结构通过指针数组引用。

Question

我正在尝试设置指向结构的指针数组。我尝试过两种方法，这两种方法都给了我编译错误。下面是代码示例。

如能提供任何协助，将不胜感激。

在第一个示例中，我得到了一个编译警告

struct stPlanets *pPlanets = &planets;

警告是：

警告:从不兼容的指针类型‘struct stPlanets ()5’初始化‘struct stPlanets’-warning兼容-指针类型结构stPlanets *pPlanets =和行星；

示例01。

#include <stdio.h>

int main(void)
{
    struct stPlanets {
        char    *planet;
        char    *deity[10];
    };  

    struct stPlanets planets[] = { 
        {"Mercury", {"Messenger"}},
        {"Venus", {"Love, Beauty"}},
        {"Earth", {"Earthly things"}},
        {"Mars", {"War"}},
        {"Jupiter", {"Sky, Thunder, King of gods"}}
    };  

    struct stPlanets *pPlanets = &planets;

    int     i = 0;
    int     sPlanets = sizeof(planets) / sizeof(struct stPlanets);

    printf("There are %d planets.\n", sPlanets);

    for (i = 0; i < sPlanets; i++)
    {   
        printf("%-10s : %s.\n", planets[i].planet, planets[i].deity[0]);
    }   

    return 0;
}

示例02。

#include <stdio.h>

int main(void)
{
    struct stPlanets {
        char    *planet;
        char    *deity[10];
    };  

    struct stPlanets *planets[] = { 
        {"Mercury", {"Messenger"}},
        {"Venus", {"Love, Beauty"}},
        {"Earth", {"Earthly things"}},
        {"Mars", {"War"}},
        {"Jupiter", {"Sky, Thunder, King of gods"}}
    };  

    int     i = 0;
    int     sPlanets = sizeof(planets) / sizeof(struct stPlanets);

    printf("There are %d planets.\n", sPlanets);

    for (i = 0; i < sPlanets; i++)
    {   
        printf("%-10s : %s.\n", planets[i]->planet, planets[i]->deity[0]); //abbreviated to just print first instance.
    }   

    return 0;
}

在第二个示例中，我得到以下内容：

planets-06a.c:11:9: warning: braces around scalar initializer
         {"Mercury", {"Messenger"}},
         ^
planets-06a.c:11:9: note: (near initialization for ‘planets[0]’)
planets-06a.c:11:10: warning: initialization of ‘struct stPlanets *’ from incompatible pointer type ‘char *’ [-Wincompatible-pointer-types]
         {"Mercury", {"Messenger"}},
          ^~~~~~~~~
planets-06a.c:11:10: note: (near initialization for ‘planets[0]’)
planets-06a.c:11:9: warning: braces around scalar initializer
         {"Mercury", {"Messenger"}},

Answer 1

在第一个代码片段中，您必须编写

struct stPlanets *pPlanets = planets;

而不是

struct stPlanets *pPlanets = &planets;

或者你可以写

struct stPlanets ( *pPlanets )[5] = &planets;

也就是说，用作初始化器的数组指示符被隐式转换为指向其第一个元素的指针。

所以在这份声明中

struct stPlanets *pPlanets = planets;

声明的对象和初始化器(在隐式转换之后)具有相同的struct stPlanets *类型。

在本声明中

struct stPlanets *pPlanets = &planets;

初始化程序的类型为struct stPlanets ( * )[5]。因此，编译器发出消息，声明的对象和初始化程序具有不兼容的类型。

在第二个代码片段中，这个声明

struct stPlanets *planets[] = { 
    {"Mercury", {"Messenger"}},
    {"Venus", {"Love, Beauty"}},
    {"Earth", {"Earthly things"}},
    {"Mars", {"War"}},
    {"Jupiter", {"Sky, Thunder, King of gods"}}
};

不正确，因为首先需要定义将由指针数组的元素指向的结构对象。

这是你最新的程序。

#include <stdio.h>

int main(void)
{
    struct stPlanets {
        char    *planet;
        char    *deity[10];
    };  

    struct stPlanets planets[] = { 
        {"Mercury", {"Messenger"}},
        {"Venus", {"Love, Beauty"}},
        {"Earth", {"Earthly things"}},
        {"Mars", {"War"}},
        {"Jupiter", {"Sky, Thunder, King of gods"}}
    };  

    struct stPlanets *pPlanets = planets;

    size_t     sPlanets = sizeof(planets) / sizeof( *planets );

    printf("There are %zu planets.\n", sPlanets);

    for ( size_t i = 0; i < sPlanets; i++ )
    {   
        printf("%-10s : %s.\n", pPlanets[i].planet, pPlanets[i].deity[0]);
    }   

    return 0;
}

其输出是

Mercury    : Messenger.
Venus      : Love, Beauty.
Earth      : Earthly things.
Mars       : War.
Jupiter    : Sky, Thunder, King of gods.