Integer promotion... what I'm missing? (It's monday...)

[Kirk McDonald]

Max Samuha wrote:
> On Tue, 27 Jun 2006 12:33:32 -0700, Kirk McDonald
> <kirklin.mcdonald at gmail.com> wrote:
> 
> 
>>Alexander Panek wrote:
>>
>>>If you take a look at how comparison works, you'll know why this one fails.
>>>
>>>Lets take an uint a = 16; as in your example:
>>>00000000 00000000 00000000 00010000
>>>
>>>And now a signed integer with the value -1:
>>>10000000 00000000 00000000 00000001
>>>
>>
>>Your point still stands, but -1 is represented as:
>>11111111 11111111 11111111 11111111
>>
>>http://en.wikipedia.org/wiki/Two%27s_complement
>>
>>
>>>You might guess which number is bigger, when our comparison is done 
>>>binary (and after all, that's what the processor does) :)
>>>
>>>Regards,
>>>Alex
>>>
>>>Paolo Invernizzi wrote:
>>>
>>>
>>>>Hi all,
>>>>
>>>>What I'm missing?
>>>>
>>>>    uint a = 16;
>>>>    int b = -1;
>>>>    assert( b < a ); // this fails! I was expecting that -1 < 16
>>>>
>>>>Thanks
>>>>
>>>>---
>>>>Paolo
> 
> 
> Maybe it's not a bug but it is very confusing, no matter how integer
> operations work internally. Compiler should give at least a warning
> about incompatibe types, or try to cast uint to int implicitly or
> require an explicit cast.

It's worth noting that this behavior (of a being less than b) follows 
the implicit conversion rules exactly:

http://www.digitalmars.com/d/type.html

[snipped non-applicable checks...]
5. Else the integer promotions are done on each operand, followed by:

    1. If both are the same type, no more conversions are done.
    2. If both are signed or both are unsigned, the smaller type is 
converted to the larger.
    3. If the signed type is larger than the unsigned type, the unsigned 
type is converted to the signed type.
    4. The signed type is converted to the unsigned type.

So the int is implicitly converted to the uint, and (apparently) it 
simply compares 2**32-1 to 16.

So I wouldn't call this a bug, just a potential oddity. Maybe it should 
detect and throw an overflow if a negative signed integer is converted 
to an unsigned type? Or not: I'd consider this an edge case. It's 
probably considered bad practice to promiscuously mix signed and 
unsigned types.

-Kirk McDonald