Address conversion

[JackMorris]

I would like to do following things:

1. Allocate the memory to store 10 integer values.
2. Store the 10 random values in the allocated memory.
3. Get the address of the first value stored. ( The address will be in hexadecimal).
4. Convert the hexadecimal address value into decimal value.
5. Add 4 in the calculated decimal value of the address.
6. The sum should be the address of the second value stored (since the integer occupies 4 bytes) in the allocated memory. The address is still in the decimal value.
7. convert the decimal value into the hexadecimal.
8. Print the value in this address ( as I said earlier this should be the second value in the allocated memory).

Could anyone suggest me how the conversion thing over here could be performed correctly such that I get the right content of the second value in the allocated memory space, if I follow the steps i mentioned above.

thanx in advance.
Jack

 

[JOLESEN]

You don't need to do any conversion. Hex / Decimal is only invented because of human eyes/brains : It's the same number hidden behind in the 32 bits.

An address is an address. (Period !).

Add 4 to the address returned and you will have the next item.

/JOlesen

 

[JOLESEN]

Just trying to solve your next problem ;-) :

If you add a number to a pointer, the pointer will be incremented with a value of number * sizeof(object).

If it's an integer pointer and you add 4 :
The pointer will point to the 4. item.
Address will be increment with 4 * sizeof(int).

That's why people often convert a pointer to a char pointer before doing math on the pointer (ie when the programmer don't want the compiler to be that 'intelligent').

/JOlesen