Why use double pointer? or Why use pointers to poi

问题:

When should a double pointer be used in C? Can anyone explain with a example?

What I know is that a double pointer is a pointer to a pointer. Why would I need a pointer to a pointer?

回答1:

If you want to have a list of characters (a word), you can use char *word

If you want a list of words (a sentence), you can use char **sentence

If you want a list of sentences (a monologue), you can use char ***monologue

If you want a list of monologues (a biography), you can use char ****biography

If you want a list of biographies (a bio-library), you can use char *****biolibrary

If you want a list of bio-libraries (a ??lol), you can use char ******lol

... ...

^{yes, I know these might not be the best data structures}

---

Usage example with a very very very boring lol

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int wordsinsentence(char **x) {
    int w = 0;
    while (*x) {
        w += 1;
        x++;
    }
    return w;
}

int wordsinmono(char ***x) {
    int w = 0;
    while (*x) {
        w += wordsinsentence(*x);
        x++;
    }
    return w;
}

int wordsinbio(char ****x) {
    int w = 0;
    while (*x) {
        w += wordsinmono(*x);
        x++;
    }
    return w;
}

int wordsinlib(char *****x) {
    int w = 0;
    while (*x) {
        w += wordsinbio(*x);
        x++;
    }
    return w;
}

int wordsinlol(char ******x) {
    int w = 0;
    while (*x) {
        w += wordsinlib(*x);
        x++;
    }
    return w;
}

int main(void) {
    char *word;
    char **sentence;
    char ***monologue;
    char ****biography;
    char *****biolibrary;
    char ******lol;

    //fill data structure
    word = malloc(4 * sizeof *word); // assume it worked
    strcpy(word, \"foo\");

    sentence = malloc(4 * sizeof *sentence); // assume it worked
    sentence[0] = word;
    sentence[1] = word;
    sentence[2] = word;
    sentence[3] = NULL;

    monologue = malloc(4 * sizeof *monologue); // assume it worked
    monologue[0] = sentence;
    monologue[1] = sentence;
    monologue[2] = sentence;
    monologue[3] = NULL;

    biography = malloc(4 * sizeof *biography); // assume it worked
    biography[0] = monologue;
    biography[1] = monologue;
    biography[2] = monologue;
    biography[3] = NULL;

    biolibrary = malloc(4 * sizeof *biolibrary); // assume it worked
    biolibrary[0] = biography;
    biolibrary[1] = biography;
    biolibrary[2] = biography;
    biolibrary[3] = NULL;

    lol = malloc(4 * sizeof *lol); // assume it worked
    lol[0] = biolibrary;
    lol[1] = biolibrary;
    lol[2] = biolibrary;
    lol[3] = NULL;

    printf(\"total words in my lol: %d\\n\", wordsinlol(lol));

    free(lol);
    free(biolibrary);
    free(biography);
    free(monologue);
    free(sentence);
    free(word);
}

Output:

total words in my lol: 243

回答2:

One reason is you want to change the value of the pointer passed to a function as the function argument, to do this you require pointer to a pointer.

In simple words, Use ** when you want to preserve (OR retain change in) the Memory-Allocation or Assignment even outside of a function call. (So, Pass such function with double pointer arg.)

This may not be a very good example, but will show you the basic use:

void allocate(int** p)
{
  *p = (int*)malloc(sizeof(int));
}

int main()
{
  int* p = NULL;
  allocate(&p);
  *p = 42;
  free(p);
}

回答3:

Here is a SIMPLE answer!!!!

• lets say you have a pointer that its value is an address.
• but now you want to change that address.
• you could, by doing pointer1 = pointer2, and pointer1 would now have the address of pointer2.

• BUT! if you want a function to do that for you, and you want the result to persist after the function is done, you need do some extra work, you need a new pointer3 just to point to pointer1, and pass pointer3 to the function.

• here is a fun example (take a look at the output bellow first, to understand!):

#include <stdio.h>

int main()
{

    int c = 1;
    int d = 2;
    int e = 3;
    int * a = &c;
    int * b = &d;
    int * f = &e;
    int ** pp = &a;  // pointer to pointer \'a\'

    printf(\"\\n a\'s value: %x \\n\", a);
    printf(\"\\n b\'s value: %x \\n\", b);
    printf(\"\\n f\'s value: %x \\n\", f);
    printf(\"\\n can we change a?, lets see \\n\");
    printf(\"\\n a = b \\n\");
    a = b;
    printf(\"\\n a\'s value is now: %x, same as \'b\'... it seems we can, but can we do it in a function? lets see... \\n\", a);
    printf(\"\\n cant_change(a, f); \\n\");
    cant_change(a, f);
    printf(\"\\n a\'s value is now: %x, Doh! same as \'b\'...  that function tricked us. \\n\", a);

    printf(\"\\n NOW! lets see if a pointer to a pointer solution can help us... remember that \'pp\' point to \'a\' \\n\");
     printf(\"\\n change(pp, f); \\n\");
    change(pp, f);
    printf(\"\\n a\'s value is now: %x, YEAH! same as \'f\'...  that function ROCKS!!!. \\n\", a);
    return 0;
}

void cant_change(int * x, int * z){
    x = z;
    printf(\"\\n ----> value of \'a\' is: %x inside function, same as \'f\', BUT will it be the same outside of this function? lets see\\n\", x);
}

void change(int ** x, int * z){
    *x = z;
    printf(\"\\n ----> value of \'a\' is: %x inside function, same as \'f\', BUT will it be the same outside of this function? lets see\\n\", *x);
}

• and here is the output:

 a\'s value: bf94c204

 b\'s value: bf94c208 

 f\'s value: bf94c20c 

 can we change a?, lets see 

 a = b 

 a\'s value is now: bf94c208, same as \'b\'... it seems we can, but can we do it in a function? lets see... 

 cant_change(a, f); 

 ----> value of \'a\' is: bf94c20c inside function, same as \'f\', BUT will it be the same outside of this function? lets see

 a\'s value is now: bf94c208, Doh! same as \'b\'...  that function tricked us. 

 NOW! lets see if a pointer to a pointer solution can help us... remember that \'pp\' point to \'a\' 

 change(pp, f); 

 ----> value of \'a\' is: bf94c20c inside function, same as \'f\', BUT will it be the same outside of this function? lets see

 a\'s value is now: bf94c20c, YEAH! same as \'f\'...  that function ROCKS!!!. 

回答4:

Adding to Asha\'s response, if you use single pointer to the example bellow (e.g. alloc1() ) you will loose the reference to the memory allocated inside the function.

void alloc2(int** p) {
   *p = (int*)malloc(sizeof(int));
   **p = 10;
}

void alloc1(int* p) {
   p = (int*)malloc(sizeof(int));
   *p = 10;
}

int main(){
   int *p;
   alloc1(p);
   //printf(\"%d \",*p);//value is undefined
   alloc2(&p);
   printf(\"%d \",*p);//will print 10
   free(p);
   return 0;
}

The reason it occurs like this is that in alloc1 the pointer is passed in by value. So, when it is reassigned to the result of the malloc call inside of alloc1, the change does not pertain to code in a different scope.

回答5:

1. Basic Concept -

When you declare as follows : -

1. char *ch - (called character pointer)
- ch contains the address of a single character.
- (*ch) will dereference to the value of the character..

2. char **ch -
\'ch\' contains the address of an Array of character pointers. (as in 1)
\'*ch\' contains the address of a single character. (Note that it\'s different from 1, due to difference in declaration).
(**ch) will dereference to the exact value of the character..

Adding more pointers expand the dimension of a datatype, from character to string, to array of strings, and so on... You can relate it to a 1d, 2d, 3d matrix..

So, the usage of pointer depends upon how you declare it.

Here is a simple code..

int main()
{
    char **p;
    p = (char **)malloc(100);
    p[0] = (char *)\"Apple\";      // or write *p, points to location of \'A\'
    p[1] = (char *)\"Banana\";     // or write *(p+1), points to location of \'B\'

    cout << *p << endl;          //Prints the first pointer location until it finds \'\\0\'
    cout << **p << endl;         //Prints the exact character which is being pointed
    *p++;                        //Increments for the next string
    cout << *p;
}

2. Another Application of Double Pointers -
(this would also cover pass by reference)

Suppose you want to update a character from a function. If you try the following : -

void func(char ch)
{
    ch = \'B\';
}

int main()
{
    char ptr;
    ptr = \'A\';
    printf(\"%c\", ptr);

    func(ptr);
    printf(\"%c\\n\", ptr);
}

The output will be AA. This doesn\'t work, as you have \"Passed By Value\" to the function.

The correct way to do that would be -

void func( char *ptr)        //Passed by Reference
{
    *ptr = \'B\';
}

int main()
{
    char *ptr;
    ptr = (char *)malloc(sizeof(char) * 1);
    *ptr = \'A\';
    printf(\"%c\\n\", *ptr);

    func(ptr);
    printf(\"%c\\n\", *ptr);
}

Now extend this requirement for updating a string instead of character.
For this, you need to receive the parameter in the function as a double pointer.

void func(char **str)
{
    strcpy(str, \"Second\");
}

int main()
{
    char **str;
    // printf(\"%d\\n\", sizeof(char));
    *str = (char **)malloc(sizeof(char) * 10);          //Can hold 10 character pointers
    int i = 0;
    for(i=0;i<10;i++)
    {
        str = (char *)malloc(sizeof(char) * 1);         //Each pointer can point to a memory of 1 character.
    }

    strcpy(str, \"First\");
    printf(\"%s\\n\", str);
    func(str);
    printf(\"%s\\n\", str);
}

In this example, method expects a double pointer as a parameter to update the value of a string.

回答6:

I saw a very good example today, from this blog post, as I summarize below.

Imagine you have a structure for nodes in a linked list, which probably is

typedef struct node
{
    struct node * next;
    ....
} node;

Now you want to implement a remove_if function, which accepts a removal criterion rm as one of the arguments and traverses the linked list: if an entry satisfies the criterion (something like rm(entry)==true), its node will be removed from the list. In the end, remove_if returns the head (which may be different from the original head) of the linked list.

You may write

for (node * prev = NULL, * curr = head; curr != NULL; )
{
    node * const next = curr->next;
    if (rm(curr))
    {
        if (prev)  // the node to be removed is not the head
            prev->next = next;
        else       // remove the head
            head = next;
        free(curr);
    }
    else
        prev = curr;
    curr = next;
}

as your for loop. The message is, without double pointers, you have to maintain a prev variable to re-organize the pointers, and handle the two different cases.

But with double pointers, you can actually write

// now head is a double pointer
for (node** curr = head; *curr; )
{
    node * entry = *curr;
    if (rm(entry))
    {
        *curr = entry->next;
        free(entry);
    }
    else
        curr = &entry->next;
}

You don\'t need a prev now because you can directly modify what prev->next pointed to.

To make things clearer, let\'s follow the code a little bit. During the removal:

1. if entry == *head: it will be *head (==*curr) = *head->next -- head now points to the pointer of the new heading node. You do this by directly changing head\'s content to a new pointer.
2. if entry != *head: similarly, *curr is what prev->next pointed to, and now points to entry->next.

No matter in which case, you can re-organize the pointers in a unified way with double pointers.

回答7:

Pointers to pointers also come in handy as \"handles\" to memory where you want to pass around a \"handle\" between functions to re-locatable memory. That basically means that the function can change the memory that is being pointed to by the pointer inside the handle variable, and every function or object that is using the handle will properly point to the newly relocated (or allocated) memory. Libraries like to-do this with \"opaque\" data-types, that is data-types were you don\'t have to worry about what they\'re doing with the memory being pointed do, you simply pass around the \"handle\" between the functions of the library to perform some operations on that memory ... the library functions can be allocating and de-allocating the memory under-the-hood without you having to explicitly worry about the process of memory management or where the handle is pointing.

For instance:

#include <stdlib.h>

typedef unsigned char** handle_type;

//some data_structure that the library functions would work with
typedef struct 
{
    int data_a;
    int data_b;
    int data_c;
} LIB_OBJECT;

handle_type lib_create_handle()
{
    //initialize the handle with some memory that points to and array of 10 LIB_OBJECTs
    handle_type handle = malloc(sizeof(handle_type));
    *handle = malloc(sizeof(LIB_OBJECT) * 10);

    return handle;
}

void lib_func_a(handle_type handle) { /*does something with array of LIB_OBJECTs*/ }

void lib_func_b(handle_type handle)
{
    //does something that takes input LIB_OBJECTs and makes more of them, so has to
    //reallocate memory for the new objects that will be created

    //first re-allocate the memory somewhere else with more slots, but don\'t destroy the
    //currently allocated slots
    *handle = realloc(*handle, sizeof(LIB_OBJECT) * 20);

    //...do some operation on the new memory and return
}

void lib_func_c(handle_type handle) { /*does something else to array of LIB_OBJECTs*/ }

void lib_free_handle(handle_type handle) 
{
    free(*handle);
    free(handle); 
}


int main()
{
    //create a \"handle\" to some memory that the library functions can use
    handle_type my_handle = lib_create_handle();

    //do something with that memory
    lib_func_a(my_handle);

    //do something else with the handle that will make it point somewhere else
    //but that\'s invisible to us from the standpoint of the calling the function and
    //working with the handle
    lib_func_b(my_handle); 

    //do something with new memory chunk, but you don\'t have to think about the fact
    //that the memory has moved under the hood ... it\'s still pointed to by the \"handle\"
    lib_func_c(my_handle);

    //deallocate the handle
    lib_free_handle(my_handle);

    return 0;
}

Hope this helps,

Jason

回答8:

Strings are a great example of uses of double pointers. The string itself is a pointer, so any time you need to point to a string, you\'ll need a double pointer.

回答9:

The following is a very simple C++ example that shows that if you want to use a function to set a pointer to point to an object, you need a pointer to a pointer. Otherwise, the pointer will keep reverting to null.

(A C++ answer, but I believe it\'s the same in C.)

(Also, for reference: Google(\"pass by value c++\") = \"By default, arguments in C++ are passed by value. When an argument is passed by value, the argument\'s value is copied into the function\'s parameter.\")

So we want to set the pointer b equal to the string a.

#include <iostream>
#include <string>

void Function_1(std::string* a, std::string* b) {
  b = a;
  std::cout << (b == nullptr);  // False
}

void Function_2(std::string* a, std::string** b) {
  *b = a;
  std::cout << (b == nullptr);  // False
}

int main() {
  std::string a(\"Hello!\");
  std::string* b(nullptr);
  std::cout << (b == nullptr);  // True

  Function_1(&a, b);
  std::cout << (b == nullptr);  // True

  Function_2(&a, &b);
  std::cout << (b == nullptr);  // False
}

// Output: 10100

What happens at the line Function_1(&a, b);?

• The \"value\" of &main::a (an address) is copied into the parameter std::string* Function_1::a. Therefore Function_1::a is a pointer to (i.e. the memory address of) the string main::a.

• The \"value\" of main::b (an address in memory) is copied into the parameter std::string* Function_1::b. Therefore there are now 2 of these addresses in memory, both null pointers. At the line b = a;, the local variable Function_1::b is then changed to equal Function_1::a (= &main::a), but the variable main::b is unchanged. After the call to Function_1, main::b is still a null pointer.

What happens at the line Function_2(&a, &b);?

• The treatment of the a variable is the same: within the function, Function_2::a is the address of the string main::a.

• But the variable b is now being passed as a pointer to a pointer. The \"value\" of &main::b (the address of the pointer main::b) is copied into std::string** Function_2::b. Therefore within Function_2, dereferencing this as *Function_2::b will access and modify main::b . So the line *b = a; is actually setting main::b (an address) equal to Function_2::a (= address of main::a) which is what we want.

If you want to use a function to modify a thing, be it an object or an address (pointer), you have to pass in a pointer to that thing. The thing that you actually pass in cannot be modified (in the calling scope) because a local copy is made.

(An exception is if the parameter is a reference, such as std::string& a. But usually these are const. Generally, if you call f(x), if x is an object you should be able to assume that f won\'t modify x. But if x is a pointer, then you should assume that f might modify the object pointed to by x.)

回答10:

Simple example that you probably have seen many times before

int main(int argc, char **argv)

In the second parameter you have it: pointer to pointer to char.

Note that the pointer notation (char* c) and the array notation (char c[]) are interchangeable in function arguments. So you could also write char *argv[]. In other words char *argv[] and char **argv are interchangeable.

What the above represents is in fact an array of character sequences (the command line arguments that are given to a program at startup).

See also this answer for more details about the above function signature.

回答11:

For example, you might want to make sure that when you free the memory of something you set the pointer to null afterwards.

void safeFree(void** memory) {
    if (*memory) {
        free(*memory);
        *memory = NULL;
    }
}

When you call this function you\'d call it with the address of a pointer

void* myMemory = someCrazyFunctionThatAllocatesMemory();
safeFree(&myMemory);

Now myMemory is set to NULL and any attempt to reuse it will be very obviously wrong.

回答12:

For instance if you want random access to noncontiguous data.

p -> [p0, p1, p2, ...]  
p0 -> data1
p1 -> data2

-- in C

T ** p = (T **) malloc(sizeof(T*) * n);
p[0] = (T*) malloc(sizeof(T));
p[1] = (T*) malloc(sizeof(T));

You store a pointer p that points to an array of pointers. Each pointer points to a piece of data.

If sizeof(T) is big it may not be possible to allocate a contiguous block (ie using malloc) of sizeof(T) * n bytes.

回答13:

One thing I use them for constantly is when I have an array of objects and I need to perform lookups (binary search) on them by different fields.
I keep the original array...

int num_objects;
OBJECT *original_array = malloc(sizeof(OBJECT)*num_objects);

Then make an array of sorted pointers to the objects.

int compare_object_by_name( const void *v1, const void *v2 ) {
  OBJECT *o1 = *(OBJECT **)v1;
  OBJECT *o2 = *(OBJECT **)v2;
  return (strcmp(o1->name, o2->name);
}

OBJECT **object_ptrs_by_name = malloc(sizeof(OBJECT *)*num_objects);
  int i = 0;
  for( ; i<num_objects; i++)
    object_ptrs_by_name[i] = original_array+i;
  qsort(object_ptrs_by_name, num_objects, sizeof(OBJECT *), compare_object_by_name);

You can make as many sorted pointer arrays as you need, then use a binary search on the sorted pointer array to access the object you need by the data you have. The original array of objects can stay unsorted, but each pointer array will be sorted by their specified field.

回答14:

As said one of the application of double poinnter is to update the string so that changes made are reflected back.

#include <iostream>
#include <cstring>  // for using strcpy
using namespace std;

void change(char **temp)
{
   strcpy(temp[0],\"new\");
   strcpy(temp[1],\"value\");
}

int main()
{
   char **str;
   str = (char **)malloc(sizeof(char *)*3);
   str[0]=(char *)malloc(10);
   str[1]=(char *)malloc(10);
   strcpy(str[0],\"old\");
   strcpy(str[1],\"name\");
   char **temp = str;  // always use the temporary variable
   while(*temp!=NULL)
{
    cout<<*temp<<endl;
    temp++;
}
temp = str;   // making it point it to the original head because we have changed the address in while loop above
change(str);
while(*temp!=NULL)
{
   cout<<*temp<<endl;
   temp++;
}

free(temp);
free(str[0]);
free(str[1]);
free(str);

回答15:

I have used double pointers today while I was programming something for work, so I can answer why we had to use them (it\'s the first time I actually had to use double pointers). We had to deal with real time encoding of frames contained in buffers which are members of some structures. In the encoder we had to use a pointer to one of those structures. The problem was that our pointer was being changed to point to other structures from another thread. In order to use the current structure in the encoder, I had to use a double pointer, in order to point to the pointer that was being modified in another thread. It wasn\'t obvious at first, at least for us, that we had to take this approach. A lot of address were printed in the process :)).

You SHOULD use double pointers when you work on pointers that are changed in other places of your application. You might also find double pointers to be a must when you deal with hardware that returns and address to you.

回答16:

Why double pointers?

The objective is to change what studentA points to, using a function.

#include <stdio.h>
#include <stdlib.h>


typedef struct Person{
    char * name;
} Person; 

/**
 * we need a ponter to a pointer, example: &studentA
 */
void change(Person ** x, Person * y){
    *x = y; // since x is a pointer to a pointer, we access its value: a pointer to a Person struct.
}

void dontChange(Person * x, Person * y){
    x = y;
}

int main()
{

    Person * studentA = (Person *)malloc(sizeof(Person));
    studentA->name = \"brian\";

    Person * studentB = (Person *)malloc(sizeof(Person));
    studentB->name = \"erich\";

    /**
     * we could have done the job as simple as this!
     * but we need more work if we want to use a function to do the job!
     */
    // studentA = studentB;

    printf(\"1. studentA = %s (not changed)\\n\", studentA->name);

    dontChange(studentA, studentB);
    printf(\"2. studentA = %s (not changed)\\n\", studentA->name);

    change(&studentA, studentB);
    printf(\"3. studentA = %s (changed!)\\n\", studentA->name);

    return 0;
}

/**
 * OUTPUT:
 * 1. studentA = brian (not changed)
 * 2. studentA = brian (not changed)
 * 3. studentA = erich (changed!)
 */

回答17:

Hopefully the following example will clear some concepts regarding pointers and double pointers , their differences and usage in common scenarios.

    int* setptr(int *x)
    {
        printf(\"%u\\n\",&x);
        x=malloc(sizeof(int));
        *x=1;
        return x;
    }

In the above function setptr we can manipulate x either
1. by taking fn arg as int *x , doing  malloc and setting value of x and return x 
Or
    2. By taking arg as int ** and malloc and then set  **x value to some value.
Note: we cant set any general pointer directly without doing  malloc.Pointer indicates that it is a type of variable which can hold address of any data type.Now either we define a variable and give reference to it or we declare a pointer(int *x=NULL) and allocate some memory to it inside the called function where we pass x or a reference to it .. In either case we need to have address of a memory in the  pointer and in the case pointer initially points  to NULL or it is defined like int *x where it points  to any random address then we need to assign a valid memory address to pointer 

    1. either we need to allocate memory to it by malloc

    int *x=NULL means its address is 0.
    Now we need to either o following
    1.



    void main()
        {
            int *x;
            x=malloc
            *x=some_val;
        }
        Or
        void main()
        {
            int *x
            Fn(x);
        }

        void Fn(int **x)
        {
            *x=malloc;
            **x=5;
        }
        OR
        int * Fn(int *x)
        {
            x=malloc();
            *x=4;
            Return x;
        }


        2. Or we need to point it to a valid memory like a defined variable inside the function where pointer is defined.


        OR
        int main()
        {
            int a;
            int *x=&a;
            Fn(x);
            printf(\"%d\",*x);
        }
        void Fn(int *x)
        {
            *x=2;
        }


     in both cases value pointed by x is changed inside fn

    But suppose if we do like


    int main()
    {
        int *x=NULL;
        printf(\"%u\\n\",sizeof(x));
        printf(\"%u\\n\",&x);
        x=setptr(x);
        //*x=2;
        printf(\"%d\\n\",*x);
        return 0;
    }

/* output
4
1
*/

#include<stdio.h>
void setptr(int *x)
{
    printf(\"inside setptr\\n\");
    printf(\"x=%u\\n\",x);
    printf(\"&x=%u\\n\",&x);
    //x=malloc(sizeof(int));
    *x=1;
    //return x;
}
int main()
{
    int *x=NULL;
    printf(\"x=%u\\n\",x);
    printf(\"&x=%u\\n\",&x);
    int a;
    x=&a;
    printf(\"x=%u\\n\",x);
    printf(\"&a=%u\\n\",&a);
    printf(\"&x=%u\\n\",&x);
    setptr(x);
    printf(\"inside main again\\n\");

    //*x=2;
    printf(\"x=%u\\n\",x);
    printf(\"&x=%u\\n\",&x);
    printf(\"*x=%d\\n\",*x);
    printf(\"a=%d\\n\",a);
    return 0;
}

回答18:

application of double pointer as shown by Bhavuk Mathur seems to be wrong. Here following example is the valid one

void func(char **str)
{
     strcpy(str[0],\"second\");
}

int main(){

    char **str;
    str = (char **)malloc(sizeof(char*)*1); // allocate 1 char* or string
    str[0] = (char *)malloc(sizeof(char)*10);      // allocate 10 character
    strcpy(str[0],\"first\");            // assign the string
    printf(\"%s\\n\",*str);
    func(str);            
    printf(\"%s\\n\",*str);           
    free(str[0]); 
    free(str);
}

回答19:

the following example, that i am giving will give an insight or an intuition about how double pointers work, i will go through the steps

1) try to understand the following statements
   char **str ;

   a) str is of type char ** whose value is an address of another pointer.
   b) *str is of type char * whose value is an address of variable or (it is a string itself).
   c) **str is of type char ,gives the value stored, in this case a character.

the following is the code to which you can relate to above points(a,b,c) to understand

str = (char **)malloc(sizeof(char *) *2); // here i am assigning the mem for two char *
       str[0]=(char *)\"abcdefghij\"; // assignin the value
       str[1]=(char *)\"xyzlmnopqr\"; 

now to print the value i.e strings in the array, just look at point b, In case of string the value as well as address is same, so no need to dereference it again.

cout<<*str<<endl;   // abcdefghij;

now to print the next string, come out of one dereference i.e (*) from *str to str and then increment , like shown below

str++;

now print the string

cout<<*str<<endl;        //xyzlmnopqr

now to print only characters in a string, refer to point c)

cout<<**str<<endl;  // prints the first character i.e \"a\"

now to print the next character of a string i.e \"b\" come out of 1 dereference operator and increment it i.e going from **str to *str and do *str++

*str++;

now print the character

cout<<**str<<endl;  // prints the second character i.e \"b\"

since the two arrays(\"abcdefghij\",\"xylmnopqr\") are stored in continous block of memory if same thing is done of incrementing the address, all characters of two strings will get printed