Re: RFC: booleans and the kernel

Rick Stevens
2002-01-25T23:08:14Z

Ragnar Hojland Espinosa wrote:

> On Fri, Jan 25, 2002 at 04:44:38PM -0600, Timothy Covell wrote:
>
>>On Thursday 24 January 2002 16:38, Robert Love wrote:
>>
>>>On Fri, 2002-01-25 at 17:30, Timothy Covell wrote:
>>>
>>>>On Thursday 24 January 2002 16:19, Robert Love wrote:
>>>>
>>>>>how is "if (x)" any less legit if x is an integer ?
>>>>>
>>>>What about
>>>>
>>>>{
>>>> char x;
>>>>
>>>> if ( x )
>>>> {
>>>> printf ("\n We got here\n");
>>>> }
>>>> else
>>>> {
>>>> // We never get here
>>>> printf ("\n We never got here\n");
>>>> }
>>>>}
>>>>
>>>>
>>>>That's not what I want. It just seems too open to bugs
>>>>and messy IHMO.
>>>>
>>>When would you ever use the above code? Your reasoning is "you may
>>>accidentally check a char for a boolean value." In other words, not
>>>realize it was a char. What is to say its a boolean? Or not? This
>>>isn't an argument. How does having a boolean type solve this? Just use
>>>an int.
>>>
>>> Robert Love
>>>
>>It would fix this because then the compiler would refuse to compile
>>"if (x)" when x is not a bool. That's what I would call type safety.
>>But I guess that you all are arguing that C wasn't built that way and
>>that you don't want it.
>>
>
> It would actually break this. if is supposed (and expected) to evaluate
> an expression, whatever it will be. Maybe a gentle warning could be in
> place, but refusing to compile is a plain broken C compiler.

Granted. "if (x)" is true if "x" is non-zero, regardless of type and
shoudn't even generate a warning if "x" is scalar.

Either printf() will occur depending on whether automatics are
initialized to zero or not. The first one will most likely print
since there's 255 to 1 odds that "x" will be non-zero if not
initialized and I don't think gcc initializes automatics.
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