Passing char matrix to C function - c

Question

I have to work with a given function that is declared like this:
void someFunc (char** characters, int from, int to);
The problem is, i cant find a way to pass a matrix to it, for example:
char matrix[4][4] = { { 'a', 'e' , 'i', 'm'},
{ 'b', 'f' , 'j', 'n'},
{ 'c', 'g' , 'k', 'p'},
{ 'd', 'h' , 'l', 's'} };
someFunc(matrix, 2, 3); //doesnt work
someFunc(&matrix, 2, 3); //doesnt work
The compiler complains, but i cant change the signature of someFunc, so how i can do to pass a char matrix to the function?

Answer 1

AS #IgorTandetnik said
define like this
char* data[]= {matrix[0], matrix[1], matrix[2], matrix[3]};
someFunc(data, 2, 3); // works fine
array of character pointers initialized with the each row of matrix and then passed this as argument
test code:
#include<stdio.h>
void someFunc (char** characters, int from, int to);
char matrix[4][4] = { { 'a', 'e' , 'i', 'm'},
{ 'b', 'f' , 'j', 'n'},
{ 'c', 'g' , 'k', 'p'},
{ 'd', 'h' , 'l', 's'} };
main()
{
char* data[]= {matrix[0], matrix[1], matrix[2], matrix[3]};
someFunc(data, 2, 3);
}
void someFunc (char** c , int from, int to)
{
int i=0,j=0;
for(i=0;i<4;i++)
for(j=0;j<4;j++)
printf("%c\n",c[i][j]); //printf("%c",*(c[i]+j));
}

Answer 2

instead of using an automatic matrix just malloc a pointer to pointer the code goes like this
main(void)
{
int i;
char **ptr = NULL;
ptr = (char**)malloc(sizeof(char) * 4);
for(i = 0 ; i < 4 ; i++)
ptr[i] = (char*)malloc(sizeof(char) * 4);
//initialize your matrix here
someFunc(ptr , 2 , 3);
}
or , if you are ready to modify the function prototype (just a tiny bit) use this one instead void sumFunc(char matrix[][4) , int from , int to) and you won't have to change any thing else

Answer 3

The below code will work. But you may need to access the array using one index instead of two index in the function. When you declare a double index array, the data is not store as a "pointer to pointer". So you may not pass char foo0[ ][ ] using char **foo1.
void someFunc (char* characters)
{
char c = character[0]; // c = 'a'
c = character[4] // c = 'b'
c = character[8] // c = 'c'
c = character[12] // c = 'd'
}
void main(void)
{
char matrix[4][4] = {
{ 'a', 'e' , 'i', 'm'},
{ 'b', 'f' , 'j', 'n'},
{ 'c', 'g' , 'k', 'p'},
{ 'd', 'h' , 'l', 's'} };
someFunc((char*)matrix);
}

Answer 4

An example of char **variable would be:
char *array[]={"this","is","char","pointer"};
Used like:
someFunc(array, 2, 3);
An example of an array of the previous would have to be something like:
char *array[][5] = {
{"this","is","char","element","one"},
{"this","is","char","element","two"},
{"this","is","char","element","three"}};
But that is not what your prototype requires. Use the previous example to see that that does for you.
[EDIT] per your question and my comment, here is how to access the content of an array. (it is random and can be printed in any format you need).
#include <windows.h>
#include <ansi_c.h>
int main(void)
{
char *array[]={"this","is","char","pointer"};
int i,j;
//print in columns
printf("Columns:\n");
for(i=0;i<7; i++)
{
for(j=0;j<4;j++)
{
if(i> strlen(array[j]))
printf(" ");
else
printf("%c", array[j][i]);
}
printf("\n");
}
//print in rows (easier)
printf("\n\nRows:\n");
for(i=0;i<4;i++)
{
printf("%s\n", array[i]);
}
getchar();
return 0;
}

Answer 5

I had a similar experience once and I found out that using char matrixes kinda give some issues ... this is my actual suggestion, I hope you'll understand it. (I know it's late, but maybe it will be to aid for somebody else)
void someFunc(char* matrix, int from, int to){/*function declaration*/}
int main(){
int x = 4; //first matrix's dimension
int y = 4; //second matrix's dimension
char matrix[x*y];
//printing the matrix all in a single row
for(int i=0, i<x, i++)
for(int j=0, j<y, j++)
printf("%c", matrix[i+j*x]);
//it is just way simpler, reduces a ton of complexity, and having a string is much more flexible than having a matrix, especially with chars
return 0
}