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From: Lew Pitcher <lew.pitcher@digitalfreehold.ca>
Newsgroups: comp.lang.c
Subject: Re: else ladders practice
Date: Sat, 16 Nov 2024 15:37:24 -0000 (UTC)
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On Sat, 16 Nov 2024 09:42:49 +0000, 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!
> 
>   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?".
> 
>   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 ); }

If I read your code correctly, you have actually included not one,
but TWO curveballs.  Well done!

-- 
Lew Pitcher
"In Skills We Trust"