如何从分配函数外部访问结构指针的值？

Question

我使用声明和初始化结构的文件

typedef struct {
    uint8_t * ringBuffer;
    size_t head;
    size_t tail;
    size_t capacity;
    bool full;
} RingBuffer;

static inline void ringBufferInit(RingBuffer* ringBuffer)
{

    ringBuffer = (RingBuffer *) malloc(sizeof(RingBuffer));

    //ringBuffer->ringBuffer = buffer;
    ringBuffer->capacity = 128;
    printf("%d \n", ringBuffer->capacity);
}

以及我调用此函数的文件：

int main()
{
    testRingBufferPushPopSingle();
}

void testRingBufferPushPopSingle()
{
    RingBuffer ringBuffer;
    ringBufferInit(&ringBuffer);
    size_t a = (ringBuffer->capacity);
    printf("%d \n", a);
}

问题是指针似乎没有在主函数中更新，因为程序打印

128

2125407052

只有 128 个值（在分配块中读取是正确的）。

知道我做错了什么吗？谢谢！

Answer 1

函数ringBufferInit没有意义。在函数 testRingBufferPushPopSingle 中，您已经创建了一个 RingBuffer.

类型的对象

RingBuffer ringBuffer;

那么函数ringBufferInit要做的就是初始化已经创建的对象的数据成员ringBuffer

可以这样看

static inline void ringBufferInit(RingBuffer* ringBuffer)
{
    //ringBuffer->ringBuffer = buffer;
    ringBuffer->capacity = 128;
    printf("%d \n", ringBuffer->capacity);
}

否则在函数内

static inline void ringBufferInit(RingBuffer* ringBuffer)
{

    ringBuffer = (RingBuffer *) malloc(sizeof(RingBuffer));

    //ringBuffer->ringBuffer = buffer;
    ringBuffer->capacity = 128;
    printf("%d \n", ringBuffer->capacity);
}

你修改了它的局部变量ringBuffer（指针类型RingBuffer *）退出函数后不会存活

如何从分配函数外部访问结构指针的值？

How to access the value of a struct pointer from outsite of the allocatiing function?

c

memory

malloc

buffer

pointers