在C中通过引用将指针传递给struct

Question

请注意以下代码：

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

typedef struct
{
    int a;
    int b;
    int c;
}A;

A *test;

void init(A* a)
{
    a->a = 3;
    a->b = 2;
    a->c = 1;
}
int main()
{
    test = malloc(sizeof(A));
    init(test);
    printf("%d\n", test->a);
    return 0;
}

运行良好！现在想象一下，我想malloc在main自身之外使用函数而不返回指向的指针struct。我将把malloc放在里面init并通过test地址。但这似乎不起作用。

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

typedef struct
{
    int a;
    int b;
    int c;
}A;

A *test;

void init(A** a)
{
    *a = malloc(sizeof(A));
    *a->a = 3;
    *a->b = 2;
    *a->c = 1;
}
int main()
{
    init(&test);
    printf("%d\n", test->a);
    return 0;
}

当我使用指针时，它一直告诉我int a（或b/ c）不是in的成员struct A。

Answer 1

Your problem is operator precedence. The -> operator has higher precedence than the * (dereference) operator, so *a->a is read as if it is *(a->a). Change *a->a to (*a)->a:

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

typedef struct
{
    int a;
    int b;
    int c;
}A;

A *test;

void init(A** a)
{
    *a = malloc(sizeof(A));
    (*a)->a = 3;
    (*a)->b = 2;
    (*a)->c = 1;
}
int main()
{
    init(&test);
    printf("%d\n", test->a);
    return 0;
}

Answer 2

您必须添加括号：

void init(A **a)
{
    *a = malloc(sizeof(A)); // bad you don't verify the return of malloc
    (*a)->a = 3;
    (*a)->b = 2;
    (*a)->c = 1;
}

但这是一个好习惯：

void init(A **a)
{
    A *ret = malloc(sizeof *ret); // we want the size that is referenced by ret
    if (ret != NULL) { // you should check the return of malloc
        ret->a = 3;
        ret->b = 2;
        ret->c = 1;
    }
    *a = ret;
}