sizeof an array passed as function argument [duplicate]

Question

Possible Duplicate:
length of array in function argument

Hi am doing homework and I am completly stumped. We were suppose to get every order of a list an array of integers so I wrote this piece of code, based off of my teacher's pseudocode:

void permute(int v[], int curr,char letters[])
{
    if(curr >= sizeof(v)/sizeof(int))
    {
        checkit(v,letters);
    }
    for(int i = curr; i < sizeof(v)/sizeof(int); i++)
    {
        swap(i,curr,v);
        permute(v,curr + 1,letters);
        swap(v[curr],v[i]);
    }//for
}//permu

The only thing I am not sure of is if sizeof(v)/sizeof(int) is the right way to go.

Answer 1

sizeof(v)/sizeof(int) is not the way to go. Your function is exactly equivalent to:

void permute(int *v, int curr, char *letters)
{
    ...
}

i.e. v is not really an array, it's a pointer. You cannot pass arrays in C or C++.

The solution is one of the following (not exhaustive):

• add an extra argument that explicitly describes the length of the array
• add an extra argument that points at the last element of the array
• use a proper container (e.g. std::vector), which you can call size() on
• the template solution that @sehe suggests

Answer 2

One of my pet peeves: you can get C++ to deduce the array size for you

template <size_t N>
void permute(int (&v)[N], int curr,char letters[])
{
    if(curr >= N)
    {
        checkit(v,letters);
    }
    for(int i = curr; i < N; i++)
    {
        swap(i,curr,v);
        permute(v,curr + 1,letters);
        swap(v[curr],v[i]);
    }//for
}//permu