这个符号**在C语言中意味着什么

Question

Hi I'm new to the C language, can someone explains what ** symbol mean.

typedef struct _TREENODE {
        struct _TREENODE *Left, *Right;
        TCHAR key[KEY_SIZE];
        LPTSTR pData;
    } TREENODE, *LPTNODE, **LPPTNODE;

Answer 1

If x is a pointer, *x dereferences it. **x is the same as *(*x) , so **x dereferences a pointer to a pointer. (eg, it is the thing that is pointed to by the thing that x opints to).

Answer 2

** is a pointer to pointer, it is also used for dereferencing a pointer variable.

 eg: int a=10,*b,**c;

    b=&a;
    c=&b;
    printf("the a value is:%d\n",a);

    printf("the b value is:%d\n",*b);
    printf("the c value is:%d\n",**c);

just execute this code you will get the idea about pointer to pointer.

Answer 3

There are two things to know about * in C:

1. It's an operation . Doing *x on a pointer dereferences that pointer . Doing **x on a pointer can dereference a pointer to a pointer , and so on.

2. It's a type . Declaring a type of int *x means that it's a pointer to an int type. Declaring int **x means that it's a pointer to a pointer to an int type.

Example:

int main() {
   int foo = 4;
   int *bar = &foo;  // declaring a pointer to int type *bar
   int **baz = &bar;  // declaring a pointer to a pointer to int type **baz
   printf("foo: %d, *bar: %d, **baz: %d\n", foo, *bar, **baz);  // derefencing the pointer *bar and **baz
   return 0;
}

Answer 4

In you want to change a variable, you pass it by pointer (to the variable ).

And if you want to change a pointer, you also pass it by pointer (to the pointer ) which is a double pointer.

Answer 5

In a declaration, ** means pointer to a pointer. When evaluating an expression, ** dereferences a pointer to a pointer.

int** p; // Declares p to be a pointer to a pointer.

And...

**p = 10;  // Dereferences p and assigns 10 to a memory location.

One common use of pointers to pointers is to represent dynamic 2D arrays. For example, if you want to create a matrix of M rows and N columns, you could do:

 int** matrix = malloc(M*sizeof(*matrix));
 int i = 0, j = 0;
 for ( i = 0; i < M; ++i )
    matrix[i] = malloc(N*sizeof(*matrix[0]));

Usage of the double pointer:

 for ( i = 0; i < M; ++i )
    for ( j = 0; j < N; ++j )
       matrix[i][j] = 0; // Assigns a value to the element
                         // at the i-th row and j-th column.

If you want to use string pointer dereferencing, you would use:

 for ( i = 0; i < M; ++i )
    for ( j = 0; j < N; ++j )
       *(*(matrix+i)+j) = 0;

Memory allocated for the matrix has to be freed in two passes also.

 for ( i = 0; i < M; ++i )
    free(matrix[i]);
 free(matrix);

Answer 6

** means a pointer to a pointer.

Most of the time I like to think of it as "pointer(s)" to "a memory area". Which in fact may be a little redundant.

For example, suppose you have to dynamically store several words on memory, how would you do that? There are several ways to do this, but I'll provide an example that illustrate the use of ** .

Now, suppose you want to store three words: hi , hello and goodbye

hi , hello and goodbye are strings, they consume, 2, 5 and 7 bytes on memory respectively. Well, in fact it's 3, 6 and 8 bytes because of the \\0 , but lets not get into many details.

But one thing is clear, we need three memory areas to hold these strings and also three pointers to reference these memory areas later.

Note that one can just declare three pointers that points to these memory areas, but, would you be willing to declare one thousand pointers to hold one thousand words? This is where ** kicks in.

Example:

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

#define NUMBER_OF_WORDS 3

int
main(int argc, char **argv)
{
        int  i;
        char **words;

        /* pointers */
        words    = malloc(sizeof(char*)*NUMBER_OF_WORDS);

        /* memory areas*/
        words[0] = strdup("Hi");
        words[1] = strdup("Hello");
        words[2] = strdup("Goodbye");

        for(i=0; i < NUMBER_OF_WORDS; i++)
                printf("%d) %s\n", i, words[i]);

        for(i=0; i < NUMBER_OF_WORDS; i++)
                free(words[i]); /* memory area */
        free(words); /* pointers */

        return 0;
}