reference of array

Question

I have the following code

char str[] = "some string"; // legal
char const* str2 = str;     // legal
char const** str3 = &str2;  // legal
char const** str4 = &str;   // illegal (error: cannot convert 'char (*)[12]' to 'const char (*)[]' in initialization )

Why the last one is not compiled? What is the type of &str?

what is the error given on the last one? – Syntactic Fructose Mar 18 '13 at 17:21 — Syntactic Fructose, Mar 18 '13 at 17:21
*Arrays are **not** pointers* – David Rodríguez - dribeas Mar 18 '13 at 17:31 — David Rodríguez - dribeas, Mar 18 '13 at 17:31

Joseph Mansfield · Accepted Answer · 2013-03-18T17:30:27.503

7

Since str is an array, the type of &str is a "pointer to array of 12 const char". The valid declaration would be:

char const (*str4)[12] = &str;

The declaration of str3 works fine because str2 is a "pointer to const char", not an array, initialized using array-to-pointer conversion. The same array-to-pointer conversion does not occur when you take the address of an array, as in the declaration of str4.

edited Mar 18 '13 at 17:30

answered Mar 18 '13 at 17:22

Joseph Mansfield

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What can I do something to adapt this declaration for array of any length? For example, I tried `char const (*str4)[]` and I got an error. – user14416 Mar 18 '13 at 17:59
@user14416 You can't - it's a pointer to an array of specific length. – Joseph Mansfield Mar 18 '13 at 18:07

emd · Answer 2 · 2013-03-18T17:43:26.510

0

Because of the number of de-referencings. (str[] or str*) -> data, str2 -> data, str3 -> pointer(str2) -> data, str4 -> pointer -> data. The number of * simbols tells you home many imes you need to dereference it to get to the data. Each time is jumping from address to address and only the last access is asctually data.

The right way would be:

char const* str4 = &str; //or
char const* str4 = str2;

edited Mar 18 '13 at 17:43

answered Mar 18 '13 at 17:26

emd

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reference of array

2 Answers2