Initializing non-const parameter with string literal

问题

So I have this code:

class ConstTest {
public:
    explicit ConstTest(char* name) {}
};

int main() {
    ConstTest t("blarghgh");
}

It obviously compiles, even though I thought that it shouldn't. As string literals in C++ have type const char[], and ConstTest constructor requires a const-less char* — not const char*. And casting a const pointer to a non-const one isn't something usually done by C++ implicitly.

So, where I'm wrong? Why it's compiling? Can I legally modify the dereferenced pointer inside the constructor?!

回答1:

So, where I'm wrong? Why it's compiling?

It is compiling because your compiler is too permissive, and your compiler is too permissive because in C++03 the implicit conversion from a string literal to char* was only deprecated, not invalid.

The rationale was backward compatibility with legacy C APIs. Per paragraph 4.2/2 of the C++03 Standard:

A string literal (2.13.4) that is not a wide string literal can be converted to an rvalue of type “pointer to char”; a wide string literal can be converted to an rvalue of type “pointer to wchar_t”. In either case, the result is a pointer to the first element of the array. This conversion is considered only when there is an explicit appropriate pointer target type, and not when there is a general need to convert from an lvalue to an rvalue. [Note: this conversion is deprecated. See Annex D.]

In C++11, however, the implicit conversion is illegal (the above paragraph has been removed altogether).

Can I legally dereference-and-modify the pointer inside the constructor?!

You can, but you cannot modify the dereferenced object. Doing so would be undefined behavior, since the type of the object is const-qualified.

来源：https://stackoverflow.com/questions/16792240/initializing-non-const-parameter-with-string-literal