错误：一元'*'的类型参数无效（具有'int'）

Question

1.This is a part of a program to add two matrices using pointers.

2.I am getting the error error : invalid type argument of unary '*' (have 'int') .

3.Here A and B are initialized 2D arrays and c1 , r1 are no. of columns and rows respectively of matrix A and c2 , r2 are no. of columns and rows respectively of matrix B .

main() {
    int i, j, A[10][10], B[10][10], r1, r2, c1, c2;

    //Inputting Matrix A
    printf("\nFOR SET A ");
    printf("\n\nEnter number of rows : ");
    scanf("%d", &r1);
    printf("\n\nEnter Number of Columns :");
    scanf("%d", &c1);
    printf("\n\nEnter Elements of matrix A :\n\n");
    for (i = 0; i < r1; i++) {
       for (j = 0; j < c1; j++) {
           scanf("%d", (*(A + i) + j));
       }
    }

    //Inputting Matrix A
    printf("\n\nFOR SET B:");
    printf("\n\nEnter number of rows : ");
    scanf("%d", &r2);
    printf("\n\nEnter Number of Columns :");
    scanf("%d", &c2);
    printf("\n\nEnter Elements of matrix B :\n\n");
    for (i = 0; i < r2; i++) {
       for (j = 0; j < c2; j++) {
           scanf(" %d", (*(B + i) + j));
       }
    }

     //Displaying matrix A
     printf("\n\nMatrix A is :\n\n");
     for (i = 0; i < r1; i++) {
         for (j = 0; j < c1; j++) {
              printf(" %d ", *(*(A + i) + j));
         }
         printf("\n\n");
     }

     //Displaying matrix B
     printf("\n\nMatrix B is :\n\n");
     for (i = 0; i < r2; i++) {
         for (j = 0; j < c2; j++) {
              printf(" %d ", *(*(B + i) + j));
         }
         printf("\n\n");
     }

     //Calling the Addition function
     add(A, r1, c1, B, r2, c2);
 }

 void add(int **A, int r1, int c1, int **B, int r2, int c2) {
     if (r1 == r2 && c1 == c2) {
         int i, j;
         printf("\n\nThe Addition of matrix A and B is :\n\n");
         for (i = 0; i < r1; i++) {
             for (j = 0; j < c1; j++) {
                 printf(" %d ", ((*(*(A + i) + j)) + (*(*(B + i) + j))));
             }
             printf("\n\n");
         }
    } else
        printf("\n\nMatrices are not of same order !!!");
}

Answer 1

Your function does not compute the offset of the elements correctly.

You can correct the code this way:

void add(int *A, int r1, int c1, int *B, int r2, int c2) { 
    if (r1 == r2 && c1 == c2) {
        int i, j;
        printf("\n\nThe Addition of matrix A and B is :\n\n");
        for (i = 0; i < r1; i++) {
            for (j = 0; j < c1; j++) {
                printf(" %d ", A[i * c1 + j] + B[i * c2 + j]);
                // if you must use the less readable pointer syntax
                // use this strictly equivalent form instead:
                //printf(" %d ", *(A + i * c1 + j) + *(B + i * c2 + j));
            }
            printf("\n");
        }
        printf("\n");
    } else {
        printf("Matrices are not of same order !!!\n\n");
    }
}

The function can be called this way:

int mat1[3][4] = { ... };
int mat2[3][4] = { ... };

add(&mat1[0][0], 3, 4, &mat2[0][0], 3, 4);

Note that since you only handle matrices with the exact same geometry, you could further simplify the main loop this way:

    int i, j, k = 0;
    printf("\n\nThe Addition of matrix A and B is :\n\n");
    for (i = 0; i < r1; i++) {
        for (j = 0; j < c1; j++, k++) {
            printf(" %d ", A[k] + B[k]);
            k++;
        }
        printf("\n");
    }
    printf("\n");

As Felix Palmen commented, C99 introduced variable length arrays and a syntax to pass them as function arguments:

void add(int r1, int c1, int (*A)[c1], int r2, int c2, int (*B)[c2]) { 
    if (r1 == r2 && c1 == c2) {
        int i, j;
        printf("\n\nThe Addition of matrix A and B is :\n\n");
        for (i = 0; i < r1; i++) {
            for (j = 0; j < c1; j++) {
                printf(" %d ", A[i][j] + B[i][j]);
                // if you must use the less readable pointer syntax
                // use this strictly equivalent form instead:
                //printf(" %d ", *(*(A + i) + j) + *(*(B + i) + j));
            }
            printf("\n");
        }
        printf("\n");
    } else {
        printf("Matrices are not of same order !!!\n\n");
    }
}

The function can be called this way:

int mat1[3][4] = { ... };
int mat2[3][4] = { ... };

add(3, 4, mat1, 3, 4, mat2);

Note however that this feature is not supported by some mainstream C compilers and was made optional in the latest version of the Standard (C11).

EDIT: Your case is actually very different: your matrices have a fixed size of 10x10 and you want to handle 2D submatrices. You function should look like this:

void add(int A[][10], int r1, int c1, int B[][10], int r2, int c2) { 
    if (r1 == r2 && c1 == c2) {
        int i, j;
        printf("\n\nThe Addition of matrix A and B is :\n\n");
        for (i = 0; i < r1; i++) {
            for (j = 0; j < c1; j++) {
                printf(" %d ", A[i][j] + B[i][j]);
                // if you must use the less readable pointer syntax
                // use this strictly equivalent form instead:
                //printf(" %d ", *(*(A + i) + j) + *(*(B + i) + j));
            }
            printf("\n");
        }
        printf("\n");
    } else {
        printf("Matrices are not of same order !!!\n\n");
    }
}

The function can be called this way:

add(mat1, 3, 4, mat2, 3, 4);

Finally, your prototype void add(int **A, int r1, int c1, int **B, int r2, int c2) is incorrect because A and B are arrays of arrays of int , not arrays of pointers to arrays of int .

Note also that the prototype for add should appear before the code that calls it. The prototype for main without arguments is int main(void) , your syntax is obsolete and no longer supported since C99.

Answer 2

No matter how many elements you actually enter, your arrays are declared with a fixed size. Your function must know about this, it has at least to know the number of elements in one row, so it can calculate the offset to the next row correctly. In your case, use the following prototype:

void add(int (*A)[10], int r1, int c1, int (*B)[10], int r2, int c2)

Answer 3

You have an array of arrays. This means you should use

void add(int **A,int r1,int c1,int **B,int r2,int c2)

as the function prototype. Also can you show the entire the code?

edit: (full code)

int** mat_add(int** A, int **B, int r1, int c1, int r2, int c2){
int** C, i, j;
C =(int **)calloc(r1*c1, sizeof(int*));

for(i=0;i <r1;i++)
    C[i] = (int *)calloc(c1, sizeof(int));

for(i = 0;i < r1; i++)
    for(j = 0; j < c1; j++)
        C[i][j] = A[i][j] + B[i][j];
return C;
}
void print_arr(int** C, int r, int c){
    int i,j;
    for (i=0;i<r;i++){
        for(j=0;j<c;j++)
            printf("%d\t", C[i][j]);
        printf("\n");
    }
}

int main(){
    int **A, **B, r1, r2, c1, c2;
    int i;
    scanf("%d%d%d%d", &r1, &r2, &c1, &c2);
    A = (int**)calloc(r1*c1, sizeof(int*)); //used for allocating memory for pointers
    B = (int**)calloc(r2*c2, sizeof(int*));

    for(i=0;i <r1;i++)
        A[i] = (int *)calloc(c1, sizeof(int)); //used for allocating memory for each row of integers having size of length of column
    for(i=0;i <r2;i++)
        B[i] = (int *)calloc(c2, sizeof(int));

    //take matrices as input from the user here
    //...

    if(r1==r2 && c1==c2){
        int ** C = mat_add(A, B, r1, c1, r2, c2);
        print_arr(C, r1, c1);
    }
    else
        printf("MATRICES MUST BE OF SAME SIZE!");
    return 0;
}