Output of C Program | Set 23
Predict the output of following C Program.
#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; } |
Output: The program compiles fine and produces following output:
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 16
At 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.
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