模板功能给出“呼叫无匹配功能".错误 [英] Template function gives "no matching function for call" error
问题描述
关于stackoverflow的第一个问题:)我对C ++相对较新，并且从未使用过模板，因此，如果我做一些愚蠢的事情，请原谅我.我有一个模板功能，可以梳理列表并检查常规类型的指定元素.这样，我可以指定是要查找字符串还是int或其他内容.
First question on stackoverflow :) I'm relatively new to C++, and have never used templates, so forgive me if I'm doing something silly. I have a template function that combs through a list and checks for a specified element of a general type. That way I can specify whether it's looking for a string, or an int, or whatever.
template <class T>
bool inList(T match, std::string list)
{
int listlen = sizeof(list);
for (int i = 0; i <= listlen; i++) {
if (list[i] == match) return true;
else continue;
}
return false;
};
这是我对 inList()的调用. testvec 是一个包含一些元素的字符串向量，包括"test":
This is my call to inList(). testvec is a string vector with a few elements, including "test":
if (inList<string>("test", testvec))
cout << "success!";
else cout << "fail :(";
令我沮丧和困惑的是，在编译时，我被以下错误打了掌:
To my dismay and confusion, upon compiling, I am slapped with the following error:
error: no matching function for call to 'inList(const char [5], std::vector<std::basic_string<char> >&)'
我做错了什么?:(
我忽略了提及模板定义在全局名称空间中.(这是一个简单的测试程序，用于查看我的模板是否可以正常工作，而显然::)
I neglected to mention that the template definition is in the global namespace. (it's a simple test program to see if my template will work, and it does not, apparently :( )
推荐答案
这是因为无法将std :: vector转换为std :: string.相反，您需要做的是对集合概念进行抽象处理.
That's because there is no way to convert from a std::vector to a std::string. Instead, what you need to do is abstract over the concept of a collection.
您这样做是为了使人们可以传递他们想要的任何集合类型.他们可以使用数组，向量，字符串，列表，双端队列...，只要它适合集合的概念"(这就是希望c ++ 1x附带概念！).他们甚至可以使用自己的内部特别优化的收藏类型.这就是模板的美.
You do this so that people can pass in any collection type they want. They can use an array, a vector, a string, a list, a deque... as long as it fits the 'concept' of a collection (here's to hoping c++1x comes with concepts!). They can even use their own in-house specially optimized collection types. That's the beauty of templates.
使用C ++ 11(可与任何标准集合，原始数组和用户定义的类型一起使用):
Using C++11 (works with any standard collection, primitive arrays, and user-defined types):
template<class elem_t, class list_t>
bool in_list(const elem_t& elem, const list_t& list) {
for (const auto& i : list) {
if (elem == i) {
return true;
}
}
return false;
}
这是将std :: initializer_list推导为模板参数的非标准扩展，因此请提供显式覆盖:
It's a non-standard extension to deduce std::initializer_list as a template argument, so provide an explicit override:
template<class elem_t>
bool in_list(const elem_t& elem, std::initializer_list<elem_t> list) {
for (const auto& i : list) {
if (elem == i) {
return true;
}
}
return false;
}
使用此版本，您可以将其命名为:
With this version, you can call it like:
int main() {
std::vector<int> a = {1, 2, 3, 4, 5};
std::cout << in_list(3, a) << std::endl;
std::string b = "asdfg";
std::cout << in_list('d', b) << std::endl;
std::cout << in_list('d', "asdfg") << std::endl;
std::cout << in_list(3, {1, 2, 3, 4, 5}) << std::endl;
return 0;
}
对于仍然使用C ++ 98的我们来说，这将适用于字符串和向量以及某些用户定义的类型.不过，它不适用于原始数组.
And for those of us still in C++98, this will work for both strings and vectors, and some user-defined types. It will not work with raw arrays though.
template<class elem_t, class list_t>
bool in_list_98(const elem_t& elem, const list_t& list) {
list_t::const_iterator end = list.end(); //prevent recomputation of end each iteration
for (list_t::const_iterator i = list.begin(); i < end; ++i) {
if (elem == *i) {
return true;
}
}
return false;
}
或者，您可以采用STL样式:
Or, you can go STL style:
template<class elem_t, class iterator_t>
bool in_list_stl(const elem_t& elem, iterator_t begin, iterator_t end) {
for (iterator_t i = begin; i < end; ++i) {
if (elem == *i) {
return true;
}
}
return false;
}
//call like std::string s = "asdf"; in_list_stl('s', s.begin(), s.end());
如果我犯了一个错误，对不起，我现在没有正在运行我的编译器...
If I made a mistake, sorry, I don't have my compiler running right now...
这篇关于模板功能给出“呼叫无匹配功能".错误的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
