Question

我试图将一个G_Variant指针返回到一个字典，该字典将uint16作为键并将其映射到字节数组（即：＆＃34; a {qay}＆＃34;）。

这是我的尝试：

range

当我尝试在D-Bus上查询此属性时，它没有工作，我收到以下错误消息：

    #define KEY   0xDEAD
    #define BYTE1 0xBE
    #define BYTE2 0xEF
    GVariantBuilder *arrBuilder, *builder;
    GVariant *arr;

    //build array of bytes ('ay') 
    arrBuilder = g_variant_builder_new(G_VARIANT_TYPE("ay"));
    g_variant_builder_add(arrBuilder, "y", BYTE1);
    g_variant_builder_add(arrBuilder, "y", BYTE2);

    arr = g_variant_new("v",g_variant_new("ay", arrBuilder));

    //put it in a dict ('a{sv}')
    builder = g_variant_builder_new(G_VARIANT_TYPE("a{qay}"));
    g_variant_builder_add(builder, "{qay}", KEY, arr);
    return g_variant_builder_end(builder);

Answer 1

arr = g_variant_new("v",g_variant_new("ay", arrBuilder));

也许你应该结束数组构建器，以便它返回你的bytearray？

arr = g_variant_builder_end(arrBuilder);

编辑：我找到了答案：

以下代码将实现您想要做的事情，诀窍是字节数组是变体（一切都是变体）！

GVariant *arr_builder, *dict_builder;

/* Build the array of bytes */
arr_builder = g_variant_builder_new(G_VARIANT_TYPE("ay"));
g_variant_builder_add(arr_builder, "y",0xAA);
// Add some more if you want
GVariant *byte_array = g_variant_builder_end(arr_builder);
g_variant_builder_unref(arr_builder);

/* Build the dictionary, now is the trick */
dict_builder = g_variant_builder_new(G_VARIANT_TYPE("a{qv}"));
g_variant_builder_add(dict_builder, "{qv}", 0x0F0F, byte_array);
GVariant *dict = g_variant_builder_end(dict_builder);
g_variant_builder_unref(dict_builder);

/* Now you can print your variants (dict and byte_array) and print their type */
g_print(g_variant_print(dict,TRUE));
g_print(g_variant_get_type_string(dict));

我在尝试你所写的内容时遇到的错误和你做的工作相同=）

如何创建一个映射到GLib中的数组的字典？

1 个答案: