Question

Possible Duplicate:
How to find the sizeof(a pointer pointing to an array)

I have the following code

int myfunc(char *arr)
{
const int sizearr = sizeof( arr ) / sizeof(char);
}

This yields the size of the pointer, not the size of the char array[17] I passed to the function.

I know that I may use strlen and solve the problem, but is there any way to convert the char*arr pointer back to the char array[17] I passed to the function in order to use the sizeof(char array[17]) ?

Answer 1

If you want to pass an array of 17 chars, you need to pass it by reference (C++ only):

int myfunc(char (&arr)[17])
{
    unsigned int const sizearr = sizeof arr;
    // ...
}

Or make a template to deduce the size:

template <unsigned int sizearr>
int myfunc(char (&arr)[sizearr])
{
    // ...
}

In C, you cannot pass arrays as arguments at all, so it is your responsibility to communicate the array size separately.

Update: As @David suggests, you can fix the array type in C by passing a pointer to the array:

int myfunc(char (*arr)[17])
{
    unsigned int const sizearr = sizeof *arr;
    // ...
}

int main()
{
    char s[17];
    return myfunc(&s);
}