How Can Multidimensional Arrays Be Passed to Functions in C and C ?

Passing Multidimensional Arrays to Functions in C and C

In programming, arrays are essential data structures for storing collections of data. When dealing with multidimensional arrays, it's important to understand how they can be passed to functions to perform operations on their elements.

Consider the following C code:

void print(int *arr[], int s1, int s2) {
    // ...
}

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

This code aims to pass a multidimensional array a of type int[4][4] to a function print that expects an array of pointers to integers (int **). Surprisingly, this code compiles successfully in C but not in C .

Why C Raises an Error

In C , the error is clear:

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

This error signifies that C does not allow implicit conversion from a multidimensional array (int[4][4]) to an array of pointers to integers (int **). This is because these two data types are fundamentally different in structure and interpretation.

How to Fix the Error

To fix this error, you can use a technique called "array address decay" in C or explicitly convert the multidimensional array to an array of pointers in C .

C:

int main() {
    int a[4][4] = {{0}};
    print(a, 4); // Pass the base address of the multidimensional array (array address decay)
}

void print(int *arr, int s) {
    // ...
}

C :

int main() {
    int a[4][4] = {{0}};
    print(a, 4, 4); // Explicit conversion to an array of pointers
}

void print(int **arr, int s1, int s2) {
    // ...
}

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