如何初始化包含指向指针的指针的结构

Question

下面是结构，我想在C中创建这个结构的数组并初始化，但与如何初始化char **input和char **output混淆。

typedef struct _test_data_vector {
    char *api;
    char **input;
    char **output;
}vector_test_data;

下面是我尝试过的。

typedef struct _test_data_vector {
    char *api;
    char **input;
    char **output;
}vector_test_data;

vector_test_data data[] = {
    {
        "vector_push_back",
        {"1","2","3","4","5","6","7","8","9","10"},
        {"1","2","3","4","5","6","7","8","9","10"}
    }
};

Answer 1

非常接近，只需通过指定类型(从C99开始)

这是我在没有任何警告的情况下测试的内容：

typedef struct _test_data_vector {
    char *api;
    char **input;
    char **output;
}vector_test_data;

vector_test_data data[] = {
    {
        "vector_push_back",
        (char*[]){"1","2","3","4","5","6","7","8","9","10"},
        (char*[]){"1","2","3","4","5","6","7","8","9","10"}
    }
};

printf("TEST: %s", data[0].input[2]);

Otput:测试:3

Answer 2

具体一点，通常当你必须初始化一个指向一个指针的指针时，你可以这样做：

char * charPointer;//this is the char you want the input to point at;
*input = charPointer;

我不得不在项目上使用指针，但如果有什么方法可以避免这一点，那就更容易了

Answer 3

默认情况下，api、input和output将初始化为NULL。您可以直接初始化api，但必须在编译时通过定义数组或在运行时(可能通过malloc )分配输入和输出。你真的应该提供更多关于你正在尝试做什么的信息，以便获得更有帮助的答案。