Question

The following code prints -10

int x = 10;
-x;
cout << -x << endl;  // printf("%d\n", -x); 

both in C and C++ compilers (gcc 4.1.2). I was expecting a compiler error for the second line.
May be it is something fundamental, but I do not understand the behavior. Could someone please explain?

Thanks

Answer 1

Statements can be expressions. Such statements discard result of the expression, and evaluate the expression for its side effects.

-x; computes the negation of x and discards the result.

For more information read [stmt.expr] in the C++ standard.