CODEWITHSUNDEEP
c++cstoragememory-address
sandbox
sandbox

Reputation: 2679

Confusion in Memory Addressing with Arrays

Lets have an array of type int:-

        int arr[5];

Now, 

       if arr[0] is at address 100 then
       Why do we have;
       arr[1] at address 102 , 
       arr[2] at address 104 and so on.

Instead of 

       arr[1] at address 101 , 
       arr[2] at address 102 and so on.

Is it because an integer takes 2 bytes?

Does each memory block has 1 Byte capacity (whether it is 32 bit processor or 64 bit)?

Upvotes: 1

Views: 114

Answers (3)

havexz
havexz

Reputation: 9590

You are right, on your machine the sizeof int is 2, so next possible value in the array will be 2 bytes away from the previous one.

-------------------------------
|100|101|102|103|104|105|106....
-------------------------------
arr[0]  arr[1]  arr[2]

There is no guaranty regarding size of int. C++ spec just says that sizeof(int) >= sizeof(char). It depends upon processor, compiler etc.

For more info try this

Upvotes: 0

NPE
NPE

Reputation: 500157

Your first example is consistent with 16-bit ints.

As to your second example (&arr[0]==100, &arr[1]==101, &arr[2]==103), this can't possibly be a valid layout since the distance between consecutive elements varies between the first pair and the second.

Upvotes: 3

Constantinius
Constantinius

Reputation: 35039

It is because an integer takes 2 bytes?

Yes

Apparently on your system, int has the size of 2. On other systems, this might not be the case. Usually int is either sized 4 or 8 bytes, but other sizes are possible also.

Upvotes: 3

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