Re: reference to non-const temporary

Ethan Eade <>
Thu, 27 Jul 2006 12:26:11 CST
Frederick Gotham wrote:

If you pass either:

    (1) non-const reference
    (2) pointer to non-const

to a function, then that implies that the function is going to alter

Logic dictates that you only alter something if there's a reason to.

But C++ does not otherwise disallow you from doing things that have no
logical effect. I can do this:

    auto tmp = make();

even though tmp is guaranteed to disappear just as in


The problem is that I cannot see how to correctly write the "fill"
function shown below so that it can work on 'array's and 'array_slice's
(obviously this is a much simplified version of my problem):

struct array_slice {
     int* x;
     int N;
     array_slice(int* x_, int N_) : x(x_), N(N_) {}

     int size() const { return N; }
     int& operator[](int i) { return x[i]; }

template <int N> struct array {
     int x[N];

     int size() const { return N; }
     int& operator[](int i) { return x[i];}

     array_slice slice(int start, int size) {
    return array_slice(x+start, size);

template <class A> void fill(A& a) {
     for (int i=0; i<a.size(); ++i)
    a[i] = i;

int main()
     array<4> a;

     fill(a); // Works

     fill(a.slice(1,2)); // Fails


- Ethan

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