Difference between two ways of declaring an object

Question

What's the difference between the following two declarations, assuming I have not specified a copy constructor and operator= in class Beatle?

Beatle john(paul);

and

Beatle john = paul;

Edit:

In objects assignment, the operator = implicitly calls the copy constructor unless told otherwise?

Answer 1

They're different grammatical constructions. The first one is direct initialization, the second is copy initialization. They behave virtually identically, only that the second requires a non-explicit constructor.*

Neither has anything to do with the assignment operator, as both lines are initializations.

To wit: const int i = 4; is fine, but const int i; i = 4; is not.

*) More accurately: The second version does not work if the relevant constructor is declared explicit. More generally, thus, direct-initialization affords you one "free" conversion:

struct Foo { Foo(std::string) {} };

Foo x("abc");  // OK: char(&)[4] -> const char * -> std::string -> Foo
Foo y = "abd"; // Error: no one-step implicit conversion of UDTs

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To address your edit: To understand the assignment operator, just break it up into parts. Suppose Foo has the obvious operator=(const Foo & rhs). We can say x = y;, which just calls the operator directly with rhs being y. Now consider this:

x = "abc";              // error, no one-step implicit conversion
x = std::string("abc"); // fine, uses Foo(std::string), then copy
x = Foo("abc");         // fine, implicit char(&)[4] -> const char* -> std::string, then as above

Answer 2

First is Direct Initialization & second is Copy Initialization.

Direct initialization means the object is initialized using a single (possibly conversion) constructor, and is equivalent to the form T t(u);:

U u;
T t1(u); // calls T::T( U& ) or similar

Copy initialization means the object is initialized using the copy constructor, after first calling a user-defined conversion if necessary, and is equivalent to the form T t = u;:

T t2 = t1;  // same type: calls T::T( T& ) or similar
T t3 = u;   // different type: calls T::T( T(u) )
            //  or T::T( u.operator T() ) or similar

Copy Initialization does not work if the constructor is declared explicit.

References:
This entry in Herb Sutter's GOTW should be a good read.

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To answer your edited Question:
= has a different meaning in depending on how it is used.

If = is used in an expression in which the object is being created and initialized at the same time, then = is not treated as Assignment Operator but as Copy Initialization.

If = is being used to assign one object to another, after the object has been created then it results in call to Assignment Operator.