Why do C and C++ allow the expression (int) + 4*5?

Solution 1:

The + here is unary + operator, not the binary addition operator. There's no addition happening here.

Also, the syntax (int) is used for typecasting.

You can re-read that statement as

(int) (+ 4) * 5;    

which is parsed as

((int) (+ 4)) * (5);    

which says,

1. Apply the unary + operator on the integer constant value 4.
2. typecast to an int
3. multiply with operand 5

This is similar to (int) (- 4) * (5);, where the usage of the unary operator is more familiar.

In your case, the unary + and the cast to int - both are redundant.

Solution 2:

This is interpreted as ((int)(+4)) * 5. That is, an expression +4 (a unary plus operator applied to a literal 4), cast to type int with a C-style cast, and the result multiplied by 5.