C2664: Explicit cast operator do not cast as expected

Question

My test class XString has two cast operators. But the compiler do not use the explicit cast operator const wchar_t*() for fooA. Why?

class XString
{
public:
    operator const CString&();
    explicit operator const wchar_t*();
};

void fooA(const wchar_t* s);
void fooB(const CString& s);

void test()
{
    XString x;

    CString c = x; //OK

    fooA(x); //Error C2664: 'void fooA(const wchar_t *)': cannot convert argument 1 from 'XString' to 'const wchar_t *'

    fooB(x); //OK
}

Because you made it explicit? You need to explicitly cast, with, e.g., `static_cast`. — Holt, Jan 24 '19 at 08:36
The entire point of `explicit` is to disallow implicit conversion. — molbdnilo, Jan 24 '19 at 08:36
x isn't a pointer and fooA expects the address of x or a pointer. — SPlatten, Jan 24 '19 at 08:39
Note that these are **conversion** operators, not **cast** operators. A cast is something you write in your source code to tell the compiler to do a conversion. — Pete Becker, Jan 24 '19 at 15:49

score 3 · Accepted Answer · answered Jan 24 '19 at 08:36

3

Since operator const wchar_t* is explicit, the conversion will not be done implicitly. That is the point of explicit.

You can force the conversion using static_cast:

fooA(static_cast<const wchar_t*>(x));

answered Jan 24 '19 at 08:36

asu

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C2664: Explicit cast operator do not cast as expected

1 Answers1