Predict the output of following C Program.
C
Output: The program compiles fine and produces following output:
C
#include <stdio.h>
#define R 4
#define C 4
void modifyMatrix(int mat[][C])
{
mat++;
mat[1][1] = 100;
mat++;
mat[1][1] = 200;
}
void printMatrix(int mat[][C])
{
int i, j;
for (i = 0; i < R; i++)
{
for (j = 0; j < C; j++)
printf("%3d ", mat[i][j]);
printf("\n");
}
}
int main()
{
int mat[R][C] = { {1, 2, 3, 4},
{5, 6, 7, 8},
{9, 10, 11, 12},
{13, 14, 15, 16}
};
printf("Original Matrix \n");
printMatrix(mat);
modifyMatrix(mat);
printf("Matrix after modification \n");
printMatrix(mat);
return 0;
}
Original Matrix 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Matrix after modification 1 2 3 4 5 6 7 8 9 100 11 12 13 200 15 16At first look, the line "mat++;" in modifyMatrix() seems invalid. But this is a valid C line as array parameters are always pointers (see this and this for details). In modifyMatrix(), mat is just a pointer that points to block of size C*sizeof(int). So following function prototype is same as "void modifyMatrix(int mat[][C])"
void modifyMatrix(int (*mat)[C]);
When we do mat++, mat starts pointing to next row, and mat[1][1] starts referring to value 10. mat[1][1] (value 10) is changed to 100 by the statement "mat[1][1] = 100;". mat is again incremented and mat[1][1] (now value 14) is changed to 200 by next couple of statements in modifyMatrix().
The line "mat[1][1] = 100;" is valid as pointer arithmetic and array indexing are equivalent in C.
On a side note, we can't do mat++ in main() as mat is 2 D array in main(), not a pointer.
Please write comments if you find above answer/explanation incorrect, or you want to share more information about the topic discussed above