##对C（C ++）预处理器意味着什么？

I have a C program below:

我有一个C程序如下:

#define f(g,g2) g##g2
main()
{
int var12=100;
printf("%d",f(var,12));
}

when I run just the preprocessor it expands this as

当我只运行预处理器时,它将其扩展为

{
int var12=100;
printf("%d",var12);
}

which is the reason why the output is 100.

这就是输出为100的原因。

Can anybody tell me how/why the preprocessor expands var##12 to var12?

任何人都可以告诉我预处理器如何/为什么将var ## 12扩展为var12?

4 个解决方案

#1

nothing too fancy: ## tells the preprocessor to concatenate the left and right sides

没有什么太花哨的:##告诉预处理器连接左侧和右侧

see http://en.wikipedia.org/wiki/C_preprocessor#Token_concatenation

#2

because ## is a token concatenation operator for the c preprocessor.

因为##是c预处理器的标记连接运算符。

Or maybe I don't understand the question.

或许我不明白这个问题。

#3

## is Token Pasting Operator

##是令牌粘贴操作员

The double-number-sign or "token-pasting" operator (##), which is sometimes called the "merging" operator, is used in both object-like and function-like macros. It permits separate tokens to be joined into a single token and therefore cannot be the first or last token in the macro definition.

双数字符号或“令牌粘贴”运算符(##),有时称为“合并”运算符,用于类似对象和类函数的宏。它允许将单独的标记连接到单个标记中,因此不能是宏定义中的第一个或最后一个标记。

If a formal parameter in a macro definition is preceded or followed by the token-pasting operator, the formal parameter is immediately replaced by the unexpanded actual argument. Macro expansion is not performed on the argument prior to replacement.

如果宏定义中的形式参数在令牌粘贴运算符之前或之后,则形式参数将立即替换为未展开的实际参数。在更换之前不对参数执行宏扩展。

#4

#define f(g,g2) g##g2

#define f(g,g2)g ## g2

## is usued to concatenate two macros in c-preprocessor. So before compiling f(var,12) should replace by the preprocessor with var12 and hence you got the output.

##用于连接c-preprocessor中的两个宏。所以在编译之前f(var,12)应该用预处理器替换为var12,因此你得到了输出。

#1