Candidate template ignored: could not match 'const type-parameter-0-0 *' against 'char'

Question

I want to make my own library and I have some problems with template function's.

main.cpp

#include <iostream>
#include "SMKLibrary.h"

int main() {
    char a[5] = {"ASFD"};

    array_print(a,5);

    return 0;
}

SMKLibrary.h

#ifndef SMKLIBRARY_H
#define SMKLIBRARY_H

#include <iostream>

template <typename T>
void array_print(const T * array[], int size);

#endif

SMKLibrary.cpp

#include "SMKLibrary.h"

template <typename T>
void array_print(const T * array[], int size) {
    int last = size - 1;
    for (int i = 0; i < last; i++) {
        std::cout << array[i] << " ";
    }
    std::cout << array[last] << std::endl;
}

Can someone explain to me why I have this error?

Solution 1

void array_print(const T * array[], int size);

Request a pointer to an array. When the compiler looks at how the function is called it sees a which is an array not a pointer to an array. The types do not match so the template deduction fails. To fix this drop the * from the function so you have

void array_print(const T array[], int size);

Solution 2

You can make function deduce array size from its argument:

template <typename T, std::size_t size>
void array_print(T(&array)[size]) {
    int last = size - 1;
    for (int i = 0; i < last; i++) {
        std::cout << array[i] << " ";
    }
    std::cout << array[last] << std::endl;
}

int main()
{
    char a[5] = {"ASFD"};
    array_print(a);
}

also the next problem you will encounter after fixing compilation error, is linker error. As πάντα ῥεῖ said in comments, you need to move your function definition to header file.