Force function called only with certain types

I was looking at security when using char * for bool in C ++ 11 and it has been suggested that if you do

template<typename T>
void foo(T) = delete;

void foo(int f) {}

      

        
        
        
      

    

Something that foo

will only work with an explicit argument given int

. I made a test case:

template<typename T>
void foo(T) = delete;

void foo(int f) {}

int main()
{
    foo(1);
    foo(3.0);
    foo(short(5));
    foo(float(7.0));
    foo(long(9));
}

      

        
        
        
      

    

I used coliru to compile the code from g++ -std=c++14 -O2 -Wall -pedantic -pthread main.cpp && ./a.out

( live example ) and I got the following errors:

main.cpp: In function 'int main()':
main.cpp:9:12: error: use of deleted function 'void foo(T) [with T = double]'
     foo(3.0);
            ^
main.cpp:2:6: note: declared here
 void foo(T) = delete;
      ^
main.cpp:10:17: error: use of deleted function 'void foo(T) [with T = short int]'
     foo(short(5));
                 ^
main.cpp:2:6: note: declared here
 void foo(T) = delete;
      ^
main.cpp:11:19: error: use of deleted function 'void foo(T) [with T = float]'
     foo(float(7.0));
                   ^
main.cpp:2:6: note: declared here
 void foo(T) = delete;
      ^
main.cpp:12:16: error: use of deleted function 'void foo(T) [with T = long int]'
     foo(long(9));
                ^
main.cpp:2:6: note: declared here
 void foo(T) = delete;
      ^

      

        
        
        
      

    

^{compiling with clang also produced similar errors}

Now that I was reading about = delete

on cppreference it stated

If the function is overloaded, the overload resolution occurs first, and the program is only ill-formed if the removed function is selected.

So if cppreference is correct and my program is poorly formed, does that mean it won't compile or undefined or undefined behavior?