问题
I was told here that:
The order of generate is not guaranteed => depending on the implementation
I have looked up gcc's implementation of generate:
for (; __first != __last; ++__first)
*__first = __gen();
And Visual Studio implements it identically to that. This is a relief to me as using a lambda in generate that reads and writes to a capture could have undeterministic results:
int foo[] = {1, 0, 13};
vector<int> bar(3);
generate(bar.begin(), bar.end(), [&]() {
static auto i = 0;
static auto total = 0;
total += foo[i];
return foo[i] / total;
});
I expect bar to contain {1, 0, 0}.
If I am allowed to execute out of order this could even cause a divide by 0 error.
So I'd sleep easier if this question could be answered with proof that generate is required to execute sequentially.
As a note here, I know that experimental::parallel::generate will not be sequential. I'm just asking about generate.
回答1:
I was intrigued by this so have done some research.
At the bottom of this answer is a copy of the relevant section from the standard as of 2011.
You will see from the template declaration of std::generate<> that the iterator parameters must conform to the concept of a ForwardIterator and OutputIterator respectively.
A ForwardIterator does not support random access. it can only read or write sequentially. An OutputIterator is even more restrictive - its operator* implicitly includes the effect of an operator++. This is an explicit feature of the concept.
Therefore, by implication, the implementation of this function must access elements sequentially (and therefore generate values sequentially) since to not do so would break the contract implicit in the interface.
Therefore the standard does (implicitly and quietly) guarantee that std::generate initialise its elements sequentially. It would be impossible to write a well-formed implementation of std::generate that did not.
QED
25.3.7 Generate [alg.generate]
template<class ForwardIterator, class Generator>
void generate(ForwardIterator first, ForwardIterator last,
Generator gen);
template<class OutputIterator, class Size, class Generator>
OutputIterator generate_n(OutputIterator first, Size n, Generator gen);
1 Effects: The first algorithm invokes the function object gen and assigns the return value of gen through all the iterators in the range [first,last). The second algorithm invokes the function object gen and assigns the return value of gen through all the iterators in the range [first,first + n) if n is positive, otherwise it does nothing.
2 Requires: gen takes no arguments, Size shall be convertible to an integral type (4.7, 12.3).
3 Returns: generate_n returns first + n for non-negative values of n and first for negative values.
4 Complexity: Exactly last - first, n, or 0 invocations of gen and assignments, respectively.
回答2:
Here is all the standard says about it (25.7.3/1):
template<class ForwardIterator, class Generator>
void generate(ForwardIterator first, ForwardIterator last, Generator gen);
template<class OutputIterator, class Size, class Generator>
OutputIterator generate_n(OutputIterator first, Size n, Generator gen);
The first algorithm invokes the function object gen and assigns the return value of gen through all the iterators in the range [first,last) . The second algorithm invokes the function object gen and assigns the return value of gen through all the iterators in the range [first,first + n) if n is positive, otherwise it does nothing.
As you can see, it is not explicitly stated that the iterator assignments must be done sequentially.
来源：https://stackoverflow.com/questions/33283275/is-generate-guaranteed-to-be-executed-sequentially