Declaring:
int** a[3];
Can I say that 6 pointers are being declared or not?
My reasoning is that for every cell of the array I can enter it by either *a[1] or **a[1].
Is this a correct assumption of I can only say that I've declared 3 pointer to pointers to 3 integers?
Can I say that 6 pointers are being declared or not?
No, this line declares an array of three pointers. Even though each pointer could be pointing to a pointer to int, initially they are not pointing to anything.
every cell of the array I can enter it by either *a[1] or **a[1]
Each element of the array is a pointer to pointer to int - there is nothing else that could be inferred from the declaration.
You could use this declaration to make a 3-D array of integers, with each dimension having a different size, or you could stuff the entire array with NULLs. Nothing in the declaration limits the number of pointers that could be held by your array of three pointers.
No, you've declared an array of 3 int** pointers with automatic storage duration, that's all. Another 3 somethings have not been spontaneously created.
For each element to have meaning, each would have to point to a pointer to an int. The following code assigns something meaningful to the first element of the array:
int main()
{
int** a[3];
int foo;
int* bar = &foo;
a[0] = &bar;
}
Finally, note that a could decay to an int*** type if passed to a function with an int*** type as a parameter.
All you can say is you've declared an array of 3 things - you may set those things to point to other things which in turn point to other things, but those other things are not created as a result of this declaration.
Related
int array[5];
Expressions such as
array[3] gets converted to *(array+3)
Or in
void fun ( int *array[] );
*array[] gets converted to int **array
I was wondering what does the array declaration
int array[5];
Get converted to? Is it
int *(array+5)
If yes, what does this even mean? And how does one interpret it and/or read it?
array[i] gets converted to *(array+i)
Correct, given that array[i] is part of an expression, then array "decays" into a pointer to its first element, which is why the above holds true.
Void fun ( Int *array[] );
*array[] gets converted to Int **array
Yes because of the rule of function parameter adjustment ("decay"), which is similar to array decay in expressions. The first item of that array is an int* so after decay you end up with a pointer to such a type, a int**.
This is only true for functions with the specific format you posted, there is otherwise no relation between pointer-to-pointers and arrays.
I was wondering what does the array declaration
Int array[5];
Get converted to?
Nothing, declarations don't get converted. It is an array of 5 integers.
To sum this up, you actually list 3 different cases.
When an array is used as part of an expression, it "decays" into a pointer to the first element.
When an array is used as part of a function parameter declaration, it "decays" too - it actually has its type replaced by the compiler at compile-time - into a pointer to the first element. C was deliberately designed this way, so that functions would work together with arrays used in expressions.
When an array is declared normally (not part of a parameter list), nothing happens except you get an array of the specified size.
I think you are confusing two things.
*(array+i)
cannot be used for declaration, only for accessing the memory location (array being the starting address and i the offset)
also, the following declaration will create an array of 5 integers onto the stack
int array[5];
You can access any element from the array with the other notation, because values are being pushed onto the stack. The following two yielding in the same result:
int a = *(array+3);
int b = array[3];
if (a == b) printf("Same value");
else printf("Not same value");
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 ** ?
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?
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.
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.
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...
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.
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.
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;
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.
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.
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.
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); }
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;.
int ** arr = NULL;
It's tell the compiler, arr is a double pointer of an integer and assigned NULL value.
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.
I have this code:
int a[2][3]={{1,2},{1,3,4}};
int b[2][3]={{4,6},{22,33,55}};
int (*comb1[4])[3]={a,b,a,b};//works
int (*comb2[4])[2][3]={a,b,a,b};//it gives me warning: initialization from incompatible pointer type
Quoting http://cdecl.org/:
1.declare comb1 as array 4 of pointer to array 3 of int
2.declare comb2 as array 4 of pointer to array 2 of array 3 of int
I want to use the comb as 3dim array, where the first dimension selects the pointer of [2][3] and the rest identify the element.
This is achieved by comb1, is it possible to achieve something similar using in the inside of the declaration the [2][3]. (like what I have tried to do with comb2 without success. At the end I want to use e.g. comb2[0][0][0])
I'm assuming you want a pointer to a two-dimensional array. Your second declaration is correct (for an array of those), but your initializer is wrong: a and b are arrays, not pointers to arrays.
Fix:
int (*ab[4])[2][3]={&a,&b,&a,&b};
If you want to use comb[0][0][0] to access the first element of a, andcomb[1][0][0] to access the first element of b, you should use comb1 from your question:
int (*comb[4])[3] = {a, b, a, b};
This might be confusing, since comb is supposed to contain pointers to a and b, which are 2x3 matrices, not arrays of 3 ints, as this declaration seems to indicate. Shouldn't the number 2 be somewhere in there?
But remember that arrays are second-class citizens in C, and in most cases they are handled not as actual arrays, but as pointers to the first element in the array. This is also true concerning pointers to arrays. In general you don't use pointer to array, but pointer to the first element of the array. The address would be the same, but the data types are different. Here is a simpler example:
float a[17]; // This is an array of 17 floats
float *p = a; // This is just a pointer to float, not to array of floats
a[0] = 3.14; // Setting the first element of a
p[0] = 3.14; // Setting the same element, through p
Note that there is no "17" in the declaration of p.
You can use a pointer to the array, but then you need an extra level of indirection, to follow that pointer and get the right data type:
float (*pa)[17] = &a; // Pointer to array of 17 floats
(*pa)[0] = 3.14; // Setting the first element of a, through pa
You could write pa[0][0] instead of (*pa)[0], since a[0] in C is equivalent to *a, but that would be misleading, since it would give the impression of there being a two-dimensional array, when all you have is a pointer to a single-dimensional array.
What might be really confusing is that in the code above, pa, *pa and a will all be the same memory address:
printf("a = %p\n", (void*)a);
printf("pa = %p\n", (void*)pa);
printf("*pa = %p\n", (void*)(*pa));
Output when I ran it on my computer:
a = 0x7fff875a07e0
pa = 0x7fff875a07e0
*pa = 0x7fff875a07e0
Since arrays are a bit special, pointers to arrays are a bit special too. If a is an array, both a and &a (when used in most contexts) give the same address (but different data types). And in reverse: If p is a pointer to an array, both p and *p give the same address (but different data types).
In summary: If you think that since a is 2x3 matrix, the declaration of comb should somehow say both 2 and 3, and not just 3, that is misguided.
You can do so, if you absolutely want to, as melpomene has shown. But in that case you need to write an extra pointer indirection: (*comb[0])[0][0].
Up till now I was pretty much sure that
int arr[4][5];
Then arr will decay to pointer to pointer.
But this link proves me wrong.
I am not sure how did I get about arr being pointer to pointer but it seemed pretty obvious to me. Because arr[i] would be a pointer, and hence arr should be a pointer to pointer.
Am I missing out on something.
Yep you are missing out on a lot :)
To avoid another wall of text I'll link to an answer I wrote earlier today explaining multi-dimensional arrays.
With that in mind, arr is a 1-D array with 4 elements, each of which is an array of 5 ints.
When used in an expression other than &arr or sizeof arr, this decays to &arr[0]. But what is &arr[0]? It is a pointer, and importantly, an rvalue.
Since &arr[0] is a pointer, it can't decay further. (Arrays decay, pointers don't). Furthermore, it's an rvalue. Even if it could decay into a pointer, where would that pointer point? You can't point at an rvalue. (What is &(x+y) ? )
Another way of looking at it is to remember that int arr[4][5]; is a contiguous bloc of 20 ints, grouped into 4 lots of 5 within the compiler's mind, but with no special marking in memory at runtime.
If there were "double decay" then what would the int ** point to? It must point to an int * by definition. But where in memory is that int * ? There are certainly not a bunch of pointers hanging around in memory just in case this situation occurs.
A simple rule is:
A reference to an object of type array-of-T which appears in an expression decays (with three exceptions) into a pointer to its first element; the type of the resultant pointer is pointer-to-T.
When you deal with 1D array, array name converts to pointer to first element when passed to a function.
A 2D array can be think as of an array of arrays. In this case int arr[4][5];, you can think arr[] as an array name and when passed to a function then converts to a pointer to the first element of array arr. Since first element of arr is an array, arr[i] is a pointer to ith row of the array and is of type pointer to array of 5 ints.
In general, a 2-dim array is implemented as an array of pointers (which, in a sense, is a pointer to a pointer... a pointer to the first element (i.e., pointer) in the array) When you specify the first index (i.e., arr[x]), it indexes into the array of pointers and gives the pointer to the x-th row. The the second index (i.e., arr[x][y]) gives the y-th int in that row.
In the case of a static declared array (as in your example), the actual storage is allocated as a single block ... in your example, as a single contiguous block of 20 integers (80 bytes, on most platforms). In this case, there IS no array of pointers, the compiler just does the appropriate arithmetic to address the correct element of the array. Specifically, arr[x][y] is equivalent to *(arr + x * 5 + y). This automatically-adjusted-arithmetic only happens in the original scope of the array... if you pass the array to a function, the dimension information is lost (just as the dimension is lost for a 1-dim array), and you have to do the array-indexing calculations explicitly.
To avoid this, do NOT declare the array as static, but as an array of pointers, with each pointer pointed to a 1-dim array, such as in this example:
int arr0[5];
int arr1[5];
int arr2[5];
int arr3[5];
int* arr[4] = { arr0, arr1, arr2, arr3 };
Then, when you pass arr to a function, you can address it as a 2-dim array within the function as well.
I have an array of int pointers int* arr[MAX]; and I want to store its address in another variable. How do I define a pointer to an array of pointers? i.e.:
int* arr[MAX];
int (what here?) val = &arr;
The correct answer is:
int* arr[MAX];
int* (*pArr)[MAX] = &arr;
Or just:
int* arr [MAX];
typedef int* arr_t[MAX];
arr_t* pArr = &arr;
The last part reads as "pArr is a pointer to array of MAX elements of type pointer to int".
In C the size of array is stored in the type, not in the value. If you want this pointer to correctly handle pointer arithmetic on the arrays (in case you'd want to make a 2-D array out of those and use this pointer to iterate over it), you - often unfortunately - need to have the array size embedded in the pointer type.
Luckily, since C99 and VLAs (maybe even earlier than C99?) MAX can be specified in run-time, not compile time.
Should just be:
int* array[SIZE];
int** val = array;
There's no need to use an address-of operator on array since arrays decay into implicit pointers on the right-hand side of the assignment operator.
IIRC, arrays are implicitly convertible to pointers, so it would be:
int ** val = arr;
According to this source http://unixwiz.net/techtips/reading-cdecl.html, by using the "go right when you can, go left when you must" rule, we get the following 2 meanings of the declarations given in the previous answers -
int **val ==> val is a pointer to pointer to int
int* (*pArr)[MAX] ==> pArr is a pointer to an array of MAX length pointers to int.
I hope the above meanings make sense and if they don't, it would probably be a good idea to peruse the above mentioned source.
Now it should be clear that the second declaration is the one which moteutsch is looking for as it declares a pointer to an array of pointers.
So why does the first one also work? Remember that
int* arr[MAX]
is an array of integer pointers. So, val is a pointer to, the pointer to the first int declared inside the int pointer array.
#define SIZE 10
int *(*yy)[SIZE];//yy is a pointer to an array of SIZE number of int pointers
and so initialize yy to array as below -
int *y[SIZE]; //y is array of SIZE number of int pointers
yy = y; // Initialize
//or yy = &y; //Initialize
I believe the answer is simply:
int **val;
val = arr;
As far as I know there is no specific type "array of integers" in c, thus it's impossible to have a specific pointer to it. The only thing you can do is to use a pointer to the int: int*, but you should take into account a size of int and your array length.