Why Does `sizeof()` Return Unexpected Results for Array Parameters in C Functions?

Why sizeof() Doesn't Work the Same for Array Parameters in C Functions

In C , arrays decay into pointers when passed to functions, rendering the use of sizeof() unreliable for determining array size. To understand this, let's analyze the following function:

int length_of_array(int some_list[])
{
    return sizeof(some_list) / sizeof(*some_list);
}

Issue with the Array Parameter

The parameter some_list is declared as an array, but the function signature is effectively equivalent to int length_of_array(int* some_list). This is because arrays decay into pointers in function parameters.

Effect on sizeof()

Using sizeof(some_list) in this context calculates the size of the pointer, resulting in a value of 1. Dividing by sizeof(*some_list) (the size of an integer) yields 1.

Example

In the given example, despite the array num_list containing 15 elements, the function length_of_array() consistently returns 1, as seen in the output:

This is the output from direct coding in the
int main function:
15
This is the length of the array determined
by the length_of_array function:
1

Solution Using Template Functions

To determine array size in functions, one can use template functions and pass the array by reference:

template<size_t N>
int length_of_array(int (&amp;arr)[N])
{
    return N;
}

In this case, the template parameter N captures the known size of the array, allowing sizeof() to return the correct value.

The above is the detailed content of Why Does `sizeof()` Return Unexpected Results for Array Parameters in C Functions?. For more information, please follow other related articles on the PHP Chinese website!