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Existuje takova pekna uloha z rossetacode co zjistuje zda dany retezec je pangram a nebo neni. Tzn. zda obsahuje vsechny pismena abecedy.

Je to jednoducha uloha, ale uz pocet osob se nevleze do jedineho auta.

#include <stdio.h>

int is_pangram(const char *s)
{
    const char *alpha = ""
        "abcdefghjiklmnopqrstuvwxyz"
        "ABCDEFGHIJKLMNOPQRSTUVWXYZ";

    char ch, wasused[26] = {0};
    int total = 0;

    while ((ch = *s++) != '\0') {
        const char *p;
        int idx;

        if ((p = strchr(alpha, ch)) == NULL)
            continue;

        idx = (p - alpha) % 26;

        total += !wasused[idx];
        wasused[idx] = 1;
        if (total == 26)
            return 1;
    }
    return 0;
}

int main(void)
{
    int i;
    const char *tests[] = {
        "The quick brown fox jumps over the lazy dog.",
        "The qu1ck brown fox jumps over the lazy d0g."
    };

    for (i = 0; i < 2; i++)
        printf("\"%s\" is %sa pangram\n",
            tests[i], is_pangram(tests[i])?"":"not ");
    return 0;
}

ja to pojal jako benchmark tak ze udelam 32 bitove binarni pole

#include <stdio.h>

int pangram(const char * str)
{
   str--;
   unsigned char c;
   unsigned long int alphabet = 0;

   while (c = *++str) {
      c |= 32; // uppercase
      c -= 'a';
      if ( c<26 ) alphabet |= (long unsigned int) 1 << c;
   }

   return alphabet==0x3FFFFFF;
}

int main()
{
   char *str = "The five boxing wizards jump quickly.";
   unsigned int i;
   for (i = 10000; i > 0; i--)
      pangram(str);

   printf("%d\n", pangram(str));
   return 0;
}

podle vzoru

: pangram? ( addr len -- ? )
  0 -rot bounds do
    i c@ 32 or [char] a -
    dup 0 26 within if
      1 swap lshift or
    else drop then
  loop
  1 26 lshift 1- = ;

s" The five boxing wizards jump quickly." pangram? .   \ -1

Tohle funguje jen pokud ma forth 32 bitove a vetsi slova.

pseudo forth 16bit kod:

: pangram ( addr -- ? )
  1-
  0.
  BEGIN
    ROT  1+ -ROT 2OVER NIP C@ 0 C<> WHILE 2OVER NIP C@
    32 OR [CHAR] a -
    DUP 26. WITHIN IF
      DBITSET ( d index -- d )
    ELSE DROP THEN
  REPEAT
  67108863. D= NIP ;

A kdyz to zkompilujes z88dk nebo forthem, tak C je 6x pomalejsi.
Docela by me zajimalo i jak si vede C prekladac pro 6502. Z80 to trvalo 2m 51.61s a ten jeho kod uplne zkolaboval a jak ten kod pro sebe rozkouskoval na male casti a ty prevadel do asm tak je to kolikrat ze neco spocita a uklada na "stack" a pak to hned zase vraci do stejnych registru a pouzije znova...