Splitting string into words. Word contained in string. C language.

We are given 2 strings. We need to split strings into words and find word with minimal length contained in the first string, that is not contained in second string. Use function to check whether string contains a word.
#include <stdio.h>
#include <conio.h>
#include <string.h>
#include <alloc.h>
#include <ctype.h>
#include <limits.h>

char** words(char* str) {
        char** wrd;
        int i=0, j, k, l, n;
        wrd = (char**)calloc(1, sizeof(char*));

Splitting string into words. Array of pointers. C language.

We are given a string. We need to split a string into words and find a words with lenght more than 3. Use array of pointers to store words.

#include <stdio.h>
#include <conio.h>
#include <string.h>
#include <alloc.h>
#include <ctype.h>
#include <limits.h>

int main(int argc, char* argv[])
{
char *s1, *s[10];
int i=0, j, sim=0, l, k, h,  n1, n2, min;

s1 = (char*)calloc(100, sizeof(char));
puts("Enter string #1:");
gets(s1);
n1 = strlen(s1);
k = 0;
while (i < n1) {

Splitting string into words. Symmetrical words. C language.

We are given a string. We need to split a string into words and find a symmetrical words.

#include <stdio.h>
#include <conio.h>
#include <string.h>
#include <alloc.h>
#include <ctype.h>


int main(int argc, char* argv[])
{
char* s;
int i=0, j, sim=0, l, k, n;
s = (char*)calloc(100, sizeof(char));
puts("Enter string:");
gets(s);

n = strlen(s);

while (i < n) {
        while ( (i < n) && (!isalpha(s[i])) ) i++;
        j = i;

AngularJS Authorization

In this article we'll talk about quickly and "painless" access control in AngularJS. We also touch some standard routing tools provided by AngularJS kernel and will write a simple application with authorization. This article is based on Доверяй, но проверяй with several important changes. So, Angular offers us $routeProvider authorization mean from ngRoute module.

Routing scheme:

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