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From: Dan Purgert <dan@djph.net>
Newsgroups: comp.lang.c
Subject: Re: else ladders practice
Date: Wed, 20 Nov 2024 12:31:35 -0000 (UTC)
Organization: A noiseless patient Spider
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Message-ID: <slrnvjrll7.fkp.dan@djph.net>
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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);
>
>   My bad if the following instruction structure's already been hashed
>   out in this thread, but I haven't been following the whole convo!

I honestly lost the plot ages ago; not sure if it was either!

>
>   In my C 101 classes, after we've covered "if" and "else",
>   I always throw this program up on the screen and hit the newbies
>   with this curveball: "What's this bad boy going to spit out?".

Segfaults? :D

>
>   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 ); }

oooh, that's way better at making a point of the hazard than mine was.

.... almost needed to engage my rubber duckie, before I realized I was
mentally auto-correcting the 'english()' function while reading it.


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