C++ asterisk and bracket operators used together

Question

Sorry for the crappy title, but I don't know how to better describe this problem.

What is the meaning of:

int (*x)[];

and how to initialize it?

I know it isn't int *x[] since:

int a,b;
int (*x)[] = {&a, &b};

won't compile.

Thank you in advance.

score 26 · Accepted Answer · answered Apr 27 '12 at 12:27

Type declarations have an expression-like syntax, so you parse them as you would an expression:

      x       x is
     *x       a pointer
    (*x)[]    to an array of unknown dimensions
int (*x)[]    of int

The precedence rule is that the operators to the right bind tighter than those to the left, in each case, the operator closer to the element binds tighter, and finally, that parentheses can be used to change these bindings, so:

int  *x[];    is an array of pointers,
int *(x[]);   as is this.
int (*x)[];   whereas here, the parentheses change the bindings.

score 13 · Answer 2 · edited Feb 12 '17 at 22:05

13

Using cdecl, you can easily determine the type:

declare x as pointer to array of int

So you can initialize x with the address of an array:

int a[]; // conceptual: error if compiled
x = &a;

Note that the size of the array is part of its type, so you cannot assign the address of an array of size N to a pointer to array of size M, where N!=M (this is the reason for the error above: you need to know the size of the array to declare it)

int b[5];
int c[6];
int (*y)[6];
y = &b; // error
y = &c; // OK

edited Feb 12 '17 at 22:05

ZachB

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answered Apr 27 '12 at 12:23

Attila

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Uptick for mentioning `cdecl` - I'd forgotten it existed - although strictly, it only supports C syntax. – Component 10 Apr 27 '12 at 12:43
Thanks for the link, it's useful! – Chlind Apr 27 '12 at 13:53

A.Danesh · Answer 3 · 2012-04-27T13:33:43.037

3

int (*x)[] using for a pointer to an empty int array. But initialize an empty array not possible in c++. Only you can define it. You should use a non-empty array:

int a[2] = {1, 2};
int (*x)[2] = &a;

edited Apr 27 '12 at 13:33

answered Apr 27 '12 at 12:24

A.Danesh

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GCC complains it can't convert `int (*)[2]` to `int (*)[]`. Edit: Now it deviates from the question ;) – chris Apr 27 '12 at 12:25

Summer_More_More_Tea · Answer 4 · 2012-04-27T12:37:10.000

2

int (*x)[] means x is a pointer to int array. You can do like this:

int a[] = {1, 2};
int (* b)[2] = &a;

When interpreted array-typed variable, we should go from right to left. e.g.

int *x[]

indicates that x is an array, elements in the array is typed as int * i.e. pointer to int. Thus x represents an array whose elements are int pointer. However, the parentheses change precedence of the operator. Now * is first interpreted indicating x is a pointer, whose type is int []. Thus, x in int (*x)[] is a pointer to int array and you should initialize it accordingly.

edited Apr 27 '12 at 12:37

answered Apr 27 '12 at 12:24

Summer_More_More_Tea

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Does not compile: error: cannot convert 'int (*)[2]' to 'int (*)[]' in assignment – Attila Apr 27 '12 at 12:31
http://ideone.com/ - I think it uses gcc in the back, but I do not know what version/options – Attila Apr 27 '12 at 12:33
It won't compile in Visual Studio 2010: cannot convert from 'int (*)[2]' to 'int (*)[]' – Chlind Apr 27 '12 at 12:34
@Chlind I don't know whether it's an extension on `gcc`, you can modify definition of `b` to `int (*b)[2]` according to the error. Update my answer. Thanks. – Summer_More_More_Tea Apr 27 '12 at 12:38

Matthieu M. · Answer 5 · 2012-04-28T10:27:45.510

1

I used to have issues with C typing too, until I learned how it was created.

In C, the type is described in the way you would use a variable of that type.

Therefore, when you see:

int *x;

it means that the expression *x is of type int, so x is variable of type pointer-to-int.

And if you see:

int x[5];

it means that the expression x[3] is of type int, so x is a variable of type array-of-int.

So, to get to your expression:

int (*x)[];

it means that the expression *x is of type int[] (ie, an array of int of unknown size). Therefore x is a variable of type pointer-to-an-array-of-int.

edited Apr 28 '12 at 10:27

answered Apr 27 '12 at 12:41

Matthieu M.

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You are essentially correct but you could be even more clear. When you see `int *x` that means that `*x` is *a variable* of type int. The way you said it, it is not clear whether you meant that `*x` was a *value* of type int or a *variable* of type int. – Eric Lippert Apr 27 '12 at 17:14
@EricLippert: `*x` is not a variable, it's an expression. I introduced "expression" and "variable" all over the place to mark the distinction and hopefully make things clearer. – Matthieu M. Apr 28 '12 at 10:27
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Of course `*x` is a *variable*; it represents a *storage location* that *you can assign values to*, so it's a *variable*; that's what a variable *is*: a storage location you can assign values to. C programmers of course tend to call such things "lvalues" rather than "variables" which is unfortunate; "lvalue" is confusing and unnecessarily jargonish way to say "variable". – Eric Lippert Apr 28 '12 at 15:03

score 0 · Answer 6 · answered Apr 27 '12 at 12:58

0

int (*x)[ ] This means x is a pointer to an int array Initialize as: int (*x)[ ]; int a[10]; x = &a;

int *x[ ] This means x is an array of pointers to int int *x[10]; int a[10]; x[0] = a;

In my opinion, always use initialized array variables or use int **x instead of *x[ ] and then allocate memory on runtime.

answered Apr 27 '12 at 12:58

Arpan

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The first use example does not compile: error: cannot convert 'int ()[10]' to 'int ()[]' in assignment – Attila Apr 27 '12 at 13:48

C++ asterisk and bracket operators used together

6 Answers6

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