在printf中使用`operator const char *`

Question

struct MyClass
{
    operator const char* ()
    {
        return "hello";
    }
};

int main()
{
    MyClass obj;  
    std::cout << obj;    // ok
    printf("%s\n", obj); // Crash
}

Why does object with operator const char* can not be automatically converted to const char* string in printf for mapping the %s ?

Is it just because there is no type awareness in printf-like functions and %s only expect a array of char with terminal 0?

Answer 1

Why? Because of the limitations on variadic parameters in C++: they have essentially no type, you can think of them as void * (but they're not).

So knowing that, the compiler has no idea you think that should be a string. It could very well need to be an integer, or a double, or another object. Or just itself, which the compiler chooses.

Answer 2

When you call printf with:

printf("%s\n", obj);

the compiler does not use the auto conversion function to convert obj to char const* . obj is passed to printf by value and printf tries to treat that value as though it is char const* . As a consequence, your program exhibits undefined behavior. You'll have to explicitly cast obj to char const* to make your program behave predictably.

printf("%s\n", (char const*)obj);

If you turn on warning levels in your compiler, you'll probably see something to indicate that using obj in that printf call is not right. With g++ -Wall , I get:

socc.cc:16:23: warning: format ‘%s’ expects argument of type ‘char*’, but argument 2 has type ‘MyClass’ [-Wformat=]
     printf("%s\n", obj); // Crash