将C结构转换为Delphi

Question

如何将此结构转换为Delphi？

typedef struct a_config {
    union {
        struct {
            char mode[10];
            char name[10];
        } dn;

        struct {
            int r;
        } sm;

        struct {
            int r;
        } xo;
    } config;
} a_config_t;

Answer 1

联合等于case of记录中的结构.并非所有事情都可以1:1翻译,但这可以.

请注意,包装是一个单独的问题.虽然在这种情况下再次可能不是问题.

type
  a_config_t = record
    config : record
      case integer of 
        0:(dn: record 
             mode : array[0..9] of ansichar;    
             name : array[0..9] of ansichar;
           end);
        1: (sm: record
             r: integer;
           end);
        2: (xo: record
             r: integer;
           end);
    end;
  end;

// Delphi has no eq for "struct x" in "struct x {} y" construct, only for the y 
a_config = a_config_t;  

Answer 2

AC/C++联合类似于Delphi变体记录.直译是:

type
  dn_t = record
    mode, name: array [0..9] of AnsiChar;
  end;

  sm_t = record
    r: integer;
  end;

  xo_t = record
    r: integer;
  end;

  a_config = record
    case integer of
    0: (dn: dn_t);
    1: (sm: sm_t);
    2: (xo: xo_t);
  end;

阅读文档中有关变体记录的更多信息:http://docwiki.embarcadero.com/RADStudio/en/Structured_Types#Variant_Parts_in_Records

在这种情况下,您可以通过删除嵌套记录来进行更简单的翻译:

type
  a_config = record
    case integer of
    0: (mode, name: array [0..9] of AnsiChar);
    1: (r: integer);
  end;