c - Pointers to string in functions

it seems i'm stuck with some basics. Can someone explain me why next code:

#include 

void Test1(char *t)
{
    t = (char *)malloc(11);
    strcpy(t, "1234567890");

}

void Test2(char **t)
{
    *t = (char *)malloc(11);
    strcpy(*t, "1234567890");
}

void Test3(char *t)
{

    strcpy(t, "1234567890");
}

char * Test4(char *t)
{
    t = (char *)malloc(11);
    strcpy(t, "1234567890");
    return t;
}


int main()
{
    char *t1 = NULL;
    Test1(t1);
    printf("\nTest1: %s\n", t1);

    char *t2 = NULL;
    Test2(&t2);
    printf("\nTest2: %s\n", t2);


    char *t3 = (char *)malloc(11);
    Test3(t3);
    printf("\nTest3: %s\n", t3);

    char *t4 = NULL;
    t4 = Test4(t4);
    printf("\nTest4: %s\n", t4);

    return 0;
}

gives this output:

Test1: (null)

Test2: 1234567890

Test3: 1234567890


Test4: 1234567890

What's wrong with Test1 function? And why Test4, which almost similar to Test1, works?
More general question: what's the correct way to create string in function and return pointer to it?

Answer

Consider your Test1 performs the following behavior:

char * x1 = NULL;

Test1 (x1);

Test1 is done within the following:

void test1 (char * t) / / t -> x1 -> null
{
    t = (char *) malloc (11);
    // t now points a different place (the memory allocated), and the reference of x1 continues pointing to null
    strcpy (t, "1234567890"); // t value is "1234567890" but x1 continues pointing to null

    // Error, the memory that is pointed by t is never released
}

printf ("\nTest1:%s \n", t1); / / Print the value of x1 (null) and the reference of t is lost