CODEWITHSUNDEEP
c++c++11
user3335905
user3335905

Reputation: 11

Error message "deduced conflicting types for parameter 'const T'"

What I'm trying to do:

Write a specialized version of the template from the previous exercise to handle vector<const char*> and a program that uses this specialization.

I wrote the program like this:

template<typename T>
int count(vector<T> tvec, const T &t);

template<>
int count(vector<const char *> tvec, const char *const &s)
{
    int count = 0;
    for (auto c : tvec)
        if (c == s) {
            ++count;
        }
    return count;
}

template<typename T>
int count(vector<T> tvec, const T &t)
{
    int count = 0;
    for (auto c : tvec)
        if (c == t) {
            ++count;
        }
    return count;
}

cout << count(svec, "GUO");

but I get the error that says

deduced conflicting types for parameter ‘const T’ (‘std::basic_string<char>’ and ‘char [4]’)

I want to know how to handle this. and further, in the template function, it seems that an array can be changed to the pointer, why my program cannot handle it?

Upvotes: 1

Views: 3409

Answers (2)

Sebastian Redl
Sebastian Redl

Reputation: 71989

Don't deduce on both parameters, it leads to conflicts. Write this:

template <typename T>
int count(const vector<T>& tvec, const typename vector<T>::value_type& t);

Also, consider overloading instead of specializing. Specializing a function template is pretty much never what you want.

Upvotes: 1

songyuanyao
songyuanyao

Reputation: 172934

Firstly, it seems svec is defined as vector<string>, maybe it should be vector<const char*>;

Secondly, explictly define a var as const char*;

Try this:

vector<const char*> svec;
const char* chars = "GUO";
std::cout<<my_count(svec,chars);

BTW: A variable of type char array(char[]) can be used as type char pointer(char*), but they are different as type, and they are different as a template paremeter.

Upvotes: 0

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