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# C++/array

Question
int i[10];
i[10]=1;

lly char ch[10];
ch[10]='c';

Are these assignment statements valid? I am trying to put the value at the last position. In above declaration as per concept they dont allocate memory to store at the N position

No, these assignments are _not_ valid.

You are _not_ putting values in the last array slots, you are putting them in the one past the end position.

This is because in C and C++ arrays are indexed from 0 not 1. Thus all arrays have valid index values from 0 to n-1 inclusive where n is the size (or bound) of the array.

Thus for your arrays where n would be 10 the valid index values are:

0, 1, 2, 3, 4, 5, 6, 7, 8, 9

Count them – there _are_ still 10 array slots.

The reason for this is to do with efficiency, obviousness and simplicity (from the compiler’s point of view!). Arrays in C and C++ are contiguous chunks of memory. The name of an array is synonymous to a pointer to (i.e. the address of) the first element of the array. The expression array[n] is the same as *(array+n). That is: take the address of the first element, add the index to obtain the address of the nth element, then de-reference the address (meaning to obtain the value at the address).

Thus the first element is obtained by adding zero to the address of the first element and dereferencing. Using an initial index of 1 would only complicate matters for no real benefit.

The value n is scaled by the size of the type of the array elements. This scaling by the size of the pointer type is true of all pointer arithmetic in C and C++. For example, assuming a byte addressed machine, then to access the element at index 3 for an array of byte sized elements (as is likely for your ch array) would add 3*1 to the address of the initial element. But to obtain the address of an element at index 3 in an array of 4 byte integers (as is probable for your i array) would add 3*4, i.e. 12 to the address of the initial element.

C++

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