C/C++ 宏中的逗号
假设我们有一个像这样的宏
Say we have a macro like this
#define FOO(type,name) type name
我们可以像这样使用
FOO(int, int_var);
但并不总是那么简单:
FOO(std::map<int, int>, map_var); // error: macro "FOO" passed 3 arguments, but takes just 2
当然可以:
typedef std::map<int, int> map_int_int_t;
FOO(map_int_int_t, map_var); // OK
这不是很符合人体工程学.必须处理 Plus 类型的不兼容性.知道如何用宏解决这个问题吗?
which is not very ergonomic. Plus type incompatibilities have to be dealt with. Any idea how to resolve this with macro ?
推荐答案
因为尖括号也可以表示(或出现在)比较运算符<、>, <= 和 >=，宏扩展不能像在括号中那样忽略尖括号内的逗号.(这也是方括号和大括号的问题，即使它们通常作为平衡对出现.)您可以将宏参数括在括号中:
Because angle brackets can also represent (or occur in) the comparison operators <, >, <= and >=, macro expansion can't ignore commas inside angle brackets like it does within parentheses. (This is also a problem for square brackets and braces, even though those usually occur as balanced pairs.) You can enclose the macro argument in parentheses:
FOO((std::map<int, int>), map_var);
问题在于，参数在宏扩展内仍然用括号括起来，这会阻止它在大多数情况下被读取为类型.
The problem is then that the parameter remains parenthesized inside the macro expansion, which prevents it being read as a type in most contexts.
解决此问题的一个好技巧是，在 C++ 中，您可以使用函数类型从带括号的类型名称中提取类型名称:
A nice trick to workaround this is that in C++, you can extract a typename from a parenthesized type name using a function type:
template<typename T> struct argument_type;
template<typename T, typename U> struct argument_type<T(U)> { typedef U type; };
#define FOO(t,name) argument_type<void(t)>::type name
FOO((std::map<int, int>), map_var);
因为形成函数类型会忽略额外的括号，所以您可以在类型名称不包含逗号的情况下使用带或不带括号的宏:
Because forming function types ignores extra parentheses, you can use this macro with or without parentheses where the type name doesn't include a comma:
FOO((int), int_var);
FOO(int, int_var2);
当然，在 C 中，这不是必需的，因为类型名称不能包含括号外的逗号.因此，对于跨语言宏，您可以编写:
In C, of course, this isn't necessary because type names can't contain commas outside parentheses. So, for a cross-language macro you can write:
#ifdef __cplusplus__
template<typename T> struct argument_type;
template<typename T, typename U> struct argument_type<T(U)> { typedef U type; };
#define FOO(t,name) argument_type<void(t)>::type name
#else
#define FOO(t,name) t name
#endif
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