Improved method to obtain the size of an array

Question

I have made a function that recieves a pointer of an array of characters, that represent an hexadecimal number, and returns the decimal value of it.

In that function, this is what I currently use to obtain the size of the array

size_t iSize = 0;
while (s[iSize] != '\0'){   iSize++;    }

I am looking to find a better method to do this. I already tried int `iSize = sizeof(s)/sizeof(s[0]).

Possible duplicate of [Convert a hexadecimal string to an integer efficiently in C?](http://stackoverflow.com/questions/10324/convert-a-hexadecimal-string-to-an-integer-efficiently-in-c) — hookenz, Oct 30 '16 at 19:57
To find the length of a null terminated string use `strlen`. Based on your description that's what you care about, not the size of the array. — Retired Ninja, Oct 30 '16 at 19:57
Have you thought about replacing your whole function that converts a hex string to a number with one function `strtol`? — hookenz, Oct 30 '16 at 19:58

score 0 · Answer 1 · answered Oct 31 '16 at 03:11

Size of array is not passed through a function. Example:

int my_getsize(int* ptr)
{
    return sizeof(ptr);
}

int main()
{
    int arr[100];
    printf("%d\n", sizeof(arr)); //output: 400
    printf("%d\n", my_getsize(arr)); //output: 4
}

In a 32bit system my_getsize(buf) returns 4, which is the size of pointer. While sizeof(buf) returns 400 which is 100 * sizeof(int) (also 4). The only way to around this is to pass the array size to the function. For example foo(arr, sizeof(arr)/sizeof(arr[0]))

Character arrays are a special case because the string is null-terminated, you can look for the last zero in the array. Note however this will not return the array size which includes the null-terminator. It's the same as strlen. Example:

size_t my_strlen(char* ptr)
{
    size_t iSize = 0;
    while (*ptr++) iSize++;
    return iSize;
}

For char buf[] = "12345", sizeof(buf) is 6, while strlen(buf) is 5. Usually you need not worry about optimization with this sort of simple functions because the compiler does it for you.

Improved method to obtain the size of an array

1 Answers1