宏中的__VA_ARGS__是什么意思? [英] What does __VA_ARGS__ in a macro mean?

问题描述

/* Debugging */
#ifdef DEBUG_THRU_UART0
#   define DEBUG(...)  printString (__VA_ARGS__)
#else
void dummyFunc(void);
#   define DEBUG(...)  dummyFunc()   
#endif

我已经在C编程的不同头文件中看到了这种表示法，我基本上理解了它在传递参数，但是我不明白这种三点表示法"是什么意思?

有人可以举例说明还是提供有关VA Args的链接?

解决方案

这是杂色宏.这意味着您可以使用任意数量的参数来调用它.这三个...类似于C语言中 variadic函数中使用的相同构造. >

这意味着您可以使用这样的宏

DEBUG("foo", "bar", "baz");

或带有任意数量的参数.

__ VA_ARGS__再次引用宏本身中的变量参数.

#define DEBUG(...)  printString (__VA_ARGS__)
               ^                     ^
               +-----<-refers to ----+

因此DEBUG("foo", "bar", "baz");将替换为printString ("foo", "bar", "baz")

/* Debugging */
#ifdef DEBUG_THRU_UART0
#   define DEBUG(...)  printString (__VA_ARGS__)
#else
void dummyFunc(void);
#   define DEBUG(...)  dummyFunc()   
#endif

I've seen this notation in different headers of C programming, I basically understood it's passing arguments, but I didn't understand what this "three dots notation" is called?

Can someone explain it with example or provide links also about VA Args?

解决方案

It's a variadic macro. It means you can call it with any number of arguments. The three ... is similar to the same construct used in a variadic function in C

That means you can use the macro like this

DEBUG("foo", "bar", "baz");

Or with any number of arguments.

The __VA_ARGS__ refers back again to the variable arguments in the macro itself.

#define DEBUG(...)  printString (__VA_ARGS__)
               ^                     ^
               +-----<-refers to ----+

So DEBUG("foo", "bar", "baz"); would be replaced with printString ("foo", "bar", "baz")

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