Can the compiler elide a const local copy of const& vector parameter?

Question

Consider these two functions:

int foo(std::array<int, 10> const& v) {
    auto const w = v;
    int s{};
    for (int i = 0; i < v.size(); ++i) {
        s += (i % 2 == 0 ? v : w)[i];
    }
    return s;
}

int bar(std::array<int, 10> const& v) {
    return std::accumulate(v.begin(),
                           v.end(),
                           0,
                           std::plus{});
}

They do exactly the same thing, if not for the fact that foo does it awkwardly, by making a const copy of argument v, and then picking elements alternatively from v and from the copy w.

Clang can see through that, and generates the same assembly for the two functions (GCC and MSVC miss this optimization).

If I change std::array<int, 10> to std::vector<int> throughout, though, the two assembly outputs differ; specifically, I can see the calls to new+memcpy and delete performed in foo, corresponding to w initialization at auto const w = v; and destruction at } respectively.

I assume that difference in the assembly maps to foo and bar behaving differently exception-wise, in that the allocation caused by auto const w = v; could fail in foo, whereas there's no allocation going on in bar.

However, is the compiler required to emit code for foo that is different from bar's?

Or could the compiler do as it does for the std::array case, i.e. work out that "if the allocation at auto const w = v; succeeds, then the execution of the code is the same as if w was changed to v, so I might as well make this simplification and even prevent any exception at all"?

Incidentally, I've also tried another case of "useless copy that does not require allocation",

int foo(std::optional<int> const& v, bool b) {
    auto const w = v;
    return (b ? v : w).value();
}

int bar(std::optional<int> const& v) {
    return v.value();
}

and Clang generates the same assembly (except for using test in foo and cmp in bar to check if the optional has_value, but that's a non-functional difference).

Jarod42 · Accepted Answer · 2025-09-18 09:47:48Z

9

From as-if rule since C++14

new expression has another exception from the as-if rule: the compiler may remove calls to the replaceable allocation functions even if a user-defined replacement is provided and has observable side-effects.

So if the allocator of your vector use a new expression, then the optimization can be done.

Default allocator doesn't use a new expression but the replaceable allocation directly, but according to std::allocate's notes

The "unspecified when and how" wording makes it possible to combine or optimize away heap allocations made by the standard library containers, even though such optimizations are disallowed for direct calls to ::operator new. For example, this is implemented by libc++

So the optimization can be done in that case too.

edited Sep 18 at 9:47

answered Sep 18 at 9:00

Jarod42

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5 Comments

Enlico Sep 18 at 9:11

Side question: why are they called replaceable allocation functions rather than overloadable allocation functions? Is it just equivalent wording?

Jarod42 Sep 18 at 9:13

User might write a function with same signature to replace default implementation by the new one. overloads use different signatures.

Enlico Sep 18 at 9:41

Can you explain a bit more why "Default allocator doesn't use a replaceable allocation function"? If its ::allocate uses ::operator new... isn't that the same as "using a replaceable allocation function"?

Enlico Sep 18 at 9:44

I'll read through the two links at the end of "implemented by libc++".

Jarod42 Sep 18 at 9:49

Fixed wording. new expression (i.e new T) is subject to optimization, but not a direct call to operator new(..).

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