I am trying to run through a sequence of memory and display it on the command window. Here is the code:

int addr[10];              //Address addr is 32 bit
for{int i=0 ; i<10 ; i++}
  int *p = &addr[i];        //Pointing to the 32 bit addresses from 0 to 9
  cout<<p<<endl;            //Display the pointed addresses
return 0;


 002AF948             //address of addr[0]
 002AF94C            //address of addr[1]
 002AF950            //address of addr[2]
 ....upto i = 9

Question: In the output, Isn't the address of addr[1] supposed to be 002AF949 instead of 002AF94C?. Why is it skipping the memory addresses 002AF49, 002AF4A, 002AF4B and then displaying addr[1] address as 002AF94C?. The same thing happens with addr[2] also.

Why is the pointer skipping these intermediate 32 bit addresses when incremented?


Since i is declared as int it will take 4 bytes to store each value of i. Since each address corresponds to a single byte, it will take 4 locations to store an int.

In your case,

It will take memory range 002AF948 - 002AF94B to store 0 (0x00000000)

It will take memory range 002AF94C - 002AF94F to store 1 (0x00000001)

And so on...

  • \$\begingroup\$ but each address such as 002AF948 is 32 bit right?. So a transition from 002AF948 to 002AF94B is 128 bits! \$\endgroup\$ – PsychedGuy Apr 1 '15 at 8:36
  • \$\begingroup\$ Why will it take a memory range of 128 bits? \$\endgroup\$ – PsychedGuy Apr 1 '15 at 8:39
  • \$\begingroup\$ @DigitalGeeK Nop. Each address stores 1 byte. So a address range 002AF948 to 002AF94B can store only 32 bits. To store 128 bits, you require 16 memory locations. So a address range 002AF948 to 002AF957 can store 128 bits. \$\endgroup\$ – nidhin Apr 1 '15 at 8:41

Four bytes are 32 bit since one byte is (normally) 8 bit.

In the past there have been some exotic exceptions to the 8 bit per byte rule but I don't think that they exist any longer.

  • \$\begingroup\$ What! I always taught that a byte = 8 bits. \$\endgroup\$ – PsychedGuy Apr 1 '15 at 8:37

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