[Bug 24] Arithmetic operators are allowed on boolean expressions

[d-bugmail at puremagic.com]

http://d.puremagic.com/bugzilla/show_bug.cgi?id=24


walter at digitalmars.com changed:

           What    |Removed                     |Added
----------------------------------------------------------------------------
             Status|REOPENED                    |RESOLVED
         Resolution|                            |INVALID




------- Comment #5 from walter at digitalmars.com  2006-03-10 00:21 -------
First, note that the bool operands undergo integral promotion rules, so that
(true op true) is evaluated as (cast(int)true op cast(int)true), with a result
type of int. The integral promotion for bools does not apply if op is & | ^

void main()
{
    bool x1 = true + true; // fails because true+true => 2, and 2 cannot be
implicitly converted to bool

    bool x2 = true - true; // accepted because true-true=>0, and 0 can be
implicitly converted to bool

    bool x3 = true * true; // accepted because true*true => 1, and 1 can be
implicitly converted to bool

    bool x4 = true / true; // accepted because true/true => 1, and 1 can be
implicitly converted to bool

    bool x5; x5 += true;  // fails because x5+true is int, and int cannot be
implicitly converted to bool unless it is 0 or 1

    bool x6; x6 -= true; // fails because x6-true is int, and int cannot be
implicitly converted to bool unless it is 0 or 1

    bool x7; x7 *= true; // fails because x7*true is int, and int cannot be
implicitly converted to bool unless it is 0 or 1

    bool x8; x8 /= true; // fails because x8/true is int, and int cannot be
implicitly converted to bool unless it is 0 or 1
}


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