为什么不能将C ++ 11大括号初始化与宏一起使用？ [英] Why can't you use C++11 brace initialization with macros?

问题描述

我刚刚发现，在将参数传递给宏时，不能总是使用括号初始化。当ASSERT（）宏编译失败时，我发现了这一点。但是，以下示例说明了该问题：

I just discovered that you cannot always use brace initialization when passing arguments to macros. I found this when an ASSERT() macro failed to compile. However, the following example illustrates the problem:

#include <iostream>
#include <string>
using namespace std;

#define PRINT_SIZE( f ) cout << "Size=" << (f).size() << endl;

int main()
{
  PRINT_SIZE( string("ABC") );  // OK, prints: "Size=3"
  PRINT_SIZE( string{"ABC"} );  // OK, prints: "Size=3"

  PRINT_SIZE( string("ABCDEF",3) ); // OK, prints: "Size=3"
  PRINT_SIZE( string{"ABCDEF",3} ); // Error: macro 'PRINT_SIZE' passed 2 arguments, but takes just 1

   return 0;
}

是否存在无法使宏用于括号初始化的原因？

Is there a reason why macros cannot be made to work with brace initialization?

编辑：

此后我发现您还可以使用可变参数宏，并且可以完美解决问题：

I have since discovered that you can also use a variadic macro, and that solves the problem perfectly:

#include <iostream>
#include <string>
using namespace std;

#define PRINT_SIZE( ... ) cout << "Size=" << (__VA_ARGS__).size() << endl;

int main()
{
  PRINT_SIZE( string("ABC") );  // OK, prints: "Size=3"
  PRINT_SIZE( string{"ABC"} );  // OK, prints: "Size=3"

  PRINT_SIZE( string("ABCDEF",3) ); // OK, prints: "Size=3"
  PRINT_SIZE( string{"ABCDEF",3} ); // OK, prints: "Size=3"

  return 0;
}

推荐答案

列表分为几部分宏参数。

The list is split into several macro parameters. When you write

PRINT_SIZE( string{"ABCDEF",3} );

这会尝试使用以下命令扩展宏 PRINT_SIZE 两个参数，一个 string { ABCDEF 和一个 3} 失败。在许多情况下（包括您在内），可以通过添加另一对括号来解决此问题：

This attempts to expand the macro PRINT_SIZE with two parameters, one string{"ABCDEF" and one 3}, which fails. This can be worked around in many cases (including yours) by adding another pair of parentheses:

PRINT_SIZE( (string{"ABCDEF",3}) );

这些括号可防止参数分裂，因此 PRINT_SIZE 用单个参数（string { ABCDEF，3}）扩展（请注意，括号是该参数的一部分）。

These parentheses prevent the splitting of the argument, so that PRINT_SIZE is expanded with a single argument (string{"ABCDEF",3}) (note that the parentheses are part of the argument).

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