CODEWITHSUNDEEP
c++
Young-hwi
Young-hwi

Reputation: 637

Conversion from const char * to const std::string &

I think following code should generate an error:

#include <iostream>
#include <string>

static void pr(const std::string &aStr)
{
    std::cout << aStr << "\n";
}

int main(void)
{
    const char *a = "Hellu";

    pr(a);

    return 0;
}

But gcc 4.1.2 compiles it successuflly.

Is it that the constructor of std::string get in the way, creating an instance of std::string?

I believe it shouldn't, because reference is merely an alias to a variable (in this case, there is no variable of type std::string that the reference is referring to).

Is anybody out there to explain why the code compiles successfully?

Thanks in advance.

Upvotes: 8

Views: 5341

Answers (2)

Frederick Roth
Frederick Roth

Reputation: 2830

What you encounter is implicit conversion.

Here is a quote from the C++ Standard (SC22-N-4411.pdf)

1 Type conversions of class objects can be specified by constructors and by conversion functions. These conversions are called user-defined conversions and are used for implicit type conversions (Clause 4), for initialization (8.5), and for explicit type conversions (5.4, 5.2.9).

So the compiler just works as intended and calls the std::string constructor you mentioned.

Upvotes: 3

Jerry Coffin
Jerry Coffin

Reputation: 490118

Yes, given a reference to a constant, the compiler can/will synthesize a temporary (in this case of type std::string) and bind the reference to that temporary.

If the reference was not to a const object, however, that wouldn't work -- only a reference to const can bind to a temporary object like that (though at least widely used compiler allows a non-const reference to bind to a reference as well).

Upvotes: 13

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