ctor: Why does 'explicit' prevent assignment construction?

Question

I've got a class ByteArray defined like this:

class ByteArray
{
public:
    explicit ByteArray( unsigned int uiSize = 0 );
    explicit ByteArray( const char * ucSource );
    ByteArray( const ByteArray & other );

    ByteArray & operator=( const char * ucSource );
    ByteArray & operator=( const ByteArray & other );
}

While almost everything works, constructing a ByteArray by assignment doesn't compile.

ByteArray ba1( 5 );     // works
ByteArray ba2( ba1 );   // works
ByteArray ba3( "ABC" ); // works
ByteArray ba4;          // works
ba4 = "ABC";            // works
ByteArray ba5 = "ABC";  // <<<----- doesn't compile!

The compiler gives me a Cannot convert 'const char *' to 'ByteArray'.
However, the "assignment-constructor" should be the same as the copy-constuktor, ie. the ba5 line should compile just as the ba3 line--- in contrast to the construction of ba4 and subsequent assignment. So, I'm not quite sure what problem the compiler is having.

I know that a solution would be to remove the explicit in front of the 3rd ctor. I would rather understand what's going on first, though...

Edit:
The answer states it nicely: ByteArray ba5 = "ABC"; would get compiled as ByteArray ba5( ByteArray("ABC") ); --- NOT as ByteArray ba5("ABC"); as I thought it would. Obvious, but sometimes you need someone to point it out. Thanks everyone for your answers!

Why use 'explicit' anyway? Because there is an ambiguity between unsigned int and const char *. If I call ByteArray ba( 0 ); both ctors would be able to handle that, so I need to forbid the implicit conversion and make it explicit.

Answer 1

ByteArray ba5 = "ABC"; is copy initialization, not assignment.

Think of it as

ByteArray ba5(ByteArray("ABC"));

or at least that's what the compiler sees. It's illegal in your case because of the explicit property of the constructor - the compiler wants to use that conversion constructor to perform copy initialization, but it can't because you didn't explicitly use it.