Re: else ladders practice

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 ); }

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