问题描述
我在尝试做一些像
std::cout << std::vector<int>{1,2,3};
说的是
cannot bind 'std::ostream {aka std::basic_ostream<char>}' lvalue to 'std::basic_ostream<char>&&'
int main() { std::cout << std::vector<int>{1,2,3}; }
(使用 gcc-4.8.1 和 -std=c++11 测试)
(tested using gcc-4.8.1 with -std=c++11)
SO 有类似的问题,例如 重载运算符<<:无法绑定左值到'std::basic_ostream<char>&&',它是关于一些用户定义的带有嵌套类的类.还有一个解决该问题的公认答案的工作.
SO has similar questions like Overloading operator<<: cannot bind lvalue to ‘std::basic_ostream<char>&&’, which is about some user defined class with nested classes. There is also a work around the accepted answer to that question.
但我不知道这是否适用于 std::vector
.有人可以解释为什么 std::vector
会发生这个错误,以及如何解释它?
But I don't know whether this applies to std::vector
. Can someone explain why this error happens to std::vector
, and how to interpret it?
谢谢
推荐答案
与模板相关的错误消息有时会令人困惑.问题是标准库没有定义 operator <<
的重载以将 std::vector
(或任何其他容器,就此而言)插入std::ostream
.因此,编译器无法为 operator <<
找到合适的重载,并尽其所能报告此失败(不幸的是,这在您的情况下不太好/不太可读).
Template-related error messages can be confusing at times. The problem is that the standard library does not define an overload of operator <<
for inserting std::vector
(or any other container, for that matter) into a std::ostream
. So the compiler fails to find a suitable overload for operator <<
, and reports this failure as best as it's able (which is unfortunately not too good/readable in your case).
如果要流式传输整个容器,可以使用 std::ostream_iterator
为此:
If you want to stream an entire container, you can use std::ostream_iterator
for that:
auto v = std::vector<int>{1, 2, 3};
std::copy(begin(v), end(v), std::ostream_iterator<int>(std::cout, " "));
<小时>
至于为什么你会得到这个神秘的错误,它有助于分析完整的错误消息:
As for why you're getting precisely this cryptic error, it helps to analyse the full error message:
prog.cpp: In function ‘int main()’:
prog.cpp:13:37: error: cannot bind ‘std::ostream {aka std::basic_ostream<char>}’ lvalue to ‘std::basic_ostream<char>&&’
std::cout << std::vector<int>{1,2,3};
^
In file included from /usr/include/c++/4.8/iostream:39:0,
from prog.cpp:3:
/usr/include/c++/4.8/ostream:602:5: error: initializing argument 1 of ‘std::basic_ostream<_CharT, _Traits>& std::operator<<(std::basic_ostream<_CharT, _Traits>&&, const _Tp&) [with _CharT = char; _Traits = std::char_traits<char>; _Tp = std::vector<int>]’
operator<<(basic_ostream<_CharT, _Traits>&& __os, const _Tp& __x)
^
显然有一个 operator<<
的模板重载,它采用 std::ostream&&
类型的 lhs 参数和模板化类型的 rhs 参数;它的存在是为了允许插入临时流.由于它是一个模板,它成为您代码中表达式的最佳匹配.然而,std::cout
是一个左值,所以它不能绑定到 std::ostream&&
.因此出现错误.
There is apparently a template overload of operator<<
which takes a lhs argument of type std::ostream&&
and a rhs argument of the templated type; it exists to allow insertion into temporary streams. Since it's a template, it becomes the best match for the expression in your code. However, std::cout
is an lvalue, so it cannot bind to std::ostream&&
. Hence the error.
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