Gil*_*esz 8 c++ rvalue-reference c++11
让我们考虑以下代码:
class Widget{
};
int main(){
Widget w;
auto lambda = bind([](Widget&& ref){ return; }, std::move(w));
return 0;
}
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它会触发错误
no match for call to ‘(std::_Bind<main()::<lambda(Widget&&)>(Widget)>) ()’
lambda();
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我的问题是:为什么错误出现了?毕竟,我对rvalue引用进行了显式转换 - 我的意思是std::move(w)
我通过rvalue引用参数 - 我的意思是Widget&& ref
.
这是怎么回事?
而且下面的代码有效,让我更担心的是:
class Widget{
};
int main(){
Widget w;
auto lambda = bind([](Widget& ref){ return; }, std::move(w));
return 0;
}
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5go*_*der 11
如果你写下std::bind
示意图的内容,可能会更清楚.
// C++14, you'll have to write a lot of boilerplate code for C++11
template <typename FuncT, typename ArgT>
auto
bind(FuncT&& func, ArgT&& arg)
{
return
[
f = std::forward<FuncT>(func),
a = std::forward<ArgT>(arg)
]() mutable { return f(a); }; // NB: a is an lvalue here
}
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由于您可以std::bind
多次调用函数对象,因此它不能"使用"捕获的参数,因此它将作为左值引用传递.您自己传递bind
rvalue 这一事实意味着在a
初始化的行上没有复制.
如果您尝试使用bind
上面显示的原理图编译示例,您还将从编译器中获得更有用的错误消息.
main.cxx: In instantiation of ‘bind(FuncT&&, ArgT&&)::<lambda()> mutable [with FuncT = main()::<lambda(Widget&&)>; ArgT = Widget]’:
main.cxx:10:33: required from ‘struct bind(FuncT&&, ArgT&&) [with FuncT = main()::<lambda(Widget&&)>; ArgT = Widget]::<lambda()>’
main.cxx:11:31: required from ‘auto bind(FuncT&&, ArgT&&) [with FuncT = main()::<lambda(Widget&&)>; ArgT = Widget]’
main.cxx:18:59: required from here
main.cxx:11:26: error: no match for call to ‘(main()::<lambda(Widget&&)>) (Widget&)’
]() mutable { return f(a); }; // NB: a is an lvalue here
^
main.cxx:11:26: note: candidate: void (*)(Widget&&) <conversion>
main.cxx:11:26: note: conversion of argument 2 would be ill-formed:
main.cxx:11:26: error: cannot bind ‘Widget’ lvalue to ‘Widget&&’
main.cxx:18:33: note: candidate: main()::<lambda(Widget&&)> <near match>
auto lambda = bind([](Widget&&){ return; }, std::move(w));
^
main.cxx:18:33: note: conversion of argument 1 would be ill-formed:
main.cxx:11:26: error: cannot bind ‘Widget’ lvalue to ‘Widget&&’
]() mutable { return f(a); }; // NB: a is an lvalue here
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