数字转换在宏字符串 [英] convert digital to string in macro

问题描述

  

可能重复：结果
  转换一个preprocessor令牌字符串

 的#define NUM 1234
 

我要定义基于NUM是为const char *，样品中这将是：

 的#define num_str1234
 

我可以写一个宏语句来实现这一目标？
注：1234将被改变。

感谢。

解决方案

当然可以，但宏替换可以得到一点点奇怪的。双宏替换是有原因的，如果你想想看一段时间，为什么需要它，它就会变得清晰起来。

 的#define STRINGIZER_（EXP）#exp
＃定义STRINGIZER（EXP）STRINGIZER_（EXP）
NUM的#define 1234INT主（INT ARGC，CHAR *的argv []）
{
    为const char * p = STRINGIZER（NUM）;
    的printf（％S \\ n，p）的;
    返回EXIT_SUCCESS;
}
 

运行以下命令：

  1234
 

究其原因，双重替代：乍一看，人们可能认为这将解决这个问题：

 的#define STRINGIZER（EXP）#exp
NUM的#define 1234INT主（INT ARGC，CHAR *的argv []）
{
    为const char * p = STRINGIZER（NUM）;
    的printf（％S \\ n，p）的;
    返回EXIT_SUCCESS;
}
 

但是，这会产生：

  NUM
 

这是不是我们所试图做的。如果你想在NUM宏观第一的实际扩张那么这是你必须做的：迫使扩张。通过迫使preprocessor替代的第一个的通过中间膨胀（如我在这个答案的顶部）传入的宏第一次扩大，然后的字符串-ized。

边栏：：该技术是用于生成，否则定期举行字符串predefined preprocessor宏宽字符版本特别有用。例如， __ __ FILE 宏。假设你想这个宽字符版本（字符串prepended以'L'）你可能首先想到这会工作：

 的#define WIDESTR（STR）长## STR
 

但 __ __ FILE这个扩展如：

 常量WCHAR * P = WIDESTR（__ FILE__）;
 

将导致编译器错误：未定义的标识符：L__FILE __

那么，如何才能解决这个问题？用同样的方法我们在上面做了。

 的#define WIDESTR_（STR）长## STR
＃定义WIDESTR（STR）WIDESTR_（STR）INT主（INT ARGC，CHAR *的argv []）
{
    常量为wchar_t * P = WIDESTR（__ FILE__）;
    wcout＆LT;＆LT; P＆LT;＆LT; ENDL;
    返回EXIT_SUCCESS;
}
 

在我的系统，这将产生：

  /Users/craig/tmp/main/main/test.cpp
 

最后...

作为安慰奖，我们在这个答案都结合成一个巨大咕堆，我们怎么想发生在我们做的这个的：

  INT的main（）
{
    常量为wchar_t * P = WIDESTR（STRINGIZE（NUM））;
    wcout＆LT;＆LT; P＆LT;＆LT; ENDL;
    返回EXIST_SUCCESS;
}
 

Possible Duplicate:
Convert a preprocessor token to a string

#define num 1234

I want to define a "const char*" based on num, in the sample it would be:

#define num_str "1234"

Can I write a macro statement to achieve this? NB: 1234 would be changed.

Thanks.

解决方案

Yes you can, but the macro substitution can get a little strange-looking. The double-macro substitution is there for a reason, and if you think about it for awhile, it will become clear why it is needed.

#define STRINGIZER_(exp)   #exp
#define STRINGIZER(exp)    STRINGIZER_(exp)
#define NUM 1234

int main(int argc, char *argv[])
{
    const char *p = STRINGIZER(NUM);
    printf("%s\n",p);
    return EXIT_SUCCESS;
}

Running this:

1234

The reason for the double substitution: At first glance one may think this will solve the problem:

#define STRINGIZER(exp)    #exp
#define NUM 1234

int main(int argc, char *argv[])
{
    const char *p = STRINGIZER(NUM);
    printf("%s\n",p);
    return EXIT_SUCCESS;
}

But this produces:

NUM

which is not what we're trying to do. If you want the actual expansion of the NUM macro first then that is what you have to do: force the expansion. By forcing the preprocessor to substitute first through an intermediate expansion (as I show at the top of this answer) the passed-in macro is expanded first, then string-ized.

Side Bar: This technique is particularly useful for generating wide-char versions of predefined preprocessor macros that otherwise hold regular strings. For example, the __FILE__ macro. Suppose you wanted a wide-char version of this (a string prepended with 'L') You may first think this will work:

#define WIDESTR(str)    L##str

but expanding this with __FILE__ as in:

const wchar *p = WIDESTR(__FILE__);

will result in a compiler error: "Undefined identifier: L__FILE__"

So how can we address this? The same way we did above.

#define WIDESTR_(str)       L##str
#define WIDESTR(str)        WIDESTR_(str)

int main(int argc, char *argv[])
{
    const wchar_t* p = WIDESTR(__FILE__);
    wcout << p << endl;
    return EXIT_SUCCESS;
}

On my system, this produces:

/Users/craig/tmp/main/main/test.cpp

In Closing...

As a consolation prize, we combine everything in this answer into one giant goo-pile, what do we suppose happens when we do this:

int main()
{
    const wchar_t *p = WIDESTR(STRINGIZE(NUM));
    wcout << p << endl;
    return EXIST_SUCCESS;
}

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