Re: else ladders practice

On 01/11/2024 17:35, David Brown wrote:

On 01/11/2024 16:59, Bart wrote:

On 01/11/2024 14:17, fir wrote:

fir wrote:

Bart wrote:

On 01/11/2024 12:55, fir wrote:

Bart wrote:

On 01/11/2024 11:32, fir wrote:

Bart wrote:

ral clear patterns here: you're testing the same variable 'n' against
several mutually exclusive alternatives, which also happen to be
consecutive values.
>
C is short of ways to express this, if you want to keep those
'somethings' as inline code (otherwise arrays of function pointers or
even label pointers could be use

>
so in short this groupo seem to have no conclusion but is tolerant
foir various approaches as it seems
>
imo the else latder is like most proper but i dont lkie it optically,
swich case i also dont like (use as far i i remember never in my code,
for years dont use even one)
>
so i persnally would use bare ifs and maybe elses ocasionally
(and switch should be mended but its fully not clear how,
>
as to those pointer tables im not sure but im like measurad it onece
and it was (not sure as to thsi as i dont remember exactly) slow maybe
dependant on architecture so its noth wort of use (if i remember
correctly)

>
Well, personally I don't like that repetition, that's why I mentioned
the patterns. You're writing 'n' 5 times, '==' 5 times, and you're
writing out the numbers 1, 2, 3, 4, 5.
>
I also don't like the lack of exclusivity.
>
However I don't need to use C. If those 'somethings' were simple, or
were expressions, I could use syntax like this:
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    (n | s1, s2, s3, s4, s5)
>

>
on a C ground more suitable is
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{s1,s2,s3,s4,s5)[n]
>
//which is just array indexing

>
No, it's specifically not array indexing, as only one of s1 - s5 is
evaluated, or nothing is when n is not in range, eg. n is 100.
>
You could try something like that in C:
>
     int x;
>
     x = ((int[]){(puts("a"),10), (puts("b"),20), (puts("c"), 30),
(puts("d"),40)})[3];
>
     printf("X=%d\n", x);
>
The output is:
>
    a
    b
    c
    d
    X=40
>
Showing that all elements are evaluated first. If index is 100, the
result is also undefined.
>
>

:-O
what is this, first time i see such thing
>

im surprised that it work, but in fact i meant that this syntax is old c compatible but sych thing like
>
>
{printf("ONE"), printf("TWO"), printf("THREE")} [2]
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shouldn evaluate al just the one is selected
like in array tab[23] not eveluates something other than tab[23]

>
It's a 'compound literal'. It allows you to have the same {...} initialisation data format, but anywhere, not just for initialing. However it always needs a cast:
>
   (int[]){printf("ONE"), printf("TWO"), printf("THREE")}[2];
>
This prints ONETWOTHREE, it also then indexes the 3rd value of the array, which is 5, as returned by printf, so this:
>
   printf("%d\n", (int[]){printf("ONE"), printf("TWO"), printf("THREE")}[2]);
>
   prints ONETWOTHREE5
>
>

 What you have written here is all correct, but a more common method would be to avoid having three printf's :
 void shout_a_number(int n) {
     printf( (const char* []) { "ONE", "TWO", "THREE" } [n] );
}
 That's more likely to match what people would want.

I was also trying to show that all elements are evaluated, so each has to have some side-effect to illustrate that.
A true N-way-select construct (C only really has ?:) would evaluate only one, and would deal with an out-of-range condition.
(In my implementations, a default/else branch value must be provided if the whole thing is expected to return a value.)