Why Can't I Pass a Multidimensional Array to a Function as an Array of Pointers in C ?

Understanding the Inability to Pass Multidimensional Arrays to Functions

Consider the following code snippet:

#include <stdio.h>
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(a, 4, 4);
}

This code aims to pass a multidimensional array a to a function print that takes an array of int pointers int *arr[]. However, in C , this code generates the following error:

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

Explanation:

The crux of this issue lies in the incompatible data types. A C-style multidimensional array like a is not directly convertible to a pointer to an array of pointers int *arr[]. In C , the data type of a is int[4][4], which represents a two-dimensional array of integers. On the other hand, int *arr[] is a pointer to an array of integers, where each element is a pointer to an integer.

Solution:

To resolve this issue and pass a multidimensional array to a function taking an array of int pointers, you can modify the code to explicitly convert the multidimensional array to an array of pointers. This can be achieved by utilizing a technique like the one described in the accepted answer to the question "Converting multidimensional arrays to pointers in C ".

Additional Notes:

• The code presented in the initial example exhibits unexpected behavior in C due to an additional bug in the printing code.
• It's crucial to initialize the array a with non-zero values to observe the effects of the bug in array passing.
• The error in array passing is masked in C because of the printing bug.
• C error messages provide a clear indication of the conversion issue, while C compilers may issue warnings that can be mistakenly ignored.

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