将字符串文字与 char 文字连接起来

本文介绍了将字符串文字与 char 文字连接起来的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我想连接字符串文字和字符文字.由于语法不正确，"abc" 'd' "efg" 会导致编译器错误:

I want to concat a string literal and char literal. Being syntactically incorrect, "abc" 'd' "efg" renders a compiler error:

x.c:4:24: 错误:预期为 ',' 或 ';'在'd'之前

x.c:4:24: error: expected ',' or ';' before 'd'

现在我必须使用 snprift(不必要地)，尽管在编译时知道字符串文字和字符文字的值.

By now I have to use snprift (needlessly), despite the value of string literal and the char literal being know at compile time.

我试过了

#define CONCAT(S,C) ({ static const char *_r = { (S), (C) }; _r; })

但它不起作用，因为 S 的空终止符没有被剥离.(除了给出编译器警告.)

but it does not work because the null terminator of S is not stripped. (Besides of giving compiler warnings.)

有没有办法写一个宏来使用

Is there a way to write a macro to use

• "abc" MACRO('d') "efg" 或
• MACRO1(MACRO2("abc", 'd'), "efg") 或
• MACRO("abc", 'd', "efg") ?

• "abc" MACRO('d') "efg" or
• MACRO1(MACRO2("abc", 'd'), "efg") or
• MACRO("abc", 'd', "efg") ?

如果有人问我为什么要这样:char 字面量来自库，我需要将字符串作为状态消息打印出来.

In case someone asks why I want that: The char literal comes from a library and I need to print the string out as a status message.

推荐答案

如果你能接受单引号，你可以使用字符串化:

If you can live with the single quotes being included with it, you could use stringification:

#define SOME_DEF 'x' #define STR1(z) #z #define STR(z) STR1(z) #define JOIN(a,b,c) a STR(b) c int main(void) { const char *msg = JOIN("Something to do with ", SOME_DEF, "..."); puts(msg); return 0; }

取决于可能合适或不合适的上下文，但就以这种方式说服它实际上是字符串文字而言，这是在运行时无需格式化的唯一方法.

Depending on the context that may or may not be appropriate, but as far as convincing it to actually be a string literal buitl this way, it's the only way that comes to mind without formatting at runtime.