1. Си / Говнокод #6636

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    char* a(int pos, char* path)
{ 
    FILE *dbf;
      char value; 
      int i,j,k,l,m,n,o,p,r;
      int numrecs, recnum; 
      Record rec;
      char aa[158];
      char *pp;
      int ln, lp, lc, le, lw, li, ls, lph, lf;    

      numrecs = rec.length; 
      recnum = pos;
    dbf = OpenDB(path, &rec);
    ReadRecord(dbf, recnum, &rec);    
    
        ln = strlen(rec.name);
        ls = strlen(rec.surname);
        lp = strlen(rec.patronymic);
        lph = strlen(rec.phone);
        lc = strlen(rec.city);
        le = strlen(rec.edu);
        lw = strlen(rec.work);
        li = strlen(rec.interests);
        lf = strlen(rec.friends);

        itoa(recnum, &(aa[0]), 10);
        for (i = 1; i < ln; i++)
            aa[i] = rec.name[i-1];
        for (j = 0; j < ls; j++)
            aa[i+j] = rec.surname[j];
        for (k = 0; k < lp; k++)
            aa[i+j+k] = rec.patronymic[k];
        for (l = 0; l < lph; l++)
            aa[i+j+k+l] = rec.phone[l];
        for (m = 0; m < lc; m++)
            aa[i+j+k+l+m] = rec.city[m];
        for (n = 0; n < le; n++)
            aa[i+j+k+l+m+n] = rec.edu[n];
        for (o = 0; o < lw; o++)
            aa[i+j+k+l+m+n+o] = rec.work[o];
        for (p = 0; p < li; p++)
            aa[i+j+k+l+m+n+o+p] = rec.interests[p];
        for (r = 0; r < lf; r++)
            aa[i+j+k+l+m+n+o+p+r] = rec.friends[r];

    pp = strdup(aa);  
    return pp;
}

    Запостил: dotnetdeveloper, 12 Мая 2011

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