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我正在解析对“net.connman.Manager.GetServices”函数的响应，如下所示：

<method name="GetServices">
    <arg name="services" type="a(oa{sv})" direction="out"/>
</method>

这是相当复杂的结构。

到目前为止，我得到的是：

GVariant* result = ... // response containing data
GVariantIter* iter1;
g_variant_get( result, "a(oa{sv})", &iter1 );

GVariant* child = g_variant_iter_next_value( iter1 );
while ( nullptr != child )
{
    gchar* string;
    GVariant* data;
    g_variant_get( child, "(oa{sv})", &string, &data );

    // how to access inner array?

    g_variant_unref( child );
    child = g_variant_iter_next_value( iter1 );
}

g_variant_iter_free( iter1 );

那么，如何访问内部数组数据呢？

我试过这个： GVariantIter* iter2; g_variant_get（数据，“a{sv}”，&iter2）；GVariant* child2 = g_variant_iter_next_value(iter2);

但它因一些对齐错误而失败：

**
GLib:ERROR:../../glib-2.48.2/glib/gvarianttypeinfo.c:163:g_variant_type_info_check: assertion failed: (info->alignment == 0 || info->alignment == 1 || info->alignment == 3 || info->alignment == 7)
Aborted
4

1 回答 1

3

data根据GVariant 格式字符串的文档，应该具有 type GVariantIter*，而不是（您将 GVariant 格式字符串作为第二个参数传递给）。GVariant*g_variant_get()

您可以使用以下方法简化代码g_variant_iter_loop()：

/* Compile with: `gcc `pkg-config --cflags --libs glib-2.0 gio-2.0` -o test test.c`.
 * Public domain. */

#include <glib.h>
#include <gio/gio.h>

int
main (void)
{
  g_autoptr(GVariant) result = g_variant_new_parsed ("@a(oa{sv}) [('/', { 'hello': <'test'>})]");

  g_autoptr(GVariantIter) iter1 = NULL;
  g_variant_get (result, "a(oa{sv})", &iter1);

  const gchar *string;
  g_autoptr(GVariantIter) iter2 = NULL;

  while (g_variant_iter_loop (iter1, "(&oa{sv})", &string, &iter2))
    {
      const gchar *key;
      g_autoptr(GVariant) value = NULL;

      while (g_variant_iter_loop (iter2, "{&sv}", &key, &value))
        g_message ("%s, %s:%p", string, key, value);
    }

  return 0;
}
于 2017-09-30T21:11:00.977 回答