为什么某些库例程同时实现为宏?为什么是"va_arg"?宏声明为函数(没有"#define")? [英] Why are some library routines implemented as macros simultaneously? Why is "va_arg" macro declared as a function (without "#define")?

问题描述

我很难用言语清楚地表达出来.因此，让我将其分为几部分.上下文摘自Mike Banahan的C书(下面各部分提供的链接).这是我的问题，用粗体字表示:

I am struggling to put it clearly in words. So let me put it in parts. The contexts are from the C book by Mike Banahan (Links provided with each part below). Here are my questions as bullet points in bold:

• 为什么某些库函数也同时实现为宏?有什么需要这就是我从《 (第9.1.1节):

最后一个一般要点是，许多库例程可能是 实现为宏，前提是没有问题要做 有副作用(如第7章所述).标准保证 如果正常情况下将函数实现为宏，则将 提供的真正功能也可以完成相同的工作.使用真实 函数，或者用 #undef 取消定义宏名称，或将其括起来 括号中的名称，以确保不会将其视为 宏:

A last general point is that many of the library routines may be implemented as macros, provided that there will be no problems to do with side-effects (as Chapter 7 describes). The Standard guarantees that, if a function is normally implemented as a macro, there will also be a true function provided to do the same job. To use the real function, either undefine the macro name with #undef, or enclose its name in parentheses, which ensures that it won't be treated as a macro:

众所周知，
• va_start 是函数还是宏?书中的以下内容是混淆的根源，因为它暗示着彼此在同一口气中，在相邻的行中！ (第9.9节)
• va_start, as we know it, is a function or macro? The following text from the book is the source of confusion as it implies both in the same breath, in adjacent lines!! (Section 9.9)

在尝试访问变量参数列表之前， 必须调用 va_start .定义为

Before any attempt can be made to access a variable argument list, va_start must be called. It is defined as

 #include <stdarg.h>
void va_start(va_list ap, parmN); 

va_start 宏会初始化ap，以供函数随后使用 va_arg 和 va_end .

The va_start macro initializes ap for subsequent use by the functions va_arg and va_end.

• 最后，最令人困惑的部分.在下面的行中，清楚地写出 va_arg 是一个宏，然后继续说明它是如何实现的.但是，如何在没有 #define 关键字的情况下实现宏，而对于带有返回类型('type')的宏也是如此，就好像它是一个函数一样? (第9.9节)
• And finally, the most confusing part. In the following line, it's clearly written that va_arg is a macro and goes on to show how it is implemented. But how can a macro be implemented without a #define keyword, and that too with return type ('type') as if it were a function? (Section 9.9)

初始化后，提供的参数可以顺序访问 通过 va_arg 宏.这是特殊的，因为类型 返回的值由宏的参数确定.请注意，这是 不可能实现为真正的功能，只能实现为宏.它是 定义为

Once initialized, the arguments supplied can be accessed sequentially by means of the va_arg macro. This is peculiar because the type returned is determined by an argument to the macro. Note that this is impossible to implement as a true function, only as a macro. It is defined as

#include <stdarg.h>
type va_arg(va_list ap, type);

非常感谢您的回答.谢谢.

Your answers will be very much appreciated. Thank you.

推荐答案

这是两个不同的问题.

首先，保证va_start，va_arg和va_end是宏. va_arg不能是一个函数，因为它的第二个参数是类型而不是值.

Firstly, va_start, va_arg and va_end are guaranteed to be macros. va_arg cannot be a function, because its second argument is a type rather than a value.

关于为什么某些函数也是宏的原因:因为您希望内联一个调用，使调用快速进行，但是您可能还想获取其地址并将其放在函数指针中:

As to why some functions are also macros: because you want a call to be inlined, making it fast, but you might also want to take its address and put that in a function pointer:

int (*function_pointer)(int, FILE *) = &(putc);

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