How is pointer dereferencing implemented in C / C ++?

The error lies in the highlighted word of this sentence.

& (* (& n)): return the address where the number we just temporarily calculated is stored (?)

The fact is that according to, for example, the C ++ standard (5.3.1 Unary operators)

It can be used if it’s not necessary to do so.

no temporary object is created.

The same is written in the C standard (6.5.3.2 Address and indirection operators)

4 The unary * operator denotes indirection. If the function is a function designator; if it is a point, it is a designating object.

Also in this same section of standard C is written

3 The unary & operator yields the address of its operand. If the operand has type '' type '', the result has a type '' pointer to type ''. It is not clear that the operator has been omitted , except for the operator.

So the expression

 &(*(&n)) 

equivalent to the original expression

 &n 

subject to the above limitations.

Note. The limitations mentioned in the quotation are sometimes substantial. For example, the result of this C program where a bunch of operators is used &*

 #include <stdio.h> int main( void ) { int a[10]; printf( "sizeof( a ) = %zu\n", sizeof( a ) ); printf( "sizeof( &*a ) = %zu\n", sizeof( &*a ) ); return 0; } 

will be next

 sizeof( a ) = 40 sizeof( &*a ) = 4 

That is, the operand of the first sizeof operator is lvalue , while the operand of the second sizeof operator will already be rvalue due to the use of the bunch &* . [End of note]

In this regard, there is an amusing side effect. Since the functions in the expressions are converted to a pointer to the function, it is possible to write the function call, for example, in the following form

 #include <iostream> void f() { std::cout << "Hello, pointers!" << std::endl; } int main() { ( **********f )(); //^^^^^^^^^^^^^ return 0; } 

This technique is useful when your productivity is estimated by the number of characters typed in the program. :)