c++ - Different machine code for empty default constructor v. implicitly-defined one

Question

Given the following struct...

#include <type_traits>

struct C {
    long a[16]{};
    long b[16]{};

    C() = default;
};

// For godbolt
C construct() {
    static_assert(not std::is_trivial_v<C>);
    static_assert(std::is_standard_layout_v<C>);

    C c;
    return c;
}

...gcc (version 10.2 on x86-64 Linux) with enabled optimization (at all 3 levels) produces the following assembly^[1] for construct:

construct():
        mov     r8, rdi
        xor     eax, eax
        mov     ecx, 32
        rep stosq
        mov     rax, r8
        ret

Once I provide empty default constructor...

#include <type_traits>

struct C {
    long a[16]{};
    long b[16]{};

    C() {}  // <-- The only change
};

// For godbolt
C construct() {
    static_assert(not std::is_trivial_v<C>);
    static_assert(std::is_standard_layout_v<C>);

    C c;
    return c;
}

...generated assembly changes to initializing every field individually instead of single memset in the original:

construct():
        mov     rdx, rdi
        mov     eax, 0
        mov     ecx, 16
        rep stosq
        lea     rdi, [rdx+128]
        mov     ecx, 16
        rep stosq
        mov     rax, rdx
        ret

Apparently, both structs are equivalent in terms of not being trivial, but being standard layout. Is it just gcc missing an optimization opportunity, or is there more to it from the C++-the-language perspective?

---

The example is a stripped down version of production code where this did have material difference in performance.

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[1] Godbolt: https://godbolt.org/z/8n1Mae

question from:https://stackoverflow.com/questions/65871288/different-machine-code-for-empty-default-constructor-v-implicitly-defined-one

Answer 1

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