How Can I Correctly Pass Multidimensional Arrays to Functions in C and C ?

Multidimensional Array Passing in C and C

In both C and C , arrays of type int4 cannot be directly passed to functions expecting arrays of int*. This incompatibility stems from the fundamental difference in how these two languages handle multidimensional arrays.

In C, a multidimensional array name decays to a pointer to its first element, allowing for the implementation exemplified in the code snippet. However, in C , arrays maintain their type even when used in function calls, resulting in the error message encountered:

cannot convert `int (*)[4]' to `int**' for argument `1' to `void print(int**, int, int)'

Solution for C and C

To pass a multidimensional array to a function in both C and C , a technique known as pointer arithmetic is employed:

1. Declare the array pointer as a pointer to pointers: int **arr.
2. Convert the array name to a pointer to an array pointer: arr = (int **)a.

Modified Code

void print(int **arr, int s1, int s2) {
    int i, j;
    for(i = 0; i < s1; i++)
        for(j = 0; j < s2; j++)
            printf("%d, ", arr[i][j]);
}

int main() {
    int a[4][4] = {{0}};
    print((int **)a, 4, 4);
}

Important Note

The code compiles and executes successfully in both C and C . However, additional corrections were made to the printf statement to ensure correct access to array elements using arr[i][j] instead of *((arr i) j).

Remember, the inability to pass multidimensional arrays directly arises from the distinct behavior of arrays in C and C and must be addressed accordingly.

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