Invoke function with 2D array as parameter [duplicate] - c

Question

This question already has answers here:
Passing a multidimensional array of variable size
(2 answers)
How to pass 2D array (matrix) in a function in C?
(4 answers)
Closed 4 years ago.
I'm Java developer, not a C programmer. I'm trying to invoke some function that receives a 2D array, tried like:
#include <stdio.h>
int foo(int mat[][N])
{
printf("%d",N); /* <--- here error */
}
int main()
{
int a[5][5] =
{
{1, 2, 3, 4,5},
{1, 5, 5, 5,2},
{1, 4, 30,4,2},
{1, 2, 2,2, 2},
{1,1, 1,1, 2 }
};
foo(a);
return 0;
}
But the compiler says:
'N' undeclared (first use in this function)
I must have the signature: int foo(int mat[][N])
How should I pass the 2D array then?

Answer 1

You can use macros to define constant numbers, i.e. #define N 5. At compile time each appearance of the defined macro name will be replaced with the given value. In our case each N would be replaced with 5.
But this would not solve your problem, because it would print 5 every time, even if you change the dimensions of your array.
Pass the dimensions of the array as parameters.
You can pass the 2D array as a pointer.
I created a printing function, with what I can show you accessing the elements from the array. So the function's prototype would look like:
void print2DMatrix(int *matrix2D, const int rowLength, const int columnLength);
where matrix2D is a pointer, there will be passed the address of the array. The other two parameters tell us the dimensions of the array.
We can access the elements in a selected row and in a selected column with matrix2D[selectedRow * maxRowLength + selectedColumn].
When you call the function, you can pass the name of the array, like print2DMatrix(*myMatrix, myRowLength, myColumnLength);, of course you shall declare and initialize it before using.
print2DMatrix(myMatrix, myRowLength, myColumnLength); would cause warning, because the passed argument and the waited one have different types, but it will run correctly. The passed one is a double pointer, when the waited one is a single pointer. Because of that you have to use print2DMatrix(*myMatrix, myRowLength, myColumnLength);, where *myMatrix will point to the first row of our array.
I would like to mention that myMatrix and *myMatrix point to the same address, the difference is: myMatrix is looked as a double pointer to an integer, when *myMatrix is looked as a pointer to an integer. Run printf("%d %d %d", myMatrix, *myMatrix, **myMatrix); to see the result. **myMatrix will point to the first element of the first row, what is 11. See the entire code below...
#include <stdio.h>
#define ROW_LENGTH 5
#define COLUMN_LENGTH 5
void print2DMatrix(int *matrix2D, const int rowLength, const int columnLength)
{
int i;
for (i = 0; i < rowLength; i++)
{
int j;
for (j = 0; j < columnLength; j++)
{
printf("%d ", matrix2D[i * rowLength + j]);
}
printf("\n");
}
}
int main(void)
{
const int myRowLength = ROW_LENGTH;
const int myColumnLength = COLUMN_LENGTH;
int myMatrix[ROW_LENGTH][COLUMN_LENGTH] =
{
{11, 12, 13, 14, 15},
{21, 22, 23, 24, 25},
{31, 32, 33, 34, 35},
{41, 42, 43, 44, 45},
{51, 52, 53, 54, 55}
};
print2DMatrix(*myMatrix, myRowLength, myColumnLength);
return 0;
}

Answer 2

You are trying to print a non existent value. N is not a declared variable.
The function foo is receiving a pointer. Unlike java, C deals with pointers, not objects, it does not have any information about what it received, except that it is a pointer that contains addresses to ints. Therefore, the function does not know (or care) about your array being of size N (regardless of you mentioning N) .
When you call printf there's nothing in memory called N, which explains the error.
You could always send a second argument which is an int specifying the size and print that one.
void foo(int[][] mat, int matSize){
printf("%d\n",matSize);
}
Notice how I used int[][] this is to simplify and illustrate the fact that is just a pointer.