C++ 如果返回值未分配给变量，则未解析函数重载

我编写了一个模板函数

foo

，它返回一个

std:：string

auto s = foo(...) << std::endl;
std::cout << s << std::endl;

---

auto s=foo（…）您有一个输入错误。换行-

std::cout << (foo(v.begin())) << std::end;

std:：cout
std:：cout
#include <iostream>
#include <string>
#include <iterator>
#include <vector>
#include <type_traits>

// checks if It is an iterator type
template<typename It, typename Base = std::input_iterator_tag>
struct is_iterator : public std::is_base_of<Base, typename std::iterator_traits<It>::iterator_category> {};

// dummy function that takes in iterator
template<typename It, typename = typename std::enable_if_t<is_iterator<It>::value>>
std::string foo(It it) { return "foo it"; }

int main() {
    std::vector<int> v{1};

    // works
    auto s = foo(v.begin());
    std::cout << s << std::endl;

    // doesn't
    std::cout << (foo(v.begin())) << std::end;
}

std::cout << (foo(v.begin())) << std::end;

std::cout << (foo(v.begin())) << std::endl;