在 C 中初始化结构数组的编译器错误(数组必须用大括号括起来的初始化器初始化)

Question

编译器错误“错误:数组必须用大括号括起来的初始化器初始化”让我很困惑。这个问题的其他例子似乎都与此无关。我有大约 14 年没有接触 C 了，所以我认为“生锈”这个词有点大方。我确定我只是错过了一些愚蠢的事情。

typedef uint8_t DeviceAddress[8];
DeviceAddress Probe01 = { 0x28, 0xFF, 0x87, 0x5A, 0x91, 0x15, 0x04, 0xE0 }; 
DeviceAddress Probe02 = { 0x28, 0xFF, 0x97, 0x5E, 0x91, 0x15, 0x04, 0x92 };
DeviceAddress Probe03 = { 0x28, 0xFF, 0xCD, 0x81, 0x91, 0x15, 0x01, 0x1E };
DeviceAddress Probe04 = { 0x28, 0xFF, 0xA6, 0x69, 0x91, 0x15, 0x04, 0x15 };
DeviceAddress Probe05 = { 0x28, 0xFF, 0xD8, 0x7E, 0x91, 0x15, 0x04, 0x64 };

struct DeviceInfo {
  DeviceAddress addr;
  const char * name;
};

struct DeviceInfo devices[5] = {
  {.addr = Probe01, .name = "Pump1"},
  {.addr = Probe02, .name = "Pump2"},
  {.addr = Probe03, .name = "Pump3"},
  {.addr = Probe04, .name = "Pump4"},
  {.addr = Probe05, .name = "Pump5"}
};

Answer 1

struct DeviceInfo devices[5] = {
  {.addr = Probe01, .name = "Pump1"},
  {.addr = Probe02, .name = "Pump2"},
  {.addr = Probe03, .name = "Pump3"},
  {.addr = Probe04, .name = "Pump4"},
  {.addr = Probe05, .name = "Pump5"}
};

这里，addr 是DeviceAddress 类型，它只是一个uint8_t 数组。你不能在 C 中赋值给数组，所以编译器告诉你将 Probe1 赋值给 addr 字段是无效的；它在那里需要自己的大括号括起来的数组初始值设定项。

您有几个选择。您可以完全摆脱 Probe01、Probe02 等，并按照编译器的建议初始化数组:

struct DeviceInfo devices[5] = {
  {.addr = { 0x28, 0xFF, 0x87, 0x5A, 0x91, 0x15, 0x04, 0xE0 }, .name = "Pump1" },
  ...
};

另一个有点迂回的选择是有两个 typedef:

typedef uint8_t DeviceAddress[8];
typedef uint8_t * DeviceAddress_P;

struct DeviceInfo {
  DeviceAddress_P addr;
  const char * name;
};

并在结构中使用指针类型，将其初始化为指向您创建的数组的第一个元素:

DeviceAddress Probe01 = { 0x28, 0xFF, 0x87, 0x5A, 0x91, 0x15, 0x04, 0xE0 };

struct DeviceInfo devices[5] = {
  {.addr = Probe01, .name = "Pump1"},
  ...
};

但是，通过这种方式，结构只是指向一个外部数组，这可能是不可取的。