Length of array in function argument

Question

This is well known code to compute array length in C:

sizeof(array)/sizeof(type)

But I can\'t seem to find out the length of the array pass

Answer 1

This is a old question, and the OP seems to mix C++ and C in his intends/examples. In C, when you pass a array to a function, it's decayed to pointer. So, there is no way to pass the array size except by using a second argument in your function that stores the array size:

void func(int A[]) 
// should be instead: void func(int * A, const size_t elemCountInA)

They are very few cases, where you don't need this, like when you're using multidimensional arrays:

void func(int A[3][whatever here]) // That's almost as if read "int* A[3]"

Using the array notation in a function signature is still useful, for the developer, as it might be an help to tell how many elements your functions expects. For example:

void vec_add(float out[3], float in0[3], float in1[3])

is easier to understand than this one (although, nothing prevent accessing the 4th element in the function in both functions):

void vec_add(float * out, float * in0, float * in1)

If you were to use C++, then you can actually capture the array size and get what you expect:

template 
void vec_add(float (&out)[N], float (&in0)[N], float (&in1)[N])
{
    for (size_t i = 0; i < N; i++) 
        out[i] = in0[i] + in1[i];
}

In that case, the compiler will ensure that you're not adding a 4D vector with a 2D vector (which is not possible in C without passing the dimension of each dimension as arguments of the function). There will be as many instance of the vec_add function as the number of dimensions used for your vectors.