假设我有一个类似的功能:
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;
}
但是我不清楚在第一种情况下是否允许编译器执行NRVO。
最佳答案
OP的As requested,这是my comment的改编版
我实际上在想同样的事情(尤其是因为标准不需要“复制省略”),所以我通过用Widget结构替换std::vector
在在线编译器中对其进行了快速测试:
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");
}
};
使用
build_vector()
的 V1 :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
使用
nrvo_friendly_build_vector()
的 V2 :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false(至少在clang和gcc中,使用Coliru中的-std=c++11
和-O2
,则V1调用move构造函数)此外,与you have noticed一样,似乎在
-O3
中也会发生相同的行为。高温超导
ps:在学习复制省略时,您可能会发现Abseil's ToW #11很有趣;)
关于c++ - 返回默认构造对象是否可以防止NRVO?,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/53823211/