pmg*_*pmg 454
如果你想要一个字符列表(一个单词),你可以使用 char *word
如果你想要一个单词列表(一个句子),你可以使用 char **sentence
如果你想要一个句子列表(一个独白),你可以使用 char ***monologue
如果你想要一个独白列表(传记),你可以使用 char ****biography
如果您想要一份传记列表(生物图书馆),您可以使用 char *****biolibrary
如果你想要一个生物库列表(一个??大声笑),你可以使用 char ******lol
......
是的,我知道这些可能不是最好的数据结构
使用示例非常非常无聊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);
}
输出:
total words in my lol: 243
Ash*_*sha 162
一个原因是你想要将传递给函数的指针的值更改为函数参数,为此需要指向指针的指针.
简单地说,使用**时要保留(或保留在变化)的内存分配或分配甚至外面的函数调用.(所以,用双指针arg传递这样的函数.)
这可能不是一个很好的例子,但会告诉你基本用法:
void allocate(int** p)
{
*p = (int*)malloc(sizeof(int));
}
int main()
{
int* p = NULL;
allocate(&p);
*p = 42;
free(p);
}
Bri*_*nos 78
这是一个简单的答案!!!!
但!如果你想要一个函数为你做这个,并且你希望结果在函数完成后保持,你需要做一些额外的工作,你需要一个新的指针3只是指向pointer1,并将指针3传递给函数.
这是一个有趣的例子(首先看看输出波纹管,了解!):
#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);
}
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!!!.
Sil*_*iu 46
添加Asha的响应,如果你使用指向示例下的单个指针(例如alloc1()),你将失去对函数内部分配的内存的引用.
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 = NULL;
alloc1(p);
//printf("%d ",*p);//undefined
alloc2(&p);
printf("%d ",*p);//will print 10
free(p);
return 0;
}
它发生的原因是alloc1指针是按值传入的.因此,当它被重新分配给malloc内部调用的结果时alloc1,更改不属于不同范围内的代码.
ziy*_*ang 22
今天我在这篇博文中看到了一个很好的例子,我在下面总结一下.
想象一下,你有一个链表中的节点结构,可能是
typedef struct node
{
struct node * next;
....
} node;
现在,您要实现一个remove_if函数,该函数接受删除条件rm作为参数之一并遍历链表:如果条目满足条件(类似rm(entry)==true),则其节点将从列表中删除.最后,remove_if返回链表的头部(可能与原始头部不同).
你可以写
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;
}
作为你的for循环.消息是,没有双指针,你必须维护一个prev变量来重新组织指针,并处理两种不同的情况.
但是使用双指针,你实际上可以写
// 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;
}
你现在不需要,prev因为你可以直接修改prev->next指向的内容.
为了使事情更清楚,让我们稍微遵循一下代码.在删除期间:
entry == *head:它将*head (==*curr) = *head->next- head现在指向新标题节点的指针.您可以通过直接将head内容更改为新指针来完成此操作.entry != *head:同样,*curr是prev->next指向的,现在指向entry->next.无论在哪种情况下,您都可以使用双指针以统一的方式重新组织指针.
Bha*_*hur 21
1.基本概念 -
当您声明如下: -
1. char*ch - (称为字符指针)
- ch包含单个字符的地址.
- (*ch)将取消引用字符的值..
2. char**ch -
'ch'包含一个字符指针数组的地址.(如1)
'*ch'包含单个字符的地址.(请注意,由于声明不同,它与1不同).
(**ch)将取消引用该字符的确切值.
添加更多指针会扩展数据类型的维度,从字符到字符串,再到字符串数组等等......您可以将它与1d,2d,3d矩阵相关联.
因此,指针的使用取决于您如何声明它.
这是一个简单的代码..
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.双指针的另一种应用 -
(这也包括通过引用传递)
假设您要从函数更新字符.如果您尝试以下方法: -
void func(char ch)
{
ch = 'B';
}
int main()
{
char ptr;
ptr = 'A';
printf("%c", ptr);
func(ptr);
printf("%c\n", ptr);
}
输出将是AA.这不起作用,因为你有"通过值传递"功能.
正确的方法是 -
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);
}
现在扩展此更新字符串而不是字符的要求.
为此,您需要在函数中接收参数作为双指针.
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);
}
在此示例中,方法需要双指针作为参数来更新字符串的值.
Jas*_*son 15
指针的指针也可以作为内存的"句柄"派上用场,你想要在函数之间传递一个"句柄"来重新定位内存.这基本上意味着函数可以更改句柄变量内指针所指向的内存,并且使用句柄的每个函数或对象都将正确指向新重定位(或分配)的内存.图书馆喜欢用"不透明"的数据类型来做这件事,那就是数据类型你不必担心他们正在做的事情与你所做的事情有关,你只需要绕过"处理器"之间的"句柄".库的函数在该内存上执行某些操作...库函数可以在内部分配和解除分配内存,而无需明确担心内存管理过程或句柄指向的位置.
例如:
#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;
}
希望这可以帮助,
贾森
int main(int argc, char **argv)
在第二个参数中你有它:指向char的指针.
请注意,指针表示法(char* c)和数组表示法(char c[])在函数参数中是可互换的.所以你也可以写char *argv[].换句话说char *argv[],char **argv是可以互换的.
上面所代表的实际上是一个字符序列数组(在启动时给予程序的命令行参数).
有关上述函数签名的更多详细信息,请参阅此答案.
例如，您可能要确保释放内存时将指针设置为null。
void safeFree(void** memory) {
if (*memory) {
free(*memory);
*memory = NULL;
}
}
调用此函数时，将使用指针地址进行调用
void* myMemory = someCrazyFunctionThatAllocatesMemory();
safeFree(&myMemory);
现在myMemory设置为NULL，任何重用它的尝试显然都是错误的。
聚会有点晚了，但希望这会对某人有所帮助。
在 C 数组中总是在堆栈上分配内存，因此函数不能返回（非静态）数组，因为当执行到达当前块的末尾时，分配在堆栈上的内存会自动释放。当您想要处理二维数组（即矩阵）并实现一些可以更改和返回矩阵的函数时，这真的很烦人。为此，您可以使用指针到指针来实现具有动态分配内存的矩阵：
/* Initializes a matrix */
double** init_matrix(int num_rows, int num_cols){
// Allocate memory for num_rows float-pointers
double** A = calloc(num_rows, sizeof(double*));
// return NULL if the memory couldn't allocated
if(A == NULL) return NULL;
// For each double-pointer (row) allocate memory for num_cols floats
for(int i = 0; i < num_rows; i++){
A[i] = calloc(num_cols, sizeof(double));
// return NULL if the memory couldn't allocated
// and free the already allocated memory
if(A[i] == NULL){
for(int j = 0; j < i; j++){
free(A[j]);
}
free(A);
return NULL;
}
}
return A;
}
这是一个插图：
double** double* double
------------- ---------------------------------------------------------
A ------> | A[0] | ----> | A[0][0] | A[0][1] | A[0][2] | ........ | A[0][cols-1] |
| --------- | ---------------------------------------------------------
| A[1] | ----> | A[1][0] | A[1][1] | A[1][2] | ........ | A[1][cols-1] |
| --------- | ---------------------------------------------------------
| . | .
| . | .
| . | .
| --------- | ---------------------------------------------------------
| A[i] | ----> | A[i][0] | A[i][1] | A[i][2] | ........ | A[i][cols-1] |
| --------- | ---------------------------------------------------------
| . | .
| . | .
| . | .
| --------- | ---------------------------------------------------------
| A[rows-1] | ----> | A[rows-1][0] | A[rows-1][1] | ... | A[rows-1][cols-1] |
------------- ---------------------------------------------------------
双指针到双指针A指向A[0]内存块的第一个元素，其元素是双指针本身。你可以把这些双指针想象成矩阵的行。这就是为什么每个双指针都为 double 类型的 num_cols 元素分配内存的原因。此外A[i]指向第 i 行，即A[i]指向A[i][0]并且这只是第 i 行的内存块的第一个双元素。最后，您可以使用 轻松访问第 i 行和第 j 列中的元素A[i][j]。
这是演示用法的完整示例：
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
/* Initializes a matrix */
double** init_matrix(int num_rows, int num_cols){
// Allocate memory for num_rows double-pointers
double** matrix = calloc(num_rows, sizeof(double*));
// return NULL if the memory couldn't allocated
if(matrix == NULL) return NULL;
// For each double-pointer (row) allocate memory for num_cols
// doubles
for(int i = 0; i < num_rows; i++){
matrix[i] = calloc(num_cols, sizeof(double));
// return NULL if the memory couldn't allocated
// and free the already allocated memory
if(matrix[i] == NULL){
for(int j = 0; j < i; j++){
free(matrix[j]);
}
free(matrix);
return NULL;
}
}
return matrix;
}
/* Fills the matrix with random double-numbers between -1 and 1 */
void randn_fill_matrix(double** matrix, int rows, int cols){
for (int i = 0; i < rows; ++i){
for (int j = 0; j < cols; ++j){
matrix[i][j] = (double) rand()/RAND_MAX*2.0-1.0;
}
}
}
/* Frees the memory allocated by the matrix */
void free_matrix(double** matrix, int rows, int cols){
for(int i = 0; i < rows; i++){
free(matrix[i]);
}
free(matrix);
}
/* Outputs the matrix to the console */
void print_matrix(double** matrix, int rows, int cols){
for(int i = 0; i < rows; i++){
for(int j = 0; j < cols; j++){
printf(" %- f ", matrix[i][j]);
}
printf("\n");
}
}
int main(){
srand(time(NULL));
int m = 3, n = 3;
double** A = init_matrix(m, n);
randn_fill_matrix(A, m, n);
print_matrix(A, m, n);
free_matrix(A, m, n);
return 0;
}