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Re: else ladders practice

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On 2024-11-30, Rosario19 wrote:
> On Wed, 20 Nov 2024 12:31:35 -0000 (UTC), Dan Purgert  wrote:
>
>>On 2024-11-16, Stefan Ram wrote:
>>> Dan Purgert <dan@djph.net> wrote or quoted:
>>>>if (n==0) { printf ("n: %u\n",n); n++;}
>>>>if (n==1) { printf ("n: %u\n",n); n++;}
>>>>if (n==2) { printf ("n: %u\n",n); n++;}
>>>>if (n==3) { printf ("n: %u\n",n); n++;}
>>>>if (n==4) { printf ("n: %u\n",n); n++;}
>>>>printf ("all if completed, n=%u\n",n);
>
> above should be equivalent to this
>
> for(;n>=0&&n<5;++n) printf ("n: %u\n",n);
> printf ("all if completed, n=%u\n",n);

Sure, but fir's original posting in 
  MID <3deb64c5b0ee344acd9fbaea1002baf7302c1e8f@i2pn2.org>

was a contrived sequence to the effect of 
    if (n==0) { //do something }
    if (n==1) { //do something }
    if (n==2) { //do something }
    if (n==3) { //do something }
    if (n==4) { //do something }

So, I merely took the contrived sequence, and made "do something" trip
each condition.

Stefan's example from a few posts back is better: 

>>>   Well, it's a blue moon when someone nails it. Most of them fall
>>>   for my little gotcha hook, line, and sinker.
>>>
>>> #include <stdio.h>
>>>
>>> const char * english( int const n )
>>> { const char * result;
>>>   if( n == 0 )result = "zero";
>>>   if( n == 1 )result = "one";
>>>   if( n == 2 )result = "two";
>>>   if( n == 3 )result = "three";
>>>   else        result = "four";
>>>   return result; }
>>>
>>> void print_english( int const n )
>>> { printf( "%s\n", english( n )); }
>>>
>>> int main( void )
>>> { print_english( 0 );
>>>   print_english( 1 );
>>>   print_english( 2 );
>>>   print_english( 3 );
>>>   print_english( 4 ); }

-- 
|_|O|_| 
|_|_|O| Github: https://github.com/dpurgert
|O|O|O| PGP: DDAB 23FB 19FA 7D85 1CC1  E067 6D65 70E5 4CE7 2860

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Re: else ladders practice Rosario19 <Ros@invalid.invalid> - 2024-11-30 17:35 +0100
  Re: else ladders practice Dan Purgert <dan@djph.net> - 2024-12-05 10:51 +0000

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