C ++ 11“自动”语义

Question

当我使用C ++ 11时auto，关于类型归纳将解析为值还是引用的规则是什么？

例如，有时很明显：

auto i = v.begin(); // Copy, begin() returns an iterator by value

这些不太清楚：

const std::shared_ptr<Foo>& get_foo();

auto p = get_foo(); // Copy or reference?

static std::shared_ptr<Foo> s_foo;

auto sp = s_foo; // Copy or reference?

std::vector<std::shared_ptr<Foo>> c;

for (auto foo: c) { // Copy for every loop iteration?

Answer 1

规则很简单：这就是您声明的方式。int i = 5;auto a1 = i;&nbsp; &nbsp; // valueauto & a2 = i;&nbsp; // reference下一个例子证明了这一点：#include <typeinfo>#include <iostream>&nbsp; &nbsp;&nbsp;template< typename T >struct A{&nbsp; &nbsp; static void foo(){ std::cout<< "value" << std::endl; }};template< typename T >struct A< T&>{&nbsp; &nbsp; static void foo(){ std::cout<< "reference" << std::endl; }};float& bar(){&nbsp; &nbsp; static float t=5.5;&nbsp; &nbsp; return t;}int main(){&nbsp; &nbsp; int i = 5;&nbsp; &nbsp; int &r = i;&nbsp; &nbsp; auto a1 = i;&nbsp; &nbsp; auto a2 = r;&nbsp; &nbsp; auto a3 = bar();&nbsp; &nbsp; A<decltype(i)>::foo();&nbsp; &nbsp; &nbsp; &nbsp;// value&nbsp; &nbsp; A<decltype(r)>::foo();&nbsp; &nbsp; &nbsp; &nbsp;// reference&nbsp; &nbsp; A<decltype(a1)>::foo();&nbsp; &nbsp; &nbsp; // value&nbsp; &nbsp; A<decltype(a2)>::foo();&nbsp; &nbsp; &nbsp; // value&nbsp; &nbsp; A<decltype(bar())>::foo();&nbsp; &nbsp;// reference&nbsp; &nbsp; A<decltype(a3)>::foo();&nbsp; &nbsp; &nbsp; // value}输出：valuereferencevaluevaluereferencevalue

Answer 2

无论您从右侧（等于“ =”）获得什么，都永远不会成为参考。更具体地说，表达式的结果永远不会是引用。因此，请注意示例中结果之间的差异。#include <typeinfo>#include <iostream>template< typename T >struct A{&nbsp; &nbsp; static void foo(){ std::cout<< "value" << std::endl; }};template< typename T >struct A< T&>{&nbsp; &nbsp; static void foo(){ std::cout<< "reference" << std::endl; }};float& bar(){&nbsp; &nbsp; static float t=5.5;&nbsp; &nbsp; return t;}int main(){&nbsp; &nbsp;auto a3 = bar();&nbsp; &nbsp;A<decltype(bar())>::foo(); // reference&nbsp; &nbsp;A<decltype(a3)>::foo();&nbsp; &nbsp; // value}