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On Friday 17 May 2002 16:02, you wrote:
> > Ako nqkoi sega natrqska edin parser na C kolekciqta funkcionalni resheniq
> > shte e pylna :)
>
> /*
> WARNING!!! Highly insecure&&exploitable code
> */
>
> #include <stdio.h>
> #include <errno.h>
>
> int main(int argc,char *argv[])
> {
> FILE *f;
> int s,e,c=0;
> char l[100000];
>
> if (argc!=4)
> {
> printf("Usage: %s <filename> <start> <end>\n",argv[0]);
> return 1;
> }
>
> if ((f=fopen(argv[1],"r"))==NULL) {perror("fopen()");return 1;}
>
> sscanf(argv[2],"%d",&s);
> sscanf(argv[3],"%d",&e);
>
> while (fgets(l,100000,f))
> {
> c++;
> if ((s<=c)&&(c<=e)) fputs(l,stdout);
> if (c>e) break;
> }
>
> fclose(f);
> return 0;
> }
>
>
> Tva go napisah za 5 minuti just for fun, znam che sigurno e bugavo :)
> Ama poneje kolegata otpravi predizvikatelstvo :)
> Moga i na C++ s iostreams da go napisha, ama aide tova niakoi drug :)
Samo 4e mernata edinica niama da e "lines" , a "words", zashtoto
razdelitelite s koito raboti fgets() sa '\xd' '\x9' '\xa' i shpacia.
On GNU systems izpolzwai getline().
On non GNU systems: fscanf(f,"%100000[^'\n']\n",l);
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