#include<stdio.h>
void display(int *q,int,int);
int main(){
int a[3][4]={
2,3,4,5,
5,7,6,8,
9,0,1,6
};
display(a,3,4);
return 0;
}
void display(int *q,int row,int col){
int i,j;
for(i=0;i<row;i++){
for(j=0;j<col;j++){
printf("%d",*(q+i*col+j));
}
printf("\n");
}
printf("\n");
}
why this code show warning in gcc that "passing argument 1 of 'display' from incompatible pointer type display(a,3,4)"?...runs successfully anyway but curious to know about error..if anyone could tell this i would be grateful..
The rule of "array decay" means that whenever you use an array name a as part of an expression, it "decays" to a pointer to the first element.
For a 1D array, this is pretty straight forward. An array int a [10] would decay into type int*.
However, in case of two-dimensional arrays, the first element of the 2D array is a 1D array. In your case, the first element of int a[3][4] has array type int [4].
The array decay rule gives you a pointer to such an array, an array pointer, of type int (*)[4]. This type is not compatible with the type int* that your function expects.
However, by sheer luck, it would appear the the array pointer and a plain int pointer have the same representation on your system, and they happen to hold the same address, so the code works. You shouldn't rely on this though, it is not well-defined behavior and there is no guarantee it will work.
You should fix your program in the following way:
#include <stdio.h>
void display (int row, int col, int arr[row][col]);
int main()
{
int a[3][4]=
{
{2,3,4,5},
{5,7,6,8},
{9,0,1,6},
};
display(3, 4, a);
return 0;
}
void display (int row, int col, int arr[row][col])
{
for(int i=0; i<row; i++)
{
for(int j=0; j<col; j++)
{
printf("%d ", arr[i][j]);
}
printf("\n");
}
printf("\n");
}
Here the array type that is the function parameter will silently "get adjusted" by the compiler to a pointer to the first element, int(*)[4], which matches what's passed to the function from the caller.
Because they are incompatible pointer types, it just happens that int * could point to an array of ints and int[][] is an array of int arranged in contigous memory.
The important difference is, that while you can access a with a two index notation, like a[i][j] you can't do that with q.
Related
I will go straight to what I'm asking for, I also see some similar question but is not what I'm looking for...so it seems I have to ask with a new forum.
I'm preparing myself for a future examination, where is not required the pointer, but I would like to get some extra information and abilities.
Here's the code followed by the question...
I'm using Fedora 33, I know is different from some IDE on Windows (ex: Visual Studio or Dev C++)
/* It's just a simple test, if this work I will get myself into a more complicated one, as you could read in the
* forum, I'm getting ready ( just a recheck of my abilities ) for an universitary examinaton. */
#include <stdio.h>
#include <stdlib.h>
#define N 5
void casual_generation(int** mat);
void prompt_print(int** mat);
int main()
{
int **mat[N][N];
casual_generation(**mat);
prompt_print(**mat);
}
void casual_generation(int** mat)
{
int i=0,j=0;
for(i=0;i<N;i++)
for(j=0;j<N;j++)
mat[i][j] = rand() % 50;
}
void prompt_print(int** mat)
{
int i=0,j=0;
for(i=0;i<N;i++)
{
for(j=0;j<N;j++)
printf("%d ", mat[i][j]);
printf("\n");
}
}
Somebody else on the forum used malloc, struct or other stuff, as you can see in this picture, when I try to execute him it says "Segmentation fault (core dumped)"
screen error
Where is my error?
And if you want, can you also send me the version with the passed value pointer?
Thanks for whoever will give me an answer, and time dedicated.
This declaration
int **mat[N][N];
does not make a sense. It means that you have a matrix elements of which are pointers of the type int **. But you need a matrix elements of which are integer numbers of the type int. That is you need a declaration like this
int mat[N][N];
So now you have a two-dimensional array (or matrix) of integers.
As you are going to pass this two-dimensional array to functions then used as an argument expression it is converted to pointer to its first element of the type int ( * )[N].
Correspondingly the functions that accepts such an array should be declared like
void casual_generation( int mat[][N], size_t n );
void prompt_print( int mat[][N], size_t n );
or (that is fully equivalent) like
void casual_generation( int ( * mat )[N], size_t n );
void prompt_print( int ( *mat )[N], size_t n );
because the compiler adjusts function parameters having array types to pointers to array element types.
Now for example the first function can be defined the following way
void casual_generation( int ( * mat )[N], size_t n )
{
for ( size_t i = 0; i < n; i++ )
{
for ( size_t j = 0; j < N; j++ )
{
mat[i][j] = rand() % 50;
}
}
}
And the function can be called like
casual_generation( mat, N );
A similar way can be defined the function prompt_print.
Using the second parameter makes the function more general. For example it can be called for two-dimensional arrays with different numbers of rows.
Now I will explain why you are getting a segmentation fault in your original code.
You have this declaration
int **mat[N][N];
a two dimensional array of pointers of the type int **.
Then you are using the expression **mat as an argument of function calls like this
casual_generation(**mat);
Then you are applying the dereference operator like *mat the array designator is converted to pointer to its first element (row) having the type int ** ( * )[N]. So dereferencing this pointer you get the first row of your array int **[N]. Applying the second time the dereferenced operator to this expression that has an array type the used expression is again is converted to pointer to its first element of the type int **( * ). That is it points to the first element of the first row of the original two-dimensional array. Dereferencing this pointer you get the first element of the type int **. This uninitialized pointer with indeterminate value the function accepts as its argument.
Thus dereferencing this first uninitialized element of the original matrix within the function
mat[i][j] = rand() % 50;
^^^
you get a segmentation fault. The reason of the fault is the incorrect matrix and the corresponding function parameter as it was shown above in tbe beginning of the answer.
Where is my error?
The "Segmentation fault" error happens because you define the variable mat as a pointer, but don't allocate any memory for it to point to.
int **mat[N][N];
You meant to do
Int mat[N][N];
and
casual_generation(mat);
prompt_print(mat);
By passing **mat you are passing mat[0][0] that is an int, but you want to pass the whole matrix which is a pointer to pointers to int (i.e. int **)
And you may want to introduce srand() in your code.
Just to make things clear:
mat is of type int ** and it's the whole matrix (or if you want it's a pointer to the first row)
*mat is of type int * and it's the first row of the matrix (or if you want it's a pointer to the first element of the first row)
**mat is of type int and it's the first element of the first row of the matrix
int **mat[N][N];
Here, you defined a double pointer to a 2D array. You only need to use one of those - a double pointer or a 2D array, like so:
int mat[N][N];
However, the bigger problem comes from trying to interchange 2D arrays and double pointers. This isn't possible in C since the 2D array is laid out flat in memory.
You need to create an array mat_ptr of pointers yourself and then pass that to casual_generation and prompt_print.
Finally, these casual_generation and prompt_print functions expect to be given a pointer, so you shouldn't dereference the pointer with ** before calling the function.
The final working code is:
int main()
{
int mat[N][N];
int *mat_ptr[N];
for (int i = 0; i < N; i++)
mat_ptr[i] = mat[i];
casual_generation(mat_ptr);
prompt_print(mat_ptr);
}
As you can find a detailed explanation why the code crashed in the other answer I will only propose a quite elegant solution that uses a neat though little known feature from C99 called Variable Length Arrays (aka VLA).
#include <stdio.h>
#include <stdlib.h>
void casual_generation(int n, int mat[n][n]);
void prompt_print(int n, int mat[n][n]);
int main()
{
const int N = 5;
int mat[N][N];
casual_generation(N, mat);
prompt_print(N, mat);
}
void casual_generation(int n, int mat[n][n])
{
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
mat[i][j] = rand() % 50;
}
void prompt_print(int n, int mat[n][n])
{
for(int i=0;i<n;i++)
{
for(int j=0;j<n;j++)
printf("%d ", mat[i][j]);
printf("\n");
}
}
It compiles in pedantic mode with no warnings and it works like a charm.
VLA were introduced to C to simplify numerical computation over multidimensional arrays.
I am recently studying pointers and arrays.
Suppose I initialize an array like
int a[4]={6,2,3,4};
Now after reading a lot ,I understand that
1) a and &a will point to same location.But they aren't pointing to the same type.
2) a points to the first element of the array which is an integer value so the type of a is int*.
3) &a points to an entire array(i.e an integer array of size 4) therefore the type is int (*)[].
Now what I actually don't get is how to use these type??
For Example:
CODE 1:
#include<stdio.h>
void foo(int (*arr)[4],int n)
{
int i;
for(i=0;i<n;i++)
printf("%d ",*(*arr+i));
printf("\n");
}
int main()
{
int arr[4]={6,2,3,4};
foo(&arr,4);
return 0;
}
CODE 2:
#include<stdio.h>
void foo(int *arr,int n)
{
int i;
for(i=0;i<n;i++)
printf("%d ",*(arr+i));
printf("\n");
}
int main()
{
int arr[4]={6,2,3,4};
foo(&arr,4);
return 0;
}
In code 2 we are passing &arr so its type should be of int (*)[],then how come we are getting the correct output even though we are having a type int *.
I really don't understand what is the meaning of type and how to use it?
Kindly explain with some examples.
Code 2 is not valid C, because a conversion between an array pointer and a pointer to int is not a valid pointer conversion - they are not compatible types. If it works, it is because of some non-standard extension of your compiler, or possibly you just "got lucky".
Please note that the best way to pass an array to a function in modern C is this:
void foo(int n, int arr[n])
{
for(int i=0;i<n;i++)
printf("%d ",arr[i]);
printf("\n");
}
As part of a function parameter list, arr will get adjusted to a pointer to the first element of the array, type int*.
The pedantically correct version would also replace int n and int i with size_t, which is the most proper type to use when describing the size of an object.
The best way to pass arrays in to functions is a pointer and the number of elements.
Note you can pass arr in to the function as arr or &arr[0]; both resolve to the same address.
#include<stdio.h>
void foo(int *arr,int n)
{
int i;
for(i=0;i<n;i++)
printf("%d ",arr[i]);
printf("\n");
}
int main()
{
int arr[4]={6,2,3,4};
foo(arr,4);
return 0;
}
Also, it's better to index the array, rather than addition and dereference. e.g. *(arr+i) becomes arr[i].
I want to use another function to print the contents of an array.
When I run the code I get "IntelliSense: argument of type "int (*)[3][3]" is incompatible with parameter of type "int *(*)[3]"
This is my function:
void display(int *p[N_ROWS][N_COLS]) {
int i, j;
for (i = 0; i < N_ROWS; i++) {
for (j = 0; j <N _COLS; j++) {
printf("%i", p[i][j]);
}
}
}
I defined N_ROWS and N_COLS
and in my main function I declare my array and then call the function
{
int Array[N_ROWS][N_COLS] = { {1,2,3},{4,5,6},{7,8,9} };
display(&Array);
}
Aren't both my parameters type int(*)[3][3] or am I missing something?
Your prototype for display is incorrect, as well as your call syntax: you define display to take a 2D array of pointers to int, whereas you just want a 2D array if int, as you pass a pointer to the array in main where you just want to pass the array, decaying as a pointer to its first element.
Here is a simpler version:
void display(int p[N_ROWS][N_COLS]) {
int i, j;
for (i = 0; i < N_ROWS; i++) {
for (j = 0; j < N_COLS; j++) {
printf("%i", p[i][j]);
}
}
}
Note however that p above can have any number of rows, N_ROWS is ignored in the prototype, equivalent to void display(int (*p)[N_COLS]).
Note also that your printf will output the matrix values without any separation. This might not be your intent.
And from main:
{
int Array[N_ROWS][N_COLS] = { {1,2,3},{4,5,6},{7,8,9} };
display(Array);
}
Much of this has already been explained in comments to your question.
Your definition of display:
void display(int*p[N_ROWS][N_COLS])
{
This says p will be an array N_ROWS of array N_COLS of pointers to int.
When you call display:
int Array [N_ROWS][N_COLS] = { {1,2,3},{4,5,6},{7,8,9} };
display(&Array);
You are passing in the address of Array, thus it is a pointer to an array N_ROWS of array N_COLS of int.
To make the definition of display match the way you are calling it:
void display(int (*p)[N_ROWS][N_COLS])
{
The parentheses make the * associate with p first. Since p is a pointer to an array of array of int, getting to the int requires an extra dereference:
printf("%i\n", (*p)[i][j]);
Defining display to take a pointer to an array means that the size of the array is bound to the type parameter, and thus display knows exactly the dimensions of the array.
An alternative means would be to define display to take the dimension of the array as a second parameter.
void display(int p[][N_COLS], int n_rows)
{
In this case the p parameter is a pointer to an array N_COLS of int. An array of T when used in most expressions will decay to a value of type pointer to T equal to the address of its first element. Thus, you could call this second version of display like this:
int Array [N_ROWS][N_COLS] = { {1,2,3},{4,5,6},{7,8,9} };
display(Array, N_ROWS);
The advantage of this second approach is that display can work with arrays that have fewer or more than N_ROWS. The disadvantage is that it is up to the caller to pass in the right number of rows.
You might think that the following declaration would give you complete type safety:
void display(int p[N_ROWS][N_COLS])
{
But, the array syntax in C for function parameters cause the size information for p to be ignored, and becomes equivalent to int p[][N_COLS], which in turn is treated as int (*p)[N_COLS].
Having looked around I've built a function that accepts a matrix and performs whatever it is I need on it, as follows:
float energycalc(float J, int **m, int row, int col){
...
}
Within the main the size of the array is defined and filled, however I cannot passs this to the function itself:
int matrix[row][col];
...
E=energycalc(J, matrix, row, col);
This results in a warning during compilation
"project.c:149: warning: passing argument 2 of ‘energycalc’ from
incompatible pointer type project.c:53: note: expected ‘int **’ but
argument is of type ‘int (*)[(long unsigned int)(col +
-0x00000000000000001)]’
and leads to a segmentation fault.
Any help is greatly appreciated, thank you.
The function should be declared like
float energycalc( float J, int row, int col, int ( *m )[col] );
if your compiler supports variable length arrays.
Otherwise if in declaration
int matrix[row][col];
col is some constant then the function can be declared the following way
float energycalc(float J, int m[][col], int row );
provided that constant col is defined before the function.
Your function declaration
float energycalc(float J, int **m, int row, int col);
is suitable when you have an array declared like
int * matrix[row];
and each element of the array is dynamically allocated like for example
for ( int i = 0; i < row; i++ ) matrix[i] = malloc( col * sizeof( int ) );
If the array is declared as
int matrix[row][col];
then change the function declaration to
float energycalc(float J, int m[][col], int row, int col){
The name of a two dimensional array of type T does not decay to T**.
If col is a local variable, then you need to call the function with the col parameter before matrix. This is done so that col is visible in the function.
Passing two dimensional array to a function in C is often confusing for newbies.
The reason is that they assume arrays are pointers and having lack of understanding how arrays decays to pointer.
Always remember that when passed as an argument arrays converted to the pointer to its first element.
In function call
E = energycalc(J, matrix, row, col);
matrix is converted to pointer to its first element which is matrix[0]. It means that passing matrix is equivalent to passing &matrix[0]. Note that the type of &matrix[0] is int(*)[col] (pointer to an array of col int) and hence is of matrix. This suggest that the second parameter of function energycalc must be of type int(*)[col]. Change the function declaration to
float energycalc(int col, int (*m)[col], int row, float J);
and call your function as
E = energycalc(col, matrix, row, J);
I think you can do something like this in C
float function_name(int mat_width, int mat_height, float matrix[mat_width][mat_height]) {
... function body...
}
A pointer-to-pointer is not an array, nor does it point to a two-dimensional array. It has nothing to do with arrays, period, so just forget you ever heard about pointer-to-pointers and arrays together.
(The confusion likely comes from the hordes of confused would-be programming teachers/authors telling everyone that they should use pointer-to-pointer when allocating dynamic 2D arrays. Which is just plain bad advice, since it will not allocate a real array allocated in adjacent memory cells, but rather a fragmented look-up table exploded in pieces all over the heap. See this for reference about how to actually allocate such arrays.)
Assuming that your compiler isn't completely ancient, simply use a variable length array (VLA), which is a feature introduced in the C language with the C99 standard.
#include <stdio.h>
void print_matrix (size_t row, size_t col, int matrix[row][col]);
int main (void)
{
int matrix[2][3] =
{
{1,2,3},
{4,5,6}
};
print_matrix(2, 3, matrix);
return 0;
}
void print_matrix (size_t row, size_t col, int matrix[row][col])
{
for(size_t r=0; r<row; r++)
{
for(size_t c=0; c<col; c++)
{
printf("%d ", matrix[r][c]);
}
printf("\n");
}
}
Below program (a toy program to pass around arrays to a function) doesn't compile.
Please explain me, why is the compiler unable to compile(either because of technical reason or because of standard reason?)
I will also look at some book explaining pointers/multi dimensional arrays(as I am shaky on these), but any off-the-shelf pointers here should be useful.
void print2(int ** array,int n, int m);
main()
{
int array[][4]={{1,2,3,4},{5,6,7,8}};
int array2[][2]={{1,2},{3,4},{5,6},{7,8}};
print2(array,2,4);
}
void print2(int ** array,int n,int m)
{
int i,j;
for(i=0;i<n;i++)
{
for(j=0;j<m;j++)
printf("%d ",array[i][j]);
printf("\n");
}
}
This (as usual) is explained in the c faq. In a nutshell, an array decays to a pointer only once (after it decayed, the resulting pointer won't further decay).
An array of arrays (i.e. a
two-dimensional array in C) decays
into a pointer to an array, not a
pointer to a pointer.
Easiest way to solve it:
int **array; /* and then malloc */
In C99, as a simple rule for functions that receive "variable length arrays" declare the bounds first:
void print2(int n, int m, int array[n][m]);
and then your function should just work as you'd expect.
Edit:
Generally you should have a look into the order in which the dimension are specified. (and me to :)
In your code the double pointer is not suitable to access a 2D array because it does not know its hop size i.e. number of columns. A 2D array is a contiguous allotted memory.
The following typecast and 2D array pointer would solve the problem.
# define COLUMN_SIZE 4
void print2(int ** array,int n,int m)
{
// Create a pointer to a 2D array
int (*ptr)[COLUMN_SIZE];
ptr = int(*)[COLUMN_SIZE]array;
int i,j;
for(i = 0; i < n; i++)
{
for(j = 0; j < m; j++)
printf("%d ", ptr[i][j]);
printf("\n");
}
}