置字符串字面量字符字面 [英] Concatenate string literal with char literal

问题描述

我要Concat的字符串文字和char文字。作为语法不正确，ABCDEFG呈现一个编译器错误：

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'，'或';'前D

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

现在我必须使用snp​​rift（不必要的），尽管字符串字面值和字符文字之中知道在编译时。

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

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

但它不工作，因为取值的空终止，不会被剥离。 （除了给予编译器警告。）

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

有没有写宏使用方式

• ABCMACRO（'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") ?

在万一有人问，为什么我想的是：该字符文字来自图书馆，我需要打印字符串作为一个状态消息

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

如果你可以用被包含在其中的单引号生活，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;
}

根据，可能会或可能不适合，但就说服它实际上是一个字符串buitl这样的背景下，它是自带没有在运行时格式化想到的唯一办法。

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.

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