Can I access pointers the same way as arrays?

Question

When I have int *integers; and I allocate memory using
integers = (int *)malloc(sizeof(int) * 10);, can I access, say the 2nd value stored in there with integers[1], or do I have to do that with *(integers + 1)?

Thanks, so is it basically a "dynamic array" when using it that way? — hm1912, Jan 30 '17 at 13:52
@Algorithmiker - be wary of the term "dynamic". It has several meanings. Be sure you don't apply the incorrect one here. — StoryTeller - Unslander Monica, Jan 30 '17 at 13:58
Ther is harld anything useful you can do with **arrays** actually, except take their address, size and alignment. The index-operator for example is applied to a pointer only, not an array. What about pointers in your C book **specifically** is unclear? — too honest for this site, Jan 30 '17 at 14:13

pmg · Answer 1 · 2017-01-30T14:00:51.703

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Yes, you can access values through pointers as if they were arrays.

In normal circumstances (any use except after &, sizeof, or a string literal) an array gets converted to a pointer to its first element, so, in effect, using an array is converted to using a pointer, not the other way round.

edited Jan 30 '17 at 14:00

answered Jan 30 '17 at 13:54

pmg

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You cannot index an array! The array name is converted ot a pointer before the index-operator is applied. – too honest for this site Jan 30 '17 at 14:14

score 3 · Accepted Answer · answered Jan 30 '17 at 15:32

The subscript operator [] is defined in terms of pointer arithmetic. The expression a[i] is evaluated as *(a + i) - given the address a, offset i elements from that address and dereference the result¹.

So, yes, you can use the [] operator on a pointer expression as well as an array expression.

Remember that with pointer arithmetic, the size of the pointed-to type is taken into account - if a is a pointer to an int, then a + 1 yields the address of the next integer object (which can be anywhere from 2 to 4 to 8 bytes from a).

Also remember that arrays are not pointers - an array expression will be converted ("decay") to a pointer expression unless it is the operand of the sizeof or unary & operators, or is a string literal used to initialize a character array in a declaration.

^{This also means that array indexing is commutative - a[1] == 1[a]. However, you'll rarely see i[a] outside of the International Obfuscated C Code Contest.}

score 0 · Answer 3 · edited May 23 '17 at 12:33

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can I access, say the 2nd value stored in there with integers1, or do I have to do that with *(integers + 1)?

Yes, you can. Array names decay to pointers so in both cases, you can act as if you have a pointer pointing at the first element of your data.

NOTE : See this link on why you should not cast the result of malloc.

edited May 23 '17 at 12:33

Community

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answered Jan 30 '17 at 13:52

Marievi

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"Array names decay to pointers" - is wrong. They decay **for most usages**. But not always as the statement implies. And it is not "as-if", but they _are_ converted to pointers to the first element. – too honest for this site Jan 30 '17 at 14:40

Can I access pointers the same way as arrays?

3 Answers3