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");
}
}
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].
#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.
This is the C code for determinant of matrix, But it gives compilation errors.
Code is:
#include<stdio.h>
#include<math.h>
int m;
float determinant(float b[][]);
int main(void)
{
int i,j,m;
printf("enter a no: ");
scanf("%d",&m);
//printf("%d",m);
float arr[m][m];
for(i=0;i<m;i++)
{
for(j=0;j<m;j++)
{
scanf("%f",&arr[i][j]);
//printf("%f",arr[i][j]);
}
}
for(i=0;i<m;i++)
{
for(j=0;j<m;j++)
{
printf("%f ",arr[i][j]);
}
printf("\n");
}
float det = determinant(arr);
printf("Determinant= %f ", det);
}
float determinant(float b[][])
{
int i,j;
int p;
float sum = 0;
float c[m][m];
for(i=0;i<m;i++)
{
for(j=0;j<m;j++)
{
printf("%f ",b[i][j]);
}
printf("\n");
}
if(m==2)
{
printf("Determinant for m=2");
sum = b[0][0]*b[1][1] - b[0][1]*b[1][0];
return sum;
}
for(p=0;p<m;p++)
{
int h = 0,k = 0;
for(i=1;i<m;i++)
{
for( j=0;j<m;j++)
{
if(j==p)
continue;
c[h][k] = b[i][j];
k++;
if(k == m-1)
{
h++;
k = 0;
}
}
}
m=m-1;
sum = sum + b[0][p]*pow(-1,p) * determinant(c);
}
return sum;
}
And the Compilation Errors are:
det.c:5:25: error: array type has incomplete element type
det.c: In function ‘main’:
det.c:36:2: error: type of formal parameter 1 is incomplete
det.c: At top level:
det.c:45:25: error: array type has incomplete element type
det.c: In function ‘determinant’:
det.c:91:3: error: type of formal parameter 1 is incomplete
det.c:99: confused by earlier errors, bailing out
Preprocessed source stored into /tmp/cc1Kp9KD.out file, please attach this to your bug report.
I think the error is in the passing of 2-D Array. when I passing it as a pointer then it gives warnings but no errors but it does not give the right result as in always gives determinant as Zero. So I guess the array is not being passed only and when I print it in the function determinant it doesn't print also.
Please help as I am stuck because of this in my project.
in your code,
scanf("%d",&m);
//printf("%d",m);
float arr[m][m];
here arr is a 2D array with static memory allocation so you can not read m at run time and declare arr like this.
so if you are want to define the array dynamically then use dynamic memory allocation methods like malloc() in C.
When you declare the prototype of a function as
int foo(int arr[], int n);
then compiler interprets it as
int foo(int (*arr), int n); // and that's why you can omit the first dimension!
i.e, your function is expecting first argument is of type int *. Similarly, when the parameter is a multidimensional array as
int foo(int arr[][col], int n); // Only first dimension can be omitted. You need to specify the second dimension.
then the compiler interprets it as
int foo(int (*arr)[col], int n);
i.e, your function is expecting first argument is of type int (*)[col] (a pointer to int array).
Since when passed to a function (in most cases) array names decay to pointer to its first element, in your case arr will be decayed to pointer to its first element, i.e, first row. Hence its type will become float (*)[m]. Its is compatible to your function parameter if you will declare it as
float determinant(int m, float b[][m]);
and the call should be like
float det = determinant(m, arr);
You can declare array dynamically like this in C99 (variable length arrays, pointed out by haccks), but not in the earlier version:
float arr[m][m];
So, if it troubles you then instead declare a pointer and malloc memory for it:
float* arr = malloc(sizeof(float)*m*m);
Also, the definition won't work (in either case):
float determinant(float b[][]);
you need to define the columns in the array that you pass to the function.
If you declare and allocate the pointer as I have shown then you can just pass a pointer in your function:
float determinant(float *b, int size); //here size is your row dimension, in this case equal to m
And inside the function, access your elements like:
*(b + size*i + j) = value // equivalent to b[i][j];
Declare your array with explicit bounds float b[m][m]; the compiler doesn't understand empty bounds in float b[][] (empty bounds are OK only for 1-D arrays, for reasons explained in the other answers).
So your determinant function should look like this:
float determinant(int m, float b[m][m])
{
...
}
There are other ways to make your code work, but I think this way is closest to what you already have.
I agree you can't move with this syntax of array initialization, u could have used
int *p=(int*)calloc(n*n,sizeof(float));
And then access your elements as :-
*(p+j+n*i);//for p[i][j] element
Hope it helps :)
I just started programming so pointers and arrays confuse me.
This program just assigns random numbers from 0 - 9 into array and prints them out
(#include <stdio.h> #include <stdlib.h> #include <time.h>)
int function(int *num[]){
int i;
for(i=0; i<10; i++){
srand((unsigned)time(NULL));
*num[i] = rand()%10;
printf("%d", *num[i]);
}
return 0;
}
int main(){
int num[10];
function(&num); // incompatable pointer type (how do i fix this?)
return 0;
}
Thankyou
Change your code to read something like this:
int function(int num[]){
int i;
for(i=0; i<10; i++){
srand((unsigned)time(NULL));
num[i] = rand()%10;
printf("%d", num[i]);
}
return 0;
}
int main(){
int num[10];
function(num);
return 0;
}
In main(), you are allocating an array of 10 integers. The call to function(num) passes the address of this array (technically, the address of the first element of the array) to the function() function. In a parameter declaration, int num[] is almost exactly equivalent to int *num (I'll let you ponder on that one). Finally, when you access the elements of num[] from within the function, you don't need any extra *. By using num[i], you are accessing the i'th element of the array pointed to by num.
It may help to remember that in C, the following are exactly the same:
num[i]
*(num+i)
You don't need to pass in a pointer to your array. Just pass in the array. I have fixed your pointer code below.
Also, you should not reset the seed (srand) through every iteration through the for loop.
int function(int num[]){
int i;
srand((unsigned(time(NULL));
for(i=0; i<10; i++){
num[i] = rand()%10;
printf("%d", num[i]);
}
return 0;
}
int main(){
int num[10];
function(num);
return 0;
}
It all depends on what you want to achieve in the end.
(1) If your function function is intended to be used with arrays of run-time size, then you have to pass a pointer to the first element of the array. That is achieved by the following equivalent declarations
int function(int *num)
or
int function(int num[])
Inside the function you have to access the passed array as
num[i] = rand() % 10;
And you call your function as
function(num);
Of course, when the array size is a run-time value, it makes sense to pass that size to the function as well, meaning that your function should have the following interface
int function(int num[], size_t n)
And implement it for an array of size n, instead of hardcoding 10 directly into your implementation.
(2) If your function function is intended to be used with arrays of fixed compile-time size (10 in this case), then the better approach would be to pass a pointer to the entire array. It is achieved by the following declaration
int function(int (*num)[10])
Inside the function you have to access the passed array as
(*num)[i] = rand() % 10;
And you call your function as
function(&num);
So it is either (1) or (2).
What you currently have in your code looks like a mix of these two approaches. More precisely, you code looks like an attempt to implement the second approach, but it is missing some important parentheses :)
Most answers given to you so far suggest using the first approach. However, seeing that the array size is actually a compile-time constant in your case, I would suggest sticking with the second approach.
Use either int *num or int num[], don't use both together as you have right now (int *num[]). An array is passed to a function as a pointer to the first element, i.e. int *num, which is equivalent to int num[].