most likely are adjacent, you just have'nt realised it.

As DaWei said, the compiler will pack your data as it see fit. Lets look at an example ( I compiled this with Dev-C++):

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#include <iostream>

using namespace std;

typedef double Scalar;
int main()
{
	Scalar x;
	Scalar y;
	Scalar z;
	
	cout << &x << endl;
	cout << &y << endl;
	cout << &z << endl;
}

output:

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0x22ff70
0x22ff68
0x22ff60

Even though x came first in the source code, it is stored last in memory. They are however packed titely. Considering the fact that a double is eight byte, this came to no surprise.

You cant assume that data you define in your source is sequential.

Also your code is wrong to begin with.

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&x+sizeof(Scalar) != &y

WRONG!!!!!

You have to be carefull when doing arithmetic with pointers or references. Lets look at an example:

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int *ptr = 0;
cout << ptr << endl;
ptr++
cout << ptr <<endl;

output:

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0
0x4

You'd might think p was 1 after increment, but its 4.

Similarly *(p+2) is element at address 8, the same as p[2]. Its the same with references. This is how your code should be:

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&x+1 != &y

Since sizeof(Scalar) is 8, &x + 8 is the eight element of type Scalar after &x. The same as saying address of x pluss 8*8 wich is pluss 40.

Jun 2nd, 2006, 2:24 PM	#5
volatile Newbie Join Date: Nov 2005 Posts: 18 Rep Power: 0	I see I'm not the only one confused. Why do you want them to be adjacent, why do you care. Thing is, they most likely are adjacent, you just have'nt realised it. As DaWei said, the compiler will pack your data as it see fit. Lets look at an example ( I compiled this with Dev-C++): (Toggle Plain Text) #include <iostream> using namespace std; typedef double Scalar; int main() { Scalar x; Scalar y; Scalar z; cout << &x << endl; cout << &y << endl; cout << &z << endl; } #include <iostream> using namespace std; typedef double Scalar; int main() { Scalar x; Scalar y; Scalar z; cout << &x << endl; cout << &y << endl; cout << &z << endl; } output: (Toggle Plain Text) 0x22ff70 0x22ff68 0x22ff60 0x22ff70 0x22ff68 0x22ff60 Even though x came first in the source code, it is stored last in memory. They are however packed titely. Considering the fact that a double is eight byte, this came to no surprise. You cant assume that data you define in your source is sequential. Also your code is wrong to begin with. (Toggle Plain Text) &x+sizeof(Scalar) != &y &x+sizeof(Scalar) != &y WRONG!!!!! You have to be carefull when doing arithmetic with pointers or references. Lets look at an example: (Toggle Plain Text) int ptr = 0; cout << ptr << endl; ptr++ cout << ptr <<endl; int ptr = 0; cout << ptr << endl; ptr++ cout << ptr <<endl; output: (Toggle Plain Text) 0 0x4 0 0x4 You'd might think p was 1 after increment, but its 4. Similarly (p+2) is element at address 8, the same as p[2]. Its the same with references. This is how your code should be: (Toggle Plain Text) &x+1 != &y &x+1 != &y Since sizeof(Scalar) is 8, &x + 8 is the eight element of type Scalar after &x. The same as saying address of x pluss 88 wich is pluss 40.