C ++匿名结构 [英] C++ anonymous structs

问题描述

我使用以下联合来简化字节，半字节和位操作:

I use the following union to simplify byte, nibble and bit operations:

union Byte
{
  struct {
    unsigned int bit_0: 1;
    unsigned int bit_1: 1;
    unsigned int bit_2: 1;
    unsigned int bit_3: 1;
    unsigned int bit_4: 1;
    unsigned int bit_5: 1;
    unsigned int bit_6: 1;
    unsigned int bit_7: 1;
  };

  struct {
    unsigned int nibble_0: 4;
    unsigned int nibble_1: 4;
  };

  unsigned char byte;
};

效果很好，但还会生成此警告:

It works nice, but it also generates this warning:

警告:ISO C ++禁止使用匿名结构[-pedantic]

warning: ISO C++ prohibits anonymous structs [-pedantic]

好的，很高兴知道.但是...如何从我的g ++输出中得到这个警告?如果没有这个问题，是否有可能写这样的联合体?

Ok, nice to know. But... how to get this warning out of my g++ output? Is there a possibility to write something like this union without this issue?

推荐答案

gcc编译器选项

The gcc compiler option -fms-extensions will allow non-standard anonymous structs without warning.

(启用了它认为的"Microsoft扩展")

(That enables what it considers "Microsoft extensions")

使用此约定，您还可以在有效的C ++ 中实现相同的效果.

You can also achieve the same effect in valid C++ using this convention.

union Byte
{
  struct bits_type {
    unsigned int _0: 1;
    unsigned int _1: 1;
    unsigned int _2: 1;
    unsigned int _3: 1;
    unsigned int _4: 1;
    unsigned int _5: 1;
    unsigned int _6: 1;
    unsigned int _7: 1;
  } bit;
  struct nibbles_type {
    unsigned int _0: 4;
    unsigned int _1: 4;
  } nibble;
  unsigned char byte;
};

这样，您的非标准byte.nibble_0就会成为合法的byte.nibble._0

With this, your non-standard byte.nibble_0 becomes the legal byte.nibble._0

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