C++多行字符串文字

有什么方法可以在C++中使用多行纯文本,常量文字,还有Perl？也许有一些#include文件的解析技巧？我想不出一个,但男孩,那会很好.我知道它将在C++ 0x中.

嗯...排序.最简单的方法是使用编译器连接相邻字符串文字的事实:

const char *text =
  "This text is pretty long, but will be "
  "concatenated into just a single string. "
  "The disadvantage is that you have to quote "
  "each part, and newlines must be literal as "
  "usual.";

缩进无关紧要,因为它不在引号内.

您也可以这样做,只要您注意逃避嵌入式换行符.不这样做,就像我的第一个答案那样,将无法编译:

const char *text2 =
  "Here, on the other hand, I've gone crazy \
and really let the literal span several lines, \
without bothering with quoting each line's \
content. This works, but you can't indent.";

再次注意每行末尾的反斜杠,它们必须紧接在行结束之前,它们正在转换源中的换行符,以便所有内容都像新行一样.您没有在反斜杠的位置获取字符串中的换行符.使用这种形式,你显然不能缩进文本,因为缩进将成为字符串的一部分,用随机空格拼写它.

在C++ 11中,您有原始字符串文字.类似于shell中的文本和Python和Perl以及Ruby等脚本语言.

const char * vogon_poem = R"V0G0N(
             O freddled gruntbuggly thy micturations are to me
                 As plured gabbleblochits on a lurgid bee.
              Groop, I implore thee my foonting turlingdromes.   
           And hooptiously drangle me with crinkly bindlewurdles,
Or I will rend thee in the gobberwarts with my blurlecruncheon, see if I don't.

                (by Prostetnic Vogon Jeltz; see p. 56/57)
)V0G0N";

保留字符串中的所有空格和缩进以及换行符.

这些也可以是utf-8 | 16 | 32或wchar_t(使用通常的前缀).

我应该指出,这里实际上并不需要转义序列V0G0N.它的存在将允许"在弦内".换句话说,我可以放

                "(by Prostetnic Vogon Jeltz; see p. 56/57)"

(注意额外的引号)和上面的字符串仍然是正确的.否则我也可以使用

const char * vogon_poem = R"( ... )";

引号内的parens仍然需要.

#define MULTILINE(...) #__VA_ARGS__
消耗括号之间的所有内容.
用单个空格替换任意数量的连续空白字符.

输入多行字符串的一种可能方便的方法是使用宏.仅当引号和括号平衡且不包含"顶级"逗号时才有效:

#define MULTI_LINE_STRING(a) #a
const char *text = MULTI_LINE_STRING(
  Using this trick(,) you don't need to use quotes.
  Though newlines and     multiple     white   spaces
  will be replaced by a single whitespace.
);
printf("[[%s]]\n",text);

用gcc 4.6或g ++ 4.6编译,产生: [[Using this trick(,) you don't need to use quotes. Though newlines and multiple white spaces will be replaced by a single whitespace.]]

请注意,,除非它包含在括号或引号中,否则不能在字符串中.单引号是可能的,但会创建编译器警告.

编辑:如评论中所述,#define MULTI_LINE_STRING(...) #__VA_ARGS__允许使用,.

你可以这样做:

const char *text = "This is my string it is "
     "very long";

由于一盎司的经验值得大量理论,我尝试了一个小测试程序MULTILINE:

#define MULTILINE(...) #__VA_ARGS__

const char *mstr[] =
{
    MULTILINE(1, 2, 3),       // "1, 2, 3"
    MULTILINE(1,2,3),         // "1,2,3"
    MULTILINE(1 , 2 , 3),     // "1 , 2 , 3"
    MULTILINE( 1 , 2 , 3 ),   // "1 , 2 , 3"
    MULTILINE((1,  2,  3)),   // "(1,  2,  3)"
    MULTILINE(1
              2
              3),             // "1 2 3"
    MULTILINE(1\n2\n3\n),     // "1\n2\n3\n"
    MULTILINE(1\n
              2\n
              3\n),           // "1\n 2\n 3\n"
    MULTILINE(1, "2" \3)      // "1, \"2\" \3"
};

const char *longString = R""""(
This is 
a very 
long 
string
)"""";

const char *text =
  "This text is pretty long, but will be\n"
  "concatenated into just a single string.\n"
  "The disadvantage is that you have to quote\n"
  "each part, and newlines must be literal as\n"
  "usual.";