g++ compiler and implicit conversion

Question

I'm using g++ for compiling my C++ program, but I want to stop the implicit conversion between type like int and dooble for example: I have a function that use a double as parameter, but when I send in this function's parameter an int, the compilation pass without error or warning.

so that is my question, how to stop the implicit conversions??

thanks.

Answer 1

You cannot avoid implicit conversion from lower to higher type. However you can do vice-versa if your compiler supports C++0x.

void func(int x){}

int main()
{
   func({2.3}); // error: narrowing
}

Answer 2

You could try this:

#include <iostream>

template<typename T>
void func(T t);

void func(double d)
{
    std::cout << "D:" << d << "\n";
}


int main()
{
    func(2.3);   // OK
    func(2);     // Fails at compile time.
}

Answer 3

I think Martin's answer is the way to go. It can find the conversion at link time. If you have to find at compile time, you can add static_assert or a similar one to the function template:

template<typename T>
void func( T ) {
  //static_assert( sizeof( T ) == 0, "..." ); // if you can use static_assert
  int a[ (sizeof( T ) == 0) ? 1 : -1 ];
}

Hope this helps.