返回默认构造的对象会阻止 NRVO 吗？

Question

假设我有这样的功能：

using std::vector;

vector<int> build_vector(int n)
{
   if (some_condition(n)) return {};

   vector<int> out;

   for(int x : something())
   {
      out.push_back(x);
   }

   return out;
}

函数开头的return {}是否可以防止NRVO？我很好奇，因为这似乎等同于以下内容：

using std::vector;

vector<int> nrvo_friendly_build_vector(int n)
{
   vector<int> out;

   if (some_condition(n)) return out;

   for(int x : something())
   {
      out.push_back(x);
   }

   return out;
}

但我不清楚在第一种情况下是否允许编译器执行 NRVO。

Answer 1

来自https://en.cppreference.com/w/cpp/language/copy_elision

Under the following circumstances, the compilers are permitted, but not required to omit the copy and move (since C++11) construction of class objects even if the copy/move (since C++11) constructor and the destructor have observable side-effects. The objects are constructed directly into the storage where they would otherwise be copied/moved to. This is an optimization: even when it takes place and the copy/move (since C++11) constructor is not called, it still must be present and accessible (as if no optimization happened at all), otherwise the program is ill-formed:

In a return statement, when the operand is the name of a non-volatile object with automatic storage duration, which isn't a function parameter or a catch clause parameter, and which is of the same class type (ignoring cv-qualification) as the function return type. This variant of copy elision is known as NRVO, "named return value optimization".

...

早期 return 没有限制，因此两个版本都是 NRVO 的候选版本。

Answer 2

As requested by the OP, here is an adapted version of

我实际上也在想同样的事情（特别是因为复制省略不是"required"标准），所以我在在线编译器中通过替换快速测试它std::vector 通过 Widget 结构：

struct Widget
{
    int val = 0;
    Widget()              { printf("default ctor\n"); }
    Widget(const Widget&) { printf("copy ctor\n"); }
    Widget(Widget&&)      { printf("move ctor\n"); }

    Widget& operator=(const Widget&) { printf("copy assign\n"); return *this; }
    Widget& operator=(Widget&&)      { printf("move assign\n"); return *this; }

    ~Widget() { printf("dtor\n"); }

    void method(int)
    {
        printf("method\n");
    }
};

V1 使用 build_vector()：http://coliru.stacked-crooked.com/a/5e55efe46bfe32f5

#include <cstdio>
#include <array>
#include <cstdlib>

using std::array;

struct Widget
{
    int val = 0;
    Widget()              { printf("default ctor\n"); }
    Widget(const Widget&) { printf("copy ctor\n"); }
    Widget(Widget&&)      { printf("move ctor\n"); }

    Widget& operator=(const Widget&) { printf("copy assign\n"); return *this; }
    Widget& operator=(Widget&&)      { printf("move assign\n"); return *this; }

    ~Widget() { printf("dtor\n"); }

    void method(int)
    {
        printf("method\n");
    }
};

bool some_condition(int x)
{
    return (x % 2) == 0;
}

array<int, 3> something()
{
    return {{1,2,3}};
}

Widget build_vector(int n)
{
   if (some_condition(n)) return {};

   Widget out;

   for(int x : something())
   {
      out.method(x);
   }

   return out;
}

int main(int argc, char* argv[])
{
    if (argc != 2)
    {
        return -1;
    }
    const int x = atoi(argv[1]);

    printf("call build_vector\n");
    Widget w = build_vector(x);
    printf("end of call\n");
    return w.val;
}

V1的输出

call build_vector
default ctor
method
method
method
move ctor
dtor
end of call
dtor

V2 使用 nrvo_friendly_build_vector()：http://coliru.stacked-crooked.com/a/51b036c66e993d62

#include <cstdio>
#include <array>
#include <cstdlib>

using std::array;

struct Widget
{
    int val = 0;
    Widget()              { printf("default ctor\n"); }
    Widget(const Widget&) { printf("copy ctor\n"); }
    Widget(Widget&&)      { printf("move ctor\n"); }

    Widget& operator=(const Widget&) { printf("copy assign\n"); return *this; }
    Widget& operator=(Widget&&)      { printf("move assign\n"); return *this; }

    ~Widget() { printf("dtor\n"); }

    void method(int)
    {
        printf("method\n");
    }
};

bool some_condition(int x)
{
    return (x % 2) == 0;
}

array<int, 3> something()
{
    return {{1,2,3}};
}

Widget nrvo_friendly_build_vector(int n)
{
   Widget out;

   if (some_condition(n)) return out;

   for(int x : something())
   {
      out.method(x);
   }

   return out;
}

int main(int argc, char* argv[])
{
    if (argc != 2)
    {
        return -1;
    }
    const int x = atoi(argv[1]);

    printf("call nrvo_friendly_build_vector\n");
    Widget w = nrvo_friendly_build_vector(x);
    printf("end of call\n");
    return w.val;
}

V2的输出

call nrvo_friendly_build_vector
default ctor
method
method
method
end of call
dtor

如您所见，在这种特殊情况下 （some_condition 看不到构造结构的副作用），如果 some_condition() 为假 （至少在 clang 和 gcc 中，使用 -std=c++11 和 -O2，在 Coliru 中）

此外，作为 you have noticed，同样的行为似乎也发生在 -O3。

HTH

ps：在学习复制省略时，您可能会发现 Abseil's ToW #11 很有趣；）

返回默认构造的对象会阻止 NRVO 吗？

Does returning a default constructed object prevent NRVO?

c++

nrvo