Least Common Subsequence

[AnabelleRose]

I have currently been trying to code the Least Common Subsequence Algorithm in C and I am having some basic problems with pointers and arrays. First of all, how do I read in a string and then pass that string to the pointer? To be more clear I will paste a portion of my code (it's not working yet)[code=c]
#include<stdio. h>
#include<string .h>
#include<math.h >
#include<stdlib .h>
#define MAX(a,b) (a>b?a:b)


static int LCS_length(char **ptr_x,char **ptr_y,char **ptr_c,int m,int n);
static backTrack(char **ptr_c,char **ptr_x,char **ptr_y,int i,int j);
static print(char **ptr_c,char **ptr_x,char **ptr_y, int i, int j);


int main()
{
char **ptr_x,**ptr_y ,**ptr_c;
char x,y;
int m,n;
int i=0,j=0;
int index_i,index_j ,index_c;
printf("Enter the value of m\n");
scanf("%d",&m);
printf("Enter the first string\n");
scanf("%s",x);
strcpy(ptr_x,x) ;[/code] ptr_x is the pointer variable and its pointing to a 2 D array. I need to read in a string and let this ptr_x point to it.
If i was not using dynamic memory allocation, I would do it as follows:
Declare an array:[code=c]
char a[10][10];
char **ptr_x;
ptr_x=a;
printf(enter the value of a);
scanf("%s",a)[/code]
I am stuck here, please help me what and how i should use pointers and arrays together when using 2D arrays and dynamic memory allocation.

I am getting these errors when I compile
lcs.c:24: warning: passing argument 1 of âstrcpyâ from incompatible pointer type
lcs.c:24: warning: passing argument 2 of âstrcpyâ makes pointer from integer without a cast
lcs.c:35: warning: passing argument 1 of âstrcpyâ from incompatible pointer type
lcs.c:35: warning: passing argument 2 of âstrcpyâ makes pointer from integer without a cast


Thanks in advance!
Anabelle

[weaknessforcats]

Firstly, this code won't compile:

int main()
{
char a[10][10];
char **ptr_x;
ptr_x=a; <<<<<<!!!!!
printf(enter the value of a);
scanf("%s",a)
}

You cannot assgn a (an array) to a pointer-to-pointer to an int.

The reason is that the name of an array (in this case a) is the address of element 0. Element 0 if the array is an array of 10 char.

Therefore, a[0] is an array of 10 char. Therefore the name a is the address of an array of 10 char:

char (*ptr_x)[10] = a; //OK.

A char** is a pointer to a pointer to a single int.

Since you need to read in a string, why not just delcare an array large enough to hold the string:

char array[100];

and scanf() into that??

Later you can treat this array as a 10x10.

[AnabelleRose]

I am dynamically allocating memory to the array which is pointed to by the pointer ptr_x. Now my question is, how do I read in a string value from stdin and store it in the **ptr_x? I do not want to do this:

char a[10][10];
char **ptr_x;
ptr_x=(int **) (malloc sizeof(int *)) //allocate memory to ptr_x.
printf("Enter a string value for a\n");
scanf("%s",a);
ptr_x=a; // I don't think this works!
This is the part where I am stuck.
I am not sure how to accept string input from the user and store it into a 2D array pointed to by a pointer!

Thanks,
Anabelle

[weaknessforcats]

You are making this too hard.

There are no multi-dimensional arrays on C. There are only one-dimensional arrays.

Assume you have this in memory:

RosesAreRedViol etsAreBlue

There are 25 characters there.

A char array[25] in memory would look like:

RosesAreRedViol etsAreBlue

A char array[5][5] in memory would look like:

RosesAreRedViol etsAreBlue

Kinda the same, right?

So, you allocate as for a one-dimensional array. In your case, allocate 25 chars.

Read in your string.

To access the V of Violets you could:

array[11];

If this was 2D array of [5][5] then you want array[2][1] or

array[ 2* 5*sizeof(char) + 1 * 1 * sizeof(char)]

Otherwise written as array[2][1].

It's just a question of who does the pointer aritmetic.

So starting with char array[25] and you string in there

RosesAreRedViol etsAreBlue

You could declare a pointer to an array of 5 char:

char (*ptr_x)[5];

and then typecast the array name to the address of an array of 5 char.

ptr_x = (char (*)[5])array;

Now if you use array, it's one dimensional of 25
If you use ptr_x it's two-dimensional. Each element is an array of 5 char.

and the memory you are using is the same memory.

[donthi]

1. #include<stdio. h>
2. #include<string .h>
3. #include<math.h >
4. #include<stdlib .h>
5. #define MAX(a,b) (a>b?a:b)
6.
7.
8. static int LCS_length(char **ptr_x,char **ptr_y,char **ptr_c,int m,int n);
9. static backTrack(char **ptr_c,char **ptr_x,char **ptr_y,int i,int j);
10. static print(char **ptr_c,char **ptr_x,char **ptr_y, int i, int j);
11.
12.
13. int main()
14. {
15. char **ptr_x,**ptr_y ,**ptr_c;
16. char x,y;
17. int m,n;
18. int i=0,j=0;
19. int index_i,index_j ,index_c;
20. printf("Enter the value of m\n");
21. scanf("%d",&m);
22. printf("Enter the first string\n");
23. scanf("%s",x);
24. strcpy(ptr_x,x) ;






can u write reaming code for it

[qartar]

Given your code:

char a[10][10];
char **ptr_x;

Like cats said, your multi-dimensional array is essentially interpreted as an single dimensional array with a size of 100; as in it is a continuous block of 100 characters. Your ptr_x is pointing to a pointer to an array. So when you use array operators [] with ptr_x it expects to find an array of pointers which would then point to your character array. But your multi-dimensional array just has the 100 characters in it, not the array of pointers that ptr_x expects. Maybe that is helpful? :/