Related
If I have:
int A[10][20];
printf("%p",A[3]);
it will print the address of A[3][0].
However, I'd like to know if this one dimensional array A[3] containing pointers really exists, or it is calculated in some way.
The way you have defined A means that the compiler will allocate for it a contiguous block of memory large enough to hold 10 x 20 (200) integers; see here (scroll down to "Multidimesional arrays"). As I'm sure you realize, if you were to do printf("%p", A); you would see the address of the beginning of that allocated block.
Now, when the compiler sees the expression A[3], it will add what it calculates as the necessary amount of "integer sizes" to the base address (that of A, or A[0][0]); in this case, it will add "3" (the index specified) multiplied by the combined size of all the other dimensions (in this case, there's only one, which is 20).
So, in your case, there is no actual array of pointers; just a memory block that the compiler can interpret according to how you described any part(s) of it.
However, in a more versatile approach, one can actually define a 2D array in terms of an actual array of pointers, like so:
int **A;
A = malloc(10 * sizeof(int*));
for (int n = 0; n < 10; ++n) A[n] = malloc(20 * sizeof(int));
In this case, using printf("%p",A[3]); would still be valid, but it would give a very different offset value from printf("%p",A); or printf("%p",A[0]);.
It's also, perhaps, worth noting that, even though these two different declarations for A can both resolve an individual element through an expression like A[i][j] (but the compiler would evaluate the addresses differently), there is here scope for major confusion! When, for example, passing such an array to a function: if the function expects data allocated in the second form, and you give it an array defined in the first form (and vice versa), you're gonna get major undefined behaviour .
yes there is a way to calculate the position:
for A[i][j]
the position of the memory block will be
pos = A + i*(number_of_columns_in_each_row) + j
here A is the pointer to the first element of the array
However, I'd like to know if this one dimensional array A containing pointers really exists, or it is calculated in some way.
The way you defined the array A :
int A[10][20];
does not contain any pointers as elements of the array. it contains only integer elements.
if you want to make an array of pointers, which should be assigned to int-variables is defined like that:
int *A[10][20];
You also can set a pointer to the start of the array, which means element [0] [0]
by using:
int *pointer;
int *A[10][20];
pointer = &A;
You also be able to set the pointer slightly forwards according to each element by increase the pointer.
pointer++;
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.
This question already has answers here:
Why can't we use double pointer to represent two dimensional arrays?
(6 answers)
Closed 6 years ago.
int main()
{
matrix[2][4] = {{11,22,33,99},{44,55,66,110}};
int **ptr = (int**)matrix;
printf("%d%d",**matrix,*ptr);
}
But when a 2-d array is passed as a parameter it is typecasted into (*matrix)[2] ..
what type does the compiler store this array as... is it storing as a 2-d array or a double pointer or an pointer to an array .. If it is storing as an array how does it interprets differently at different situations like above. Please help me understand.
Is 2d array a double pointer?
No. This line of your program is incorrect:
int **ptr = (int**)matrix;
This answer deals with the same topic
If you want concrete image how multidimensional arrays are implemented:
The rules for multidimensional arrays are not different from those for ordinary arrays, just substitute the "inner" array type as element type. The array items are stored in memory directly after each other:
matrix: 11 22 33 99 44 55 66 110
----------- the first element of matrix
------------ the second element of matrix
Therefore, to address element matrix[x][y], you take the base address of matrix + x*4 + y (4 is the inner array size).
When arrays are passed to functions, they decay to pointers to their first element. As you noticed, this would be int (*)[4]. The 4 in the type would then tell the compiler the size of the inner type, which is why it works. When doing pointer arithmetic on a similar pointer, the compiler adds multiples of the element size, so for matrix_ptr[x][y], you get matrix_ptr + x*4 + y, which is exactly the same as above.
The cast ptr=(int**)matrix is therefore incorrect. For once, *ptr would mean a pointer value stored at address of matrix, but there isn't any. Secondly, There isn't a pointer to matrix[1] anywhere in the memory of the program.
Note: the calculations in this post assume sizeof(int)==1, to avoid unnecessary complexity.
No. A multidimensional array is a single block of memory. The size of the block is the product of the dimensions multiplied by the size of the type of the elements, and indexing in each pair of brackets offsets into the array by the product of the dimensions for the remaining dimensions. So..
int arr[5][3][2];
is an array that holds 30 ints. arr[0][0][0] gives the first, arr[1][0][0] gives the seventh (offsets by 3 * 2). arr[0][1][0] gives the third (offsets by 2).
The pointers the array decays to will depend on the level; arr decays to a pointer to a 3x2 int array, arr[0] decays to a pointer to a 2 element int array, and arr[0][0] decays to a pointer to int.
However, you can also have an array of pointers, and treat it as a multidimensional array -- but it requires some extra setup, because you have to set each pointer to its array. Additionally, you lose the information about the sizes of the arrays within the array (sizeof would give the size of the pointer). On the other hand, you gain the ability to have differently sized sub-arrays and to change where the pointers point, which is useful if they need to be resized or rearranged. An array of pointers like this can be indexed like a multidimensional array, even though it's allocated and arranged differently and sizeof won't always behave the same way with it. A statically allocated example of this setup would be:
int *arr[3];
int aa[2] = { 10, 11 },
ab[2] = { 12, 13 },
ac[2] = { 14, 15 };
arr[0] = aa;
arr[1] = ab;
arr[2] = ac;
After the above, arr[1][0] is 12. But instead of giving the int found at 1 * 2 * sizeof(int) bytes past the start address of the array arr, it gives the int found at 0 * sizeof(int) bytes past the address pointed to by arr[1]. Also, sizeof(arr[0]) is equivalent to sizeof(int *) instead of sizeof(int) * 2.
In C, there's nothing special you need to know to understand multi-dimensional arrays. They work exactly the same way as if they were never specifically mentioned. All you need to know is that you can create an array of any type, including an array.
So when you see:
int matrix[2][4];
Just think, "matrix is an array of 2 things -- those things are arrays of 4 integers". All the normal rules for arrays apply. For example, matrix can easily decay into a pointer to its first member, just like any other array, which in this case is an array of four integers. (Which can, of course, itself decay.)
If you can use the stack for that data (small volume) then you usually define the matrix:
int matrix[X][Y]
When you want to allocate it in the heap (large volume), the you usually define a:
int** matrix = NULL;
and then allocate the two dimensions with malloc/calloc.
You can treat the 2d array as int** but that is not a good practice since it makes the code less readable. Other then that
**matrix == matrix[0][0] is true
This question already has answers here:
Why can't we use double pointer to represent two dimensional arrays?
(6 answers)
Closed 6 years ago.
int main()
{
matrix[2][4] = {{11,22,33,99},{44,55,66,110}};
int **ptr = (int**)matrix;
printf("%d%d",**matrix,*ptr);
}
But when a 2-d array is passed as a parameter it is typecasted into (*matrix)[2] ..
what type does the compiler store this array as... is it storing as a 2-d array or a double pointer or an pointer to an array .. If it is storing as an array how does it interprets differently at different situations like above. Please help me understand.
Is 2d array a double pointer?
No. This line of your program is incorrect:
int **ptr = (int**)matrix;
This answer deals with the same topic
If you want concrete image how multidimensional arrays are implemented:
The rules for multidimensional arrays are not different from those for ordinary arrays, just substitute the "inner" array type as element type. The array items are stored in memory directly after each other:
matrix: 11 22 33 99 44 55 66 110
----------- the first element of matrix
------------ the second element of matrix
Therefore, to address element matrix[x][y], you take the base address of matrix + x*4 + y (4 is the inner array size).
When arrays are passed to functions, they decay to pointers to their first element. As you noticed, this would be int (*)[4]. The 4 in the type would then tell the compiler the size of the inner type, which is why it works. When doing pointer arithmetic on a similar pointer, the compiler adds multiples of the element size, so for matrix_ptr[x][y], you get matrix_ptr + x*4 + y, which is exactly the same as above.
The cast ptr=(int**)matrix is therefore incorrect. For once, *ptr would mean a pointer value stored at address of matrix, but there isn't any. Secondly, There isn't a pointer to matrix[1] anywhere in the memory of the program.
Note: the calculations in this post assume sizeof(int)==1, to avoid unnecessary complexity.
No. A multidimensional array is a single block of memory. The size of the block is the product of the dimensions multiplied by the size of the type of the elements, and indexing in each pair of brackets offsets into the array by the product of the dimensions for the remaining dimensions. So..
int arr[5][3][2];
is an array that holds 30 ints. arr[0][0][0] gives the first, arr[1][0][0] gives the seventh (offsets by 3 * 2). arr[0][1][0] gives the third (offsets by 2).
The pointers the array decays to will depend on the level; arr decays to a pointer to a 3x2 int array, arr[0] decays to a pointer to a 2 element int array, and arr[0][0] decays to a pointer to int.
However, you can also have an array of pointers, and treat it as a multidimensional array -- but it requires some extra setup, because you have to set each pointer to its array. Additionally, you lose the information about the sizes of the arrays within the array (sizeof would give the size of the pointer). On the other hand, you gain the ability to have differently sized sub-arrays and to change where the pointers point, which is useful if they need to be resized or rearranged. An array of pointers like this can be indexed like a multidimensional array, even though it's allocated and arranged differently and sizeof won't always behave the same way with it. A statically allocated example of this setup would be:
int *arr[3];
int aa[2] = { 10, 11 },
ab[2] = { 12, 13 },
ac[2] = { 14, 15 };
arr[0] = aa;
arr[1] = ab;
arr[2] = ac;
After the above, arr[1][0] is 12. But instead of giving the int found at 1 * 2 * sizeof(int) bytes past the start address of the array arr, it gives the int found at 0 * sizeof(int) bytes past the address pointed to by arr[1]. Also, sizeof(arr[0]) is equivalent to sizeof(int *) instead of sizeof(int) * 2.
In C, there's nothing special you need to know to understand multi-dimensional arrays. They work exactly the same way as if they were never specifically mentioned. All you need to know is that you can create an array of any type, including an array.
So when you see:
int matrix[2][4];
Just think, "matrix is an array of 2 things -- those things are arrays of 4 integers". All the normal rules for arrays apply. For example, matrix can easily decay into a pointer to its first member, just like any other array, which in this case is an array of four integers. (Which can, of course, itself decay.)
If you can use the stack for that data (small volume) then you usually define the matrix:
int matrix[X][Y]
When you want to allocate it in the heap (large volume), the you usually define a:
int** matrix = NULL;
and then allocate the two dimensions with malloc/calloc.
You can treat the 2d array as int** but that is not a good practice since it makes the code less readable. Other then that
**matrix == matrix[0][0] is true
Why can't my parameter be
void example(int Array[][]){ /*statements*/}
Why do I need to specify the column size of the array? Say for example, 3
void example(int Array[][3]){/*statements*/}
My professor said its mandatory, but I was coding before school started and I remembered that there was no syntactical or semantic error when I made this my parameter? Or did I miss something?
When it comes to describing parameters, arrays always decay into pointers to their first element.
When you pass an array declared as int Array[3] to the function void foo(int array[]), it decays into a pointer to the beginning of the array i.e. int *Array;. Btw, you can describe a parameter as int array[3] or int array[6] or even int *array - all these will be equivalent and you can pass any integer array without problems.
In case of arrays of arrays (2D arrays), it decays to a pointer to its first element as well, which happens to be a single dimensional array i.e. we get int (*Array)[3].
Specifying the size here is important. If it were not mandatory, there won't be any way for compiler to know how to deal with expression Array[2][1], for example.
To dereference that a compiler needs to compute the offset of the item we need in a contiguous block of memory (int Array[2][3] is a contiguous block of integers), which should be easy for pointers. If a is a pointer, then a[N] is expanded as start_address_in_a + N * size_of_item_being_pointed_by_a. In case of expression Array[2][1] inside a function (we want to access this element) the Array is a pointer to a single dimensional array and the same formula applies. The number of bytes in the last square bracket is required to find size_of_item_being_pointed_by_a. If we had just Array[][] it would be impossible to find it out and hence impossible to dereference an array element we need.
Without the size, pointers arithmetics wouldn't work for arrays of arrays. What address would Array + 2 produce: advance the address in Array 2 bytes ahead (wrong) or advance the pointer 3* sizeof(int) * 2 bytes ahead?
In C/C++, even 2-D arrays are stored sequentially, one row after another in memory. So, when you have (in a single function):
int a[5][3];
int *head;
head = &a[0][0];
a[2][1] = 2; // <--
The element you are actually accessing with a[2][1] is *(head + 2*3 + 1), cause sequentially, that element is after 3 elements of the 0 row, and 3 elements of the 1 row, and then one more index further.
If you declare a function like:
void some_function(int array[][]) {...}
syntactically, it should not be an error. But, when you try to access array[2][3] now, you can't tell which element is supposed to be accessed. On the other hand, when you have:
void some_function(int array[][5]) {...}
you know that with array[2][3], it can be determined that you are actually accessing element at the memory address *(&array[0][0] + 2*5 + 3) because the function knows the size of the second dimension.
There is one other option, as previously suggested, you can declare a function like:
void some_function(int *array, int cols) { ... }
because this way, you are calling the function with the same "information" as before -- the number of columns. You access the array elements a bit differently then: you have to write *(array + i*cols + j) where you would usually write array[i][j], cause array is now a pointer to integer (not to a pointer).
When you declare a function like this, you have to be careful to call it with the number of columns that are actually declared for the array, not only used. So, for example:
int main(){
int a[5][5];
int i, j;
for (i = 0; i < 3; ++i){
for (int j=0; j < 3; ++j){
scanf("%d", &a[i][j]);
}
}
some_function(&a[i][j], 5); // <- correct
some_function(&a[i][j], 3); // <- wrong
return 0;
}
C 2018 6.7.6.2 specifies the semantics of array declarators, and paragraph 1 gives constraints for them, including:
The element type shall not be an incomplete or function type.
In a function declaration such as void example(int Array[][]), Array[] is an array declarator. So it must satisfy the constraint that its element type must not be incomplete. Its element type in that declaration is int [], which is incomplete since the size is not specified.
There is no fundamental reason the C standard could not remove that constraint for parameters that are about to be adjusted to pointers. The resulting type int (*Array)[] is a legal declaration, is accepted by compilers, and can be used in the form (*Array)[j].
However, the declaration int Array[][] suggests that Array is at least associated with a two-dimensional array, and hence is to be used in the form Array[i][j]. Even if the declaration int Array[][] were accepted and were adjusted to int (*Array)[], using it as Array[i][j] would not be possible because the subscript operator requires that its pointer operand be a pointer to a complete type, and this requirement is not avoidable as it is needed to calculate the address of the element. Thus, keeping the constraint on the array declarator makes sense, as it is consistent with the intended expression that the argument will be a two-dimensional array, not just a pointer to one one-dimensional array.
Actually whether it is a 2d array or a 1d array, it is stored in the memory in a single line.So to say the compiler where should it break the row indicating the next numbers to be in the next rows we are supposed to provide the column size. And breaking the rows appropriately will give the size of the rows.
Let's see an example:
int a[][3]={ 1,2,3,4,5,6,7,8,9,0 };
This array a is stored in the memory as:
1 2 3 4 5 6 7 8 9 0
But since we have specified the column size as 3 the memory splits after every 3 numbers.
#include<stdio.h>
int main() {
int a[][3]={1,2,3,4,5,6},i,j;
for(i=0;i<2;i++)
{
for(j=0;j<3;j++)
{
printf("%d ",a[i][j]);
}
printf("\n");
}
}
OUTPUT:
1 2 3
4 5 6
In the other case,
int a[3][]={1,2,3,4,5,6,7,8,9,0};
The compiler only knows that there are 3 rows but it doesn't know the number of elements in each row so it cannot allocate memory and will show an error.
#include<stdio.h>
int main() {
int a[3][]={1,2,3,4,5,6},i,j;
for(i=0;i<3;i++)
{
for(j=0;j<2;j++)
{
printf("%d ",a[i][j]);
}
printf("\n");
}
}
OUTPUT:
c: In function 'main':
c:4:8: error: array type has incomplete element type 'int[]'
int a[3][]={1,2,3,4,5,6},i,j;
^
As we know, we can pass a variable as an argument(s) in a function. Similarly, we can pass two-dimensional arrays in C++.
C++ does not allow us to pass an entire array as an argument to a function. However, we can pass a pointer to an array by specifying the array's name without an index.
We can pass a 2D array to a function by specifying the size of the columns of a 2D array. One of the important things to remember here is that the size of rows is optional but the size of the column should not be left empty else the compiler will show an error. A 2D array is stored in the memory in a single line. So, to say the compiler where should it break the row indicating the following numbers to be in the next rows we are supposed to provide the column size. And breaking the rows appropriately will automatically give the size of the rows.
source: https://www.scaler.com/topics/two-dimensional-array-in-cpp/
There is a similar post regarding this. You can refer below link.
Creating Array in C and passing pointer to said array to function
Hope it helps.
On the other hand, compiler needs to the second dimension so that it can move "Array" from one pointer to next since the whole memory is arranged in a linear fashion
I thought this was a cool approach. If you take this as the formula to calculate the address of an element in the array:
a[i][j] = baseArrayAddress + (i + (colSize + elementSize)) + (j * (elementSize))
Then you can see that the only thing the compiler needs to know (which it can't otherwise infer) is the size of the column, thus you need to provide it as the programmer so the algorithm can run to calculate the offset.
The row number only acts as a multiplier and is provided by the programmer when trying to dereference an array location.
When you create a 2D array, anytype a[3][4], in memory what you actually create is 3 contiguous blocks of 4 anytype objects.
a[0][0] a[0][1] a[0][2] a[0][3] a[1][0] a[1][1] a[1][2] a[1][3] a[2][0] a[2][1] a[2][2] a[2][3]
Now the next question is, why is that so? Because, keeping with the spec and structure of the language, anytype a[3][4] actually expands out into anytype (*a)[4], because arrays decay into pointers. And in fact that also expands out into anytype (*(*a)), however, you've now completely lost the size of the 2D array. So, you must help the compiler out a bit.
If you ask the program for a[2], the program can follow the exact same steps that it does for 1D arrays. It simply can return the 3rd element of sizeof(object pointed to), the object pointed to here is of size 4 anytype objects.