c++ - Does the C++11 standard require that two iterations through a constant unordered_container visit elements in the same order?

Question

for (auto&& i : unordered_container)
{ /* ... */ }

for (auto&& i : unordered_container)
{ /* .. */ }

Does the standard require that both of these loops visit elements in the same order (assuming the container is unmodified)?

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My analysis of this question...

I read the standard and as best I can tell the answer is "no"...

Since iterators of containers are forward, there is language that requires a==b imply that ++a==++b for forward iterators. That means two iterations will go through the same path IF they both start in the same place. This reduces the question to a different question of whether the standard requires container.begin() == container.begin(). I couldn't find any language that requires this.

question from:https://stackoverflow.com/questions/25536376/does-the-c11-standard-require-that-two-iterations-through-a-constant-unordered

Answer 1

Containers are required to implement operator==(). That is we can do:

container c;
c == c;

That relation is required to work the same as:

std::distance(a.begin(), a.end()) == std::distance(b.begin(), b.end()) &&
std::equal(a.begin(), a.end(), b.begin());

The important part here is the call to std::equal(). This call requires that two independent calls to container.begin() will produce the same sequence of values. If it didn't, then c == c would be false, and that doesn't make any sense because == is an equivalence relation.

Therefore, my answer is that we can claim that the standard requires that two passes of any container must result in the same ordering. Obviously this requirement breaks if you do anything that changes the container or invalidates iterators.

Citations:

• C++ 2011 Table 96 — Container requirements