C pointer clarification

/*
WK Shearer, April 2012. This is a revision of pointers in C. Being largely a C++ guy myself,
I ran into some confusion with pointer handling in C, which does not use dereferencing like C++.
Here it is for clarificaction, in a pure C file.
*/
#include <stdio.h>

struct test
{
    int i_snum;
};

void func_1(int*);
void func_2(int*);
void func_3(struct test*);
void func_4(struct test**);
//void func_5(int&);

void func_1(int *ptr)
{
    *ptr += 10;
    printf("i_num is %d\n", *ptr);
    func_2(ptr);
}

void func_2(int *ptr1)
{
    *ptr1 += 10;
}

void func_3(struct test *p_struct)
{
    p_struct->i_snum += 10;
}

void func_4(struct test **p_struct)
{
    (*p_struct)->i_snum += 100;
    printf("struct_test.i_snum = %d\n", (*p_struct)->i_snum);
    (**p_struct).i_snum += 50;
}

// This is the C++ style that DOES NOT work.
/*
void func_5(int &num)
{
        num += 1;
}
*/

int main()
{
    int i_num;
    int *p_num;

    p_num = &i_num;

    *p_num = 14;

    printf("i_num is %d\n", i_num);
    func_1(p_num);
    printf("i_num is %d\n", i_num);
    func_1(&i_num);
    printf("i_num is %d\n", i_num);
//    func_5(i_num);
//    printf("i_num is %d\n", i_num);

    struct test struct_test;
    struct test *p_test;

    p_test = &struct_test;

    p_test->i_snum = 100;
    printf("struct_test.i_snum = %d\n", struct_test.i_snum);

    func_3(p_test);
    printf("struct_test.i_snum = %d\n", struct_test.i_snum);

    func_3(&struct_test);
    printf("struct_test.i_snum = %d\n", struct_test.i_snum);

    func_4(&p_test);
    printf("struct_test.i_snum = %d\n", struct_test.i_snum);

    return 0;
}