Alias for a std::vector that does not take space

[Peter Olcott]

True. But in the example I gave, most compilers would optimize it
away.


You've jumped to an erroneous conclusion. When sizeof is applied to a
reference, it returns the size of the aliased object. Apparently, on
your implementation sizeof(std:vector<int>) is the same as
sizeof(void*). That's just coincidence. Try the following:

#include <iostream>

struct Foo {
double x;
long y;
void* z;
};

int main()
{
using std::cout; using std::endl;
Foo foo;
Foo& bar = foo;
cout << "sizeof foo: " << sizeof(foo) << endl;
cout << "sizeof bar: " << sizeof(bar) << endl;
cout << "sizeof void* " << sizeof(void*) << endl;
}

Please run this program on your system, then report back on the
results. Do you still believe that sizeof a reference must be the same
as sizeof a pointer?

Best regards,

Tom

16, 16, 4 with maximum space optimization turned on.

 

[Squeamizh]

Unless someone comes up with a better way, that is what I will do.

Knock yourself out...