Why Does \"Hello World\" Print as \"True\" in an Overloaded C Print Method?

String Literal Ambiguity in Overloaded Methods

In C , when overloading methods that accept multiple types, it's essential to be aware of potential ambiguity. As illustrated below:

<code class="cpp">class Output
{
public:
    static void Print(bool value)
    {
        std::cout << (value ? "True" : "False");
    }

    static void Print(std::string value)
    {
        std::cout << value;
    }
};</code>

If we attempt to call the Print method with a string literal as follows:

<code class="cpp">Output::Print("Hello World");</code>

Unexpectedly, the output is "True" instead of "Hello World". Why?

String Literals vs. User-Defined Conversions

In C , string literals like "Hello World" are not of type std::string but rather an array of constant characters. However, they can be implicitly converted to a bool value. This conversion, known as a standard conversion sequence, is preferred by the compiler over the user-defined conversion constructor for std::string.

Overload Resolution and Standard Conversions

During overload resolution, the compiler determines the best function to call for each argument. The preference is given to standard conversion sequences over user-defined conversions. In our case, the standard conversion from "Hello World" to bool is considered better than the user-defined conversion to std::string.

How to Avoid Ambiguity

To ensure that the std::string overload is used, we need to explicitly pass an std::string argument:

<code class="cpp">Output::Print(std::string("Hello World"));</code>

This resolves the ambiguity and correctly outputs "Hello World".

Conclusion

Understanding implicit conversions and their impact on overload resolution is crucial in C . By default, standard conversions take precedence over user-defined conversions. Hence, it's essential to be aware of such conversions and make explicit type conversions when necessary to avoid unexpected behavior.

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