From: bc@freeuk.com
On 01/12/2025 20:34, Keith Thompson wrote:
> bart writes:
>> The following table for example shows the rules for mixed sign
>> arithmetic: S means the result (32 or 64 bits) has signed type, and u
>> means it is unsigned:
>>
>> u8 u16 u32 u64 i8 i16 i32 i64
>>
>> u8 S S u u S S S S
>> u16 S S u u S S S S
>> u32 u u u u u u u S
>> u64 u u u u u u u u
>>
>> i8 S S u u S S S S
>> i16 S S u u S S S S
>> i32 S S u u S S S S
>> i64 S S S u S S S S
>>
>> But of course, every C programmer knows this and doesn't need such a chart!
>
> I'm not going to take the time to confirm that your chart is correct.
> It assumes that int is 32 bits; an implementation with 16-bit or
> 64-bit int would require a different chart.
>
> But the fact that you were able to generate the chart means that
> *you already understand the rules*. You just choose to express
> those rules in a way that's more confusing.
That doesn't follow; the chart was created by a C program (by submitting
64 combinations of typed variables (eg. issigned(x * y)) to the macro
below, compiled with gcc.
My own C compiler produces a quite different chart, but I'm not
interested at this point in rewriting the front-end, considering the
many other ways it doesn't conform.
Fortunately this doesn't seem to affect too many things.
As example, the above says that i8 * u32 (or u32 * i8) is unsigned; my
chart says it's signed. The difference can be demonstrated here:
signed char a = -2;
unsigned int b = 13;
printf("%f\n", (double)(a*b));
The output is:
gcc 4294967270.000000
bcc -26.000000
I don't know about you, but to me, my result looks a lot more intuitive!
So I also didn't /want/ to change it to something I didn't agree with.
-------------------------------------------------
#define issigned(x) _Generic((x),\
int8_t: "S",\
int16_t: "S",\
int32_t: "S",\
int64_t: "S",\
uint8_t: "u",\
uint16_t: "u",\
uint32_t: "u",\
uint64_t: "u",\
default: "other")
--- SoupGate-Win32 v1.05
* Origin: you cannot sedate... all the things you hate (1:229/2)
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