C前向引用结构 - 1)必须带指针?2)必须初始化?

Dav*_*eli 1 c struct pointers initialization forward-reference

我正试图在C中转发引用(嵌套)结构.

这意味着我有一个结构,在其中我引用了稍后声明的另一个结构.

如果我将嵌套结构声明为指针,并使用值初始化它,它就可以工作.

以下代码有效:

#include <stdio.h>

struct computer{
    double cost;
    int year;
    char cpu_type[16];
    struct cpu_speed *sp; //this is the question (1)
};

struct cpu_speed{
    int num;
    char type[16];
};

typedef struct computer SC;
typedef struct cpu_speed SS;

void DataR(SC *s);
void DataP(SC *s);

int main(){
   // this is question (2)
    SS speed = {4,"giga"};    
    SC model[2] = {
    { 0,1990, "intel", &speed},
    { 0,1990, "intel", &speed}
    };

    int i;
    for(i=0;i<2;i++) {
        printf("computer no. %d \n", i+1);
        DataR(&model[i]);
    }
    printf("here's what you entered: \n");
    for(i=0;i<2;i++) {
        printf("computer no. %d \n", i+1);
        DataP(&model[i]);
    }
    return 0;
}

void DataR(SC *s){
    printf("the cost of your computer: ");
    scanf("%lf", &s->cost);
    printf("the year of your computer: ");
    scanf("%d", &s->year);
    printf("the type of cpu inside your computer: ");
    scanf("%s", s->cpu_type);
    printf("the speed of the cpu: ");
    scanf("%d %s", &(s->sp->num), s->sp->type);

}

void DataP(SC *s){

    printf("the cost: %.2lf\n",s->cost); 
    printf("the year: %d\n",s->year); 
    printf("the cpu type: %s\n",s->cpu_type); 
    printf("the cpu speed: %d %s\n",s->sp->num, s->sp->type);
}
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如果我struct cpu_speed{...};在父(?)结构之前声明嵌套结构(即),我不需要使用指针,我也不需要初始化.

含义:

(1)我可以用struct cpu_speed speed;而不是struct cpu_speed *sp;.(2)我不需要给结构变量赋予初始化值.

我的问题再次出现 - 在前向引用结构中 - (1)你是否必须用指针声明它?(2)你是否必须初始化这些值?

els*_*ooo 5

结构仅由编译器用于对齐内存.因此,它需要知道struct成员的大小.

struct foo {
    struct bar *pointer;
};
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在这种情况下,sizeof(pointer)它与bar结构无关,因此编译器不需要知道任何其他内容.

但是,如果要将bar结构添加到foo结构中,则需要了解其各个成员.

struct bar {
    const char *string;
    uint64_t integer;
};

struct foo {
    struct bar sub;
};
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您需要先声明bar结构,foo因为编译器需要知道您所指的是什么.否则,它将如何知道如何处理这个(非法)代码:

struct bar {
    struct foo sub;
};

struct foo {
    struct bar sub;
};
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