Boolean equality (C++/Fortran)

Boolean equality (C++/Fortran)

[David Lecomber]

In valarith.c:value_binop, where v1 and v2 have been established to be
values of type bool, we have:

       case BINOP_EQUAL:
          v = v1 == v2;
          break;
          
        case BINOP_NOTEQUAL:
          v = v1 != v2;
	  break;

Isn't this wrong?  If you are mixing your compilers, then, at least for
Fortran, the actual value of true can vary (1 or -1 I have so far
seen).  For C++ this is less likely to happen, and so far as I can tell
changing the above would not harm anything.  

Does anyone have any comments on replacing the above with:

       case BINOP_EQUAL:
          v = !((!v1 && v2) || (v1 && !v2));
          break;
          
        case BINOP_NOTEQUAL:
          v = (!v1 && v2) || (v1 && !v2);
	  break;


d.

Re: Boolean equality (C++/Fortran)

[Jim Blandy]

David Lecomber <david@streamline-computing.com> writes:

> In valarith.c:value_binop, where v1 and v2 have been established to be
> values of type bool, we have:
> 
>        case BINOP_EQUAL:
>           v = v1 == v2;
>           break;
>           
>         case BINOP_NOTEQUAL:
>           v = v1 != v2;
> 	  break;
> 
> Isn't this wrong?  If you are mixing your compilers, then, at least for
> Fortran, the actual value of true can vary (1 or -1 I have so far
> seen).  For C++ this is less likely to happen, and so far as I can tell
> changing the above would not harm anything.  
> 
> Does anyone have any comments on replacing the above with:
> 
>        case BINOP_EQUAL:
>           v = !((!v1 && v2) || (v1 && !v2));
>           break;
>           
>         case BINOP_NOTEQUAL:
>           v = (!v1 && v2) || (v1 && !v2);
> 	  break;

The general idea looks right.  But how about:

>        case BINOP_EQUAL:
>           v = !v1 == !v2;
>           break;
>           
>         case BINOP_NOTEQUAL:
>           v = !v1 != !v2;
> 	  break;