C ++：隐式转换

Question

At first I saw a program on MSDN.

For example:

#include <iostream>

class Money
{
public:
    Money() : amount{ 0.0 } {};
    Money(double _amount) : amount{ _amount } {};

    double amount;
};

void display_balance(const Money balance)
{
    std::cout << "The balance is: " << balance.amount << std::endl;
}

int main(int argc, char* argv[])
{
    Money payable{ 79.99 };

    display_balance(payable);
    display_balance(49.95);
    display_balance(9.99f);

    return 0;
}

The explanation for it is:

On the second call to display_balance, the type of the argument, a double with a value of 49.95, is not the function expects, so a conversion is needed.

There is a conversion from the type of argument— double to Money , and what I don't know about it is why the implicit conversion happened.

Think about it more deeply, we assume that a function needs a type of object as a parameter, and the constructor of the object needs one parameter, when call to the function, whether can provide this parameter.

Answer 1

[C++14: 5.2.2/4]: When a function is called, each parameter (8.3.5) shall be initialized (8.5, 12.8, 12.1) with its corresponding argument. [..]

Money can obviously be initialized from double because you wrote a constructor to do exactly that.

It is, however, possible to prohibit such implicit conversions by adding the explicit keyword to that constructor:

explicit Money(double _amount) : amount{ _amount } {};

Now you'd have to explicitly convert (or "cast") the double to a Money in the function call expression:

display_balance(static_cast<Money>(49.95));

Answer 2

i think this example will help:

int number;
double rate;
number =2;
rate=1.0345;
number=rate;
cout<<number<<endl; //implicit conversion //narrowing conersion //gives 1 //tried to store floating point number in integer data type variable
rate=2 // implicit conversion //widening conversion // value actually assigned is 2.0 //we are not ordering compiler to convert double data type to integer. it is automatically understood by compiler.

number=(int)rate; //explicit conversion //we are ordering compiler here to take value from rate and convert it to integer and assign it to number. //we are specific here.