adresses in array

vbx_wx · 11-11-2010, 11:21 AM

Code:

	int main()
{
	double numbers[10] = {0.0,1.1,2.2,3.3,4.4,5.5,6.6,7.7,8.8,9.9};
	double* nPtr = &numbers[0];

	cout << numbers << endl;
	cout << nPtr + 1 << endl;
	cout << nPtr + 2 << endl;
	cout << nPtr + 3 << endl;
	cout << nPtr + 4 << endl;
	cout << nPtr + 5 << endl;
	cout << nPtr + 6 << endl;
	cout << nPtr + 7 << endl;
	cout << nPtr + 8 << endl;
	cout << nPtr + 9 << endl;

}

OUTPUT:

Code:

0xbf92e408
0xbf92e410
0xbf92e418
0xbf92e420
0xbf92e428
0xbf92e430
0xbf92e438
0xbf92e440
0xbf92e448
0xbf92e450

I am confuse why my memory adresses are not from 8 to 8 bytes. Can someone explain why ? Thank you

rupertwh · 11-11-2010, 11:30 AM

Quote:

Originally Posted by vbx_wx

... why my memory adresses are not from 8 to 8 bytes.

If by 'from 8 to 8 bytes' you mean what I think you mean, they are. I.e. each line's number is the previous plus 8.

graemef · 11-11-2010, 07:47 PM

Remember the address is printed in hexadecimal format (base 16) so the addresses are each 8 bytes long and are on zero or eight byte boundaries.

lesca · 11-12-2010, 11:05 PM

The size of double is 8 byes.

The output base is 16, digits are from 0 to 9 and a to f.

0x00 + 0x08 = 0x08 (8 in Dec)
0x08 + 0x08 = 0x10 (16 in Dec)

Hope my explanation may help you.