通过std算法。模板推演

Question

我试着让这个小例子起作用。

#include <iostream>
#include <vector>
#include <algorithm>

int main() {
    std::vector<char> v{'a', 'b', 'c', 'd'};
    std::transform(v.begin(), v.end(), v.begin(), std::toupper);
    std::for_each(v.begin(), v.end(), [](char & c) {
                    std::cout << c << " ";
                  });
}

我想很明显，我想要实现的是什么。然而，我得到了这些错误

    ||=== Build file: "no target" in "no project" (compiler: unknown) ===|
C:\Users\vitaly.bushaev\Documents\vk cup\1.cpp||In function 'int main()':|
C:\Users\vitaly.bushaev\Documents\vk cup\1.cpp|8|error: no matching function for call to 'transform(std::vector<char>::iterator, std::vector<char>::iterator, std::vector<char>::iterator, <unresolved overloaded function type>)'|
C:\Users\vitaly.bushaev\Documents\vk cup\1.cpp|8|note: candidates are:|
C:\Program Files (x86)\CodeBlocks\MinGW\lib\gcc\mingw32\4.9.2\include\c++\bits\stl_algo.h|4152|note: template<class _IIter, class _OIter, class _UnaryOperation> _OIter std::transform(_IIter, _IIter, _OIter, _UnaryOperation)|
C:\Program Files (x86)\CodeBlocks\MinGW\lib\gcc\mingw32\4.9.2\include\c++\bits\stl_algo.h|4152|note:   template argument deduction/substitution failed:|
C:\Users\vitaly.bushaev\Documents\vk cup\1.cpp|8|note:   couldn't deduce template parameter '_UnaryOperation'|
C:\Program Files (x86)\CodeBlocks\MinGW\lib\gcc\mingw32\4.9.2\include\c++\bits\stl_algo.h|4189|note: template<class _IIter1, class _IIter2, class _OIter, class _BinaryOperation> _OIter std::transform(_IIter1, _IIter1, _IIter2, _OIter, _BinaryOperation)|
C:\Program Files (x86)\CodeBlocks\MinGW\lib\gcc\mingw32\4.9.2\include\c++\bits\stl_algo.h|4189|note:   template argument deduction/substitution failed:|
C:\Users\vitaly.bushaev\Documents\vk cup\1.cpp|8|note:   candidate expects 5 arguments, 4 provided|
||=== Build failed: 1 error(s), 0 warning(s) (0 minute(s), 0 second(s)) ===|

所以我正在读这篇文章，我理解这个错误，但我不明白为什么会发生这种情况。当然，这可以很好地工作：

#include <iostream>
#include <vector>
#include <algorithm>

int main() {
    std::vector<char> v{'a', 'b', 'c', 'd'};
    using it = std::vector<char>::iterator;
    std::transform<it, it, int (int)>(v.begin(), v.end(), v.begin(), std::toupper);
    std::for_each(v.begin(), v.end(), [](char & c) {
                    std::cout << c << " ";
                  });
}

所以问题来了：

1)为什么c++不能在这里推导模板参数？

2)有没有办法绕过这一点，不指定类型？

Answer 1

std::toupper有多个重载。

模板推导发生在重载选择之前，因此您必须指定或将调用包装在lambda或函数对象中。

#include <iostream>
#include <vector>
#include <algorithm>



int main() {
    std::vector<char> v{'a', 'b', 'c', 'd'};

    // method 1
    struct make_upper
    {
        int operator()(int ch) const { return std::toupper(ch); }
    };
    std::transform(v.begin(), v.end(), v.begin(), make_upper());

    // method 2
    std::transform(v.begin(), v.end(), v.begin(), [](auto&& ch) { return std::toupper(ch); });
}

参考文献：

http://en.cppreference.com/w/cpp/locale/toupper http://en.cppreference.com/w/cpp/string/byte/toupper