Why Does sizeof() Behave Differently When Passing Arrays to Functions in C ?

sizeof() Anomaly When Passing Arrays to Functions in C

In C , the sizeof() operator acts differently when it comes to arrays passed to functions than when they are used in the main function. This discrepancy can be confusing for beginners.

Proper Array Passing

To properly pass an array to a function, it's essential to pass it by reference, using the syntax:

int length_of_array(int (&arr)[N]);

Where:

• arr is the array being passed to the function.
• N is an integer constant representing the size of the array.

Using Sizeof() in Functions

When using sizeof() within the function, the following is true:

sizeof(arr) == N * sizeof(int);

Explanation:

• sizeof(arr) returns the total memory size allocated to the array, which is N times the size of each element (sizeof(int)).

Problem with Incorrect Array Passing

In the provided code snippet, the function length_of_array() inadvertently passes the array by pointer, like this:

int length_of_array(int some_list[]);

This incorrect passing method causes the following issue:

• sizeof(some_list) returns the size of the pointer to the first element in the array.
• sizeof(*some_list) returns the size of a single element in the array.
• The division of these values always results in 1, regardless of the actual array size.

Workaround Using Templates

To overcome this limitation, one can use a template-based approach like this:

template<size_t N>
int length_of_array(int (&amp;arr)[N])
{
    std::cout << N << std::endl; // Will output the correct array size
    return N;
}

Key Point:

The essential difference between the two methods lies in passing arrays by reference versus passing them by pointer. The former preserves the knowledge of the array's size, while the latter does not.

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