在C语言中什么时候应该使用双重间接?谁能举个例子解释一下吗?

我所知道的是双间接是指向指针的指针。为什么我需要指向指针的指针?


当前回答

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 **句子

如果你想要一个句子列表(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 ****传记

如果你想要一个传记列表(一个生物图书馆),你可以使用char *****biolibrary

如果你想要一个生物库列表(a ??lol),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


一个非常非常非常无聊的lol的用法例子

#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

其他回答

例如,你可能想要确保当你释放某个东西的内存时,你将指针设置为空。

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

当你调用这个函数时,你会用指针的地址来调用它

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

现在myMemory被设置为NULL,任何重用它的尝试都将是非常明显的错误。

一个原因是你想要改变传递给函数的作为函数参数的指针的值,要做到这一点,你需要指针指向指针。

简单地说,当你想在函数调用之外保留(或保留)内存分配或分配的变化时,使用**。(因此,传递带有双指针arg的函数。)

这可能不是一个很好的例子,但会告诉你基本的用法:

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

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

int main()
{
    int *p = NULL;
    allocate(&p);
    *p = 42;
    printf("%d\n", *p);
    free(p);
}

简单的例子,你可能已经见过很多次了

int main(int argc, char **argv)

在第二个参数中有它:指向char的指针的指针。

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[]和char **argv是可互换的。

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,请参见此回答。

添加到Asha的响应,如果你使用单个指针指向下面的例子(例如alloc1()),你将失去对函数内部分配的内存的引用。

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

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;
}

发生这种情况的原因是在alloc1中,指针是按值传入的。因此,当它被重新分配给alloc1内部的malloc调用的结果时,更改不属于不同作用域中的代码。

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 :)).

当你处理在应用程序其他地方被更改的指针时,你应该使用双指针。在处理返回和寻址给您的硬件时,您可能还会发现双指针是必须的。