不能使用显式类型的lambda

Question

我有这个代码:

std::function<std::string&(std::string&)> change_str = [](std::string& str){
    return (str = "Hello world!");
};

std::string s;

std::cout << change_str(s) << std::endl;

它没有编译,并说:

main.cpp:8:47: error: no viable conversion from '(lambda at main.cpp:8:60)' to 'std::function<std::string &(std::string &)>'
    std::function<std::string&(std::string&)> change_str = [](std::string& str){
                                              ^            ~~~~~~~~~~~~~~~~~~~~~
/usr/include/c++/v1/functional:1448:5: note: candidate constructor not viable: no known conversion from '(lambda at main.cpp:8:60)' to 'nullptr_t' for 1st argument
    function(nullptr_t) _NOEXCEPT : __f_(0) {}
    ^
/usr/include/c++/v1/functional:1449:5: note: candidate constructor not viable: no known conversion from '(lambda at main.cpp:8:60)' to 'const std::__1::function<std::__1::basic_string<char, std::__1::char_traits<char>, std::__1::allocator<char> > &(std::__1::basic_string<char, std::__1::char_traits<char>, std::__1::allocator<char> > &)> &' for 1st argument
    function(const function&);
    ^
/usr/include/c++/v1/functional:1450:5: note: candidate constructor not viable: no known conversion from '(lambda at main.cpp:8:60)' to 'std::__1::function<std::__1::basic_string<char, std::__1::char_traits<char>, std::__1::allocator<char> > &(std::__1::basic_string<char, std::__1::char_traits<char>, std::__1::allocator<char> > &)> &&' for 1st argument
    function(function&&) _NOEXCEPT;
    ^
/usr/include/c++/v1/functional:1454:41: note: candidate template ignored: disabled by 'enable_if' [with _Fp = (lambda at main.cpp:8:60)]
                                        __callable<_Fp>::value &&
                                        ^
main.cpp:8:60: note: candidate function
    std::function<std::string&(std::string&)> change_str = [](std::string& str){
                                                           ^
1 error generated.

但是,如果我更改std::functionto 的声明auto,那么它的工作原理如下:

auto change_str = ...

为什么显式类型不适用于lambda？

Answer 1

没有返回值的lambda是auto,自动去除外部基准,这样你就不会回来string&,但只string.

只需声明功能为

std::function<std::string&(std::string&)> change_str = 
[](std::string& str) -> string&  ///<--- NOTE THIS
{
    return (str = "Hello world!");
};