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

(int) + 4*5;

Why is this (adding a type with a value) possible? (tried with g++ and gcc.)

I know that it doesn't make sense (and has no effect), but I want to know why this is possible.

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最佳答案

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.

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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.


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