Scanning to 2d array and then passing it to function - C
My program is getting segmentation fault if I allocate function as 1D array and then pass it to function. It is built for 2d array. Problem is, that I can't find out how to allocate 2d array and how to pass it correctly into function. Hope all is explained clearly. If you know what is wrong please try to lead me on correct way to fix it. Many thanks. Here is code:
int main()
{
int i, j, size;
scanf("%d", &size);
int *a;
//here i try to allocate it as 2d array
*a = (int *)malloc(size * sizeof(int));
for (i=0; i<size; i++)
{
a[i] = (int *)malloc(size * sizeof(int));
}
//here i scan value to 2d array
for (i = 0; i < size; i++)
for (j = 0; j < size; j++){
scanf("%d", &a[i][j]); }
//here i pass array and size of it into function
if (is_magic(a,size))
function header looks like:
int is_magic(int **a, int n)
c arrays 2d
asked Dec 27 '18 at 18:25
Dudo
175
add a comment |
My program is getting segmentation fault if I allocate function as 1D array and then pass it to function. It is built for 2d array. Problem is, that I can't find out how to allocate 2d array and how to pass it correctly into function. Hope all is explained clearly. If you know what is wrong please try to lead me on correct way to fix it. Many thanks. Here is code:
int main()
{
int i, j, size;
scanf("%d", &size);
int *a;
//here i try to allocate it as 2d array
*a = (int *)malloc(size * sizeof(int));
for (i=0; i<size; i++)
{
a[i] = (int *)malloc(size * sizeof(int));
}
//here i scan value to 2d array
for (i = 0; i < size; i++)
for (j = 0; j < size; j++){
scanf("%d", &a[i][j]); }
//here i pass array and size of it into function
if (is_magic(a,size))
function header looks like:
int is_magic(int **a, int n)
c arrays 2d
asked Dec 27 '18 at 18:25
Dudo
175
Do not usex = (int *)malloc...- usex = malloc..
– SergeyA
Dec 27 '18 at 18:42
add a comment |
My program is getting segmentation fault if I allocate function as 1D array and then pass it to function. It is built for 2d array. Problem is, that I can't find out how to allocate 2d array and how to pass it correctly into function. Hope all is explained clearly. If you know what is wrong please try to lead me on correct way to fix it. Many thanks. Here is code:
int main()
{
int i, j, size;
scanf("%d", &size);
int *a;
//here i try to allocate it as 2d array
*a = (int *)malloc(size * sizeof(int));
for (i=0; i<size; i++)
{
a[i] = (int *)malloc(size * sizeof(int));
}
//here i scan value to 2d array
for (i = 0; i < size; i++)
for (j = 0; j < size; j++){
scanf("%d", &a[i][j]); }
//here i pass array and size of it into function
if (is_magic(a,size))
function header looks like:
int is_magic(int **a, int n)
c arrays 2d
asked Dec 27 '18 at 18:25
Dudo
175
My program is getting segmentation fault if I allocate function as 1D array and then pass it to function. It is built for 2d array. Problem is, that I can't find out how to allocate 2d array and how to pass it correctly into function. Hope all is explained clearly. If you know what is wrong please try to lead me on correct way to fix it. Many thanks. Here is code:
int main()
{
int i, j, size;
scanf("%d", &size);
int *a;
//here i try to allocate it as 2d array
*a = (int *)malloc(size * sizeof(int));
for (i=0; i<size; i++)
{
a[i] = (int *)malloc(size * sizeof(int));
}
//here i scan value to 2d array
for (i = 0; i < size; i++)
for (j = 0; j < size; j++){
scanf("%d", &a[i][j]); }
//here i pass array and size of it into function
if (is_magic(a,size))
function header looks like:
int is_magic(int **a, int n)
c arrays 2d
c arrays 2d
asked Dec 27 '18 at 18:25
Dudo
175
asked Dec 27 '18 at 18:25
Dudo
175
asked Dec 27 '18 at 18:25
Dudo
175
asked Dec 27 '18 at 18:25
Dudo
175
asked Dec 27 '18 at 18:25
Dudo
175
175
Do not usex = (int *)malloc...- usex = malloc..
– SergeyA
Dec 27 '18 at 18:42
add a comment |
Do not usex = (int *)malloc...- usex = malloc..
– SergeyA
Dec 27 '18 at 18:42
Do not use
x = (int *)malloc... - use x = malloc..– SergeyA
Dec 27 '18 at 18:42
Do not use
x = (int *)malloc... - use x = malloc..– SergeyA
Dec 27 '18 at 18:42
add a comment |
2 Answers
2
active
oldest
votes
Scanning 2D array ? For that you need to take a as of int** type not just int* type. For e.g
int **a = malloc(NUM_OF_ROW * sizeof(int*)); /* allocate memory dynamically for n rows */
And then allocate memory for each row for e.g
for (i=0; i<size; i++){
a[i] = malloc(NUM_OF_COLUMN * sizeof(int)); /* in each row how many column, allocate that much memory dynamically */
}
answered Dec 27 '18 at 18:30
Achal
9,1202730
It worked! Thanks alot! I did not notice this little detail before hehe
– Dudo
Dec 27 '18 at 19:20
I'm glad that my answer helped you. Your welcome on stackoverflow.
– Achal
Dec 27 '18 at 19:26
add a comment |
This doesn't work:
*a = (int *)malloc(size * sizeof(int));
Because a has type int * so *a has type int, so it doesn't make sense to assign a pointer to that. You're also attempting to dereference a pointer which has not been initialized yet, invoking undefined behavior.
You need to define a as an int **:
int **a;
And assign to it directly on the first allocation, using sizeof(int *) for the element size:
a = malloc(size * sizeof(int *));
Note also that you shouldn't cast the return value of malloc.
answered Dec 27 '18 at 18:29
dbush
92.9k12101133
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Scanning 2D array ? For that you need to take a as of int** type not just int* type. For e.g
int **a = malloc(NUM_OF_ROW * sizeof(int*)); /* allocate memory dynamically for n rows */
And then allocate memory for each row for e.g
for (i=0; i<size; i++){
a[i] = malloc(NUM_OF_COLUMN * sizeof(int)); /* in each row how many column, allocate that much memory dynamically */
}
answered Dec 27 '18 at 18:30
Achal
9,1202730
It worked! Thanks alot! I did not notice this little detail before hehe
– Dudo
Dec 27 '18 at 19:20
I'm glad that my answer helped you. Your welcome on stackoverflow.
– Achal
Dec 27 '18 at 19:26
add a comment |
Scanning 2D array ? For that you need to take a as of int** type not just int* type. For e.g
int **a = malloc(NUM_OF_ROW * sizeof(int*)); /* allocate memory dynamically for n rows */
And then allocate memory for each row for e.g
for (i=0; i<size; i++){
a[i] = malloc(NUM_OF_COLUMN * sizeof(int)); /* in each row how many column, allocate that much memory dynamically */
}
answered Dec 27 '18 at 18:30
Achal
9,1202730
It worked! Thanks alot! I did not notice this little detail before hehe
– Dudo
Dec 27 '18 at 19:20
I'm glad that my answer helped you. Your welcome on stackoverflow.
– Achal
Dec 27 '18 at 19:26
add a comment |
Scanning 2D array ? For that you need to take a as of int** type not just int* type. For e.g
int **a = malloc(NUM_OF_ROW * sizeof(int*)); /* allocate memory dynamically for n rows */
And then allocate memory for each row for e.g
for (i=0; i<size; i++){
a[i] = malloc(NUM_OF_COLUMN * sizeof(int)); /* in each row how many column, allocate that much memory dynamically */
}
answered Dec 27 '18 at 18:30
Achal
9,1202730
Scanning 2D array ? For that you need to take a as of int** type not just int* type. For e.g
int **a = malloc(NUM_OF_ROW * sizeof(int*)); /* allocate memory dynamically for n rows */
And then allocate memory for each row for e.g
for (i=0; i<size; i++){
a[i] = malloc(NUM_OF_COLUMN * sizeof(int)); /* in each row how many column, allocate that much memory dynamically */
}
answered Dec 27 '18 at 18:30
Achal
9,1202730
answered Dec 27 '18 at 18:30
Achal
9,1202730
answered Dec 27 '18 at 18:30
Achal
9,1202730
answered Dec 27 '18 at 18:30
Achal
9,1202730
9,1202730
It worked! Thanks alot! I did not notice this little detail before hehe
– Dudo
Dec 27 '18 at 19:20
I'm glad that my answer helped you. Your welcome on stackoverflow.
– Achal
Dec 27 '18 at 19:26
add a comment |
It worked! Thanks alot! I did not notice this little detail before hehe
– Dudo
Dec 27 '18 at 19:20
I'm glad that my answer helped you. Your welcome on stackoverflow.
– Achal
Dec 27 '18 at 19:26
It worked! Thanks alot! I did not notice this little detail before hehe
– Dudo
Dec 27 '18 at 19:20
It worked! Thanks alot! I did not notice this little detail before hehe
– Dudo
Dec 27 '18 at 19:20
I'm glad that my answer helped you. Your welcome on stackoverflow.
– Achal
Dec 27 '18 at 19:26
I'm glad that my answer helped you. Your welcome on stackoverflow.
– Achal
Dec 27 '18 at 19:26
add a comment |
This doesn't work:
*a = (int *)malloc(size * sizeof(int));
Because a has type int * so *a has type int, so it doesn't make sense to assign a pointer to that. You're also attempting to dereference a pointer which has not been initialized yet, invoking undefined behavior.
You need to define a as an int **:
int **a;
And assign to it directly on the first allocation, using sizeof(int *) for the element size:
a = malloc(size * sizeof(int *));
Note also that you shouldn't cast the return value of malloc.
answered Dec 27 '18 at 18:29
dbush
92.9k12101133
add a comment |
This doesn't work:
*a = (int *)malloc(size * sizeof(int));
Because a has type int * so *a has type int, so it doesn't make sense to assign a pointer to that. You're also attempting to dereference a pointer which has not been initialized yet, invoking undefined behavior.
You need to define a as an int **:
int **a;
And assign to it directly on the first allocation, using sizeof(int *) for the element size:
a = malloc(size * sizeof(int *));
Note also that you shouldn't cast the return value of malloc.
answered Dec 27 '18 at 18:29
dbush
92.9k12101133
add a comment |
This doesn't work:
*a = (int *)malloc(size * sizeof(int));
Because a has type int * so *a has type int, so it doesn't make sense to assign a pointer to that. You're also attempting to dereference a pointer which has not been initialized yet, invoking undefined behavior.
You need to define a as an int **:
int **a;
And assign to it directly on the first allocation, using sizeof(int *) for the element size:
a = malloc(size * sizeof(int *));
Note also that you shouldn't cast the return value of malloc.
answered Dec 27 '18 at 18:29
dbush
92.9k12101133
This doesn't work:
*a = (int *)malloc(size * sizeof(int));
Because a has type int * so *a has type int, so it doesn't make sense to assign a pointer to that. You're also attempting to dereference a pointer which has not been initialized yet, invoking undefined behavior.
You need to define a as an int **:
int **a;
And assign to it directly on the first allocation, using sizeof(int *) for the element size:
a = malloc(size * sizeof(int *));
Note also that you shouldn't cast the return value of malloc.
answered Dec 27 '18 at 18:29
dbush
92.9k12101133
edited Dec 27 '18 at 18:44
answered Dec 27 '18 at 18:29
dbush
92.9k12101133
answered Dec 27 '18 at 18:29
dbush
92.9k12101133
answered Dec 27 '18 at 18:29
dbush
92.9k12101133
92.9k12101133
add a comment |
add a comment |
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Do not use
x = (int *)malloc...- usex = malloc..– SergeyA
Dec 27 '18 at 18:42