Differences when using ** in C

问题:

I started learning C recently, and I'm having a problem understanding pointer syntax, for example when I write the following line:

int ** arr = NULL;

How can I know if:

• arr is a pointer to a pointer of an integer

• arr is a pointer to an array of pointers to integers

• arr is a pointer to an array of pointers to arrays of integers

Isn't it all the same with int ** ?

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Another question for the same problem:

If I have a function that receives char ** s as a parameter, I want to refer to it as a pointer to an array of strings, meaning a pointer to an array of pointers to an array of chars, but is it also a pointer to a pointer to a char?

回答1:

Isn't it all the same with int **?

You've just discovered what may be considered a flaw in the type system. Every option you specified can be true. It's essentially derived from a flat view of a programs memory, where a single address can be used to reference various logical memory layouts.

The way C programmers have been dealing with this since C's inception, is by putting a convention in place. Such as demanding size parameter(s) for functions that accept such pointers, and documenting their assumptions about the memory layout. Or demanding that arrays be terminated with a special value, thus allowing "jagged" buffers of pointers to buffers.

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I feel a certain amount of clarification is in order. As you'd see when consulting the other very good answers here, arrays are most definitely not pointers. They do however decay into ones in enough contexts to warrant a decades long error in teaching about them (but I digress).

What I originally wrote refers to code as follows:

void func(int **p_buff)
{
}

//...

int a = 0, *pa = &a;
func(&pa);

//...

int a[3][10];
int *a_pts[3] = { a[0], a[1], a[2] };
func(a_pts);

//...

int **a = malloc(10 * sizeof *a);
for(int i = 0; i < 10; ++i)
  a[i] = malloc(i * sizeof *a[i]);
func(a);

Assume func and each code snippet is compiled in a separate translation unit. Each example (barring any typos by me) is valid C. The arrays will decay into a "pointer-to-a-pointer" when passed as arguments. How is the definition of func to know what exactly it was passed from the type of its parameter alone!? The answer is that it cannot. The static type of p_buff is int**, but it still allows func to indirectly access (parts of) objects with vastly different effective types.

回答2:

The declaration int **arr says: "declare arr as a pointer to a pointer to an integer". It (if valid) points to a single pointer that points (if valid) to a single integer object. As it is possible to use pointer arithmetic with either level of indirection (i.e. *arr is the same as arr[0] and **arr is the same as arr[0][0]) , the object can be used for accessing any of the 3 from your question (that is, for second, access an array of pointers to integers, and for third, access an array of pointers to first elements of integer arrays), provided that the pointers point to the first elements of the arrays...

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Yet, arr is still declared as a pointer to a single pointer to a single integer object. It is also possible to declare a pointer to an array of defined dimensions. Here a is declared as a pointer to 10-element array of pointers to arrays of 10 integers:

cdecl> declare a as pointer to array 10 of pointer to array 10 of int;
int (*(*a)[10])[10]

In practice array pointers are most used for passing in multidimensional arrays of constant dimensions into functions, and for passing in variable-length arrays. The syntax to declare a variable as a pointer to an array is seldom seen, as whenever they're passed into a function, it is somewhat easier to use parameters of type "array of undefined size" instead, so instead of declaring

void func(int (*a)[10]);

one could use

void func(int a[][10])

to pass in a a multidimensional array of arrays of 10 integers. Alternatively, a typedef can be used to lessen the headache.

回答3:

How can I know if :

• arr is a pointer to a pointer of an integer

It is always a pointer to pointer to integer.

• arr is a pointer to an array of pointers to integers
• arr is a pointer to an array of pointers to arrays of integers

It can never be that. A pointer to an array of pointers to integers would be declared like this:

int* (*arr)[n]

It sounds as if you have been tricked to use int** by poor teachers/books/tutorials. It is almost always incorrect practice, as explained here and here and ( with detailed explanation about array pointers) here.

EDIT

Finally got around to writing a detailed post explaining what arrays are, what look-up tables are, why the latter are bad and what you should use instead: Correctly allocating multi-dimensional arrays.

回答4:

Having solely the declaration of the variable, you cannot distinguish the three cases. One can still discuss if one should not use something like int *x[10] to express an array of 10 pointers to ints or something else; but int **x can - due to pointer arithmetics, be used in the three different ways, each way assuming a different memory layout with the (good) chance to make the wrong assumption.

Consider the following example, where an int ** is used in three different ways, i.e. p2p2i_v1 as a pointer to a pointer to a (single) int, p2p2i_v2 as a pointer to an array of pointers to int, and p2p2i_v3 as a pointer to a pointer to an array of ints. Note that you cannot distinguish these three meanings solely by the type, which is int** for all three. But with different initialisations, accessing each of them in the wrong way yields something unpredictable, except accessing the very first elements:

int i1=1,i2=2,i3=3,i4=4;

int *p2i = &i1;
int **p2p2i_v1 = &p2i;  // pointer to a pointer to a single int

int *arrayOfp2i[4] = { &i1, &i2, &i3, &i4 };
int **p2p2i_v2 = arrayOfp2i; // pointer to an array of pointers to int

int arrayOfI[4] = { 5,6,7,8 };
int *p2arrayOfi = arrayOfI;
int **p2p2i_v3 = &p2arrayOfi; // pointer to a pointer to an array of ints

// assuming a pointer to a pointer to a single int:
int derefi1_v1 = *p2p2i_v1[0];  // correct; yields 1
int derefi1_v2 = *p2p2i_v2[0];  // correct; yields 1
int derefi1_v3 = *p2p2i_v3[0];  // correct; yields 5

// assuming a pointer to an array of pointers to int's
int derefi1_v1_at1 = *p2p2i_v1[1];  // incorrect, yields ? or seg fault
int derefi1_v2_at1 = *p2p2i_v2[1]; // correct; yields 2
int derefi1_v3_at1 = *p2p2i_v3[1]; // incorrect, yields ? or seg fault


// assuming a pointer to an array of pointers to an array of int's
int derefarray_at1_v1 = (*p2p2i_v1)[1]; // incorrect; yields ? or seg fault;
int derefarray_at1_v2 = (*p2p2i_v2)[1]; // incorrect; yields ? or seg fault;
int derefarray_at1_v3 = (*p2p2i_v3)[1]; // correct; yields 6;

回答5:

How can I know if :

arr is a pointer to a pointer of an integer

arr is a pointer to an array of pointers to integers

arr is a pointer to an array of pointers to arrays of integers

You cannot. It can be any of those. What it ends up being depends on how you allocate / use it.

So if you write code using these, document what you're doing with them, pass size parameters to the functions using them, and generally be sure about what you allocated before using it.

回答6:

Pointers do not keep the information whether they point to a single object or an object that is an element of an array. Moreover for the pointer arithmetic single objects are considered like arrays consisting from one element.

Consider these declarations

int a;
int a1[1];
int a2[10];

int *p;

p = &a;
//...
p = a1;
//...
p = a2;

In this example the pointer p deals with addresses. It does not know whether the address it stores points to a single object like a or to the first element of the array a1 that has only one element or to the first element of the array a2 that has ten elements.

回答7:

The type of

int ** arr;

only have one valid interpretation. It is:

arr is a pointer to a pointer to an integer

If you have no more information than the declaration above, that is all you can know about it, i.e. if arr is probably initialized, it points to another pointer, which - if probably initialized - points to an integer.

Assuming proper initialization, the only guaranteed valid way to use it is:

**arr = 42;
int a = **arr;

However, C allows you to use it in multiple ways.

• arr can be used as a pointer to a pointer to an integer (i.e. the basic case)

int a = **arr;

• arr can be used as a pointer to a pointer to an an array of integer

int a = (*arr)[4];

• arr can be used as a pointer to an array of pointers to integers

int a = *(arr[4]);

• arr can be used as a pointer to an array of pointers to arrays of integers

int a = arr[4][4];

In the last three cases it may look as if you have an array. However, the type is not an array. The type is always just a pointer to a pointer to an integer - the dereferencing is pointer arithmetic. It is nothing like a 2D array.

To know which is valid for the program at hand, you need to look at the code initializing arr.

Update

For the updated part of the question:

If you have:

void foo(char** x) { .... };

the only thing that you know for sure is that **x will give a char and *x will give you a char pointer (in both cases proper initialization of x is assumed).

If you want to use x in another way, e.g. x[2] to get the third char pointer, it requires that the caller has initialized x so that it points to a memory area that has at least 3 consecutive char pointers. This can be described as a contract for calling foo.

回答8:

C syntax is logical. As an asterisk before the identifier in the declaration means pointer to the type of the variable, two asterisks mean pointer to a pointer to the type of the variable.

In this case arr is a pointer to a pointer to integer.

There are several usages of double pointers. For instance you could represent a matrix with a pointer to a vector of pointers. Each pointer in this vector points to the row of the matrix itself.

One can also create a two dimensional array using it,like this

 int **arr=(int**)malloc(row*(sizeof(int*)));  
 for(i=0;i<row;i++) {
 *(arr+i)=(int*)malloc(sizeof(int)*col); //You can use this also. Meaning of both is same. //
  arr[i]=(int*)malloc(sizeof(int)*col); }

回答9:

There is one trick when using pointers, read it from right hand side to the left hand side:

int** arr = NULL;

What do you get: arr, *, *, int, so array is a pointer to a pointer to an integer.

And int **arr; is the same as int** arr;.

回答10:

int ** arr = NULL;

It's tell the compiler, arr is a double pointer of an integer and assigned NULL value.

回答11:

There are already good answers here, but I want to mention my "goto" site for complicated declarations: http://cdecl.org/

Visit the site, paste your declaration and it will translate it to English.

For int ** arr;, it says declare arr as pointer to pointer to int.

The site also shows examples. Test yourself on them, then hover your cursor to see the answer.

(double (^)(int , long long ))foo

cast foo into block(int, long long) returning double

int (*(*foo)(void ))[3]

declare foo as pointer to function (void) returning pointer to array 3 of int

It will also translate English into C declarations, which is prety neat - if you get the description correct.