函数式宏与宏 [英] function-like macro versus macros

问题描述

gcc (GCC) 4.7.2
c89

你好

我一直在看一个测试套件，我注意到这样声明的类似于函数的宏是这样的:

I have been looking at a test suite and I noticed this function-like macro declared like this:

#define MU_SUITE_START() char *msg = NULL

但是，执行此操作有什么真正的区别:

However, is there any real difference to doing just this instead:

#define MU_SUITE_START char *msg = NULL

宏只做文本替换，所以我认为不会有任何性能问题.

Macros just do text replacement so I don't think there is going to be any performance issues.

使用cpp我得到以下结果，所以这里没什么特别的.

Using cpp I get the following results so nothing special here.

类似函数的宏

char *msg = __null;

马可

char *msg = __null;

请注意:是否值得声明输入参数中未提供的类似函数的marco? 这里我没有传递任何参数 即

Just a side note: Is it worth declaring a function-like marco that doesn't provide in input parameters? Here I am not passing any parameters i.e.

#define PRINT_MSG() printf("This is the message\n")

如果没有输入参数，为什么还要烦扰类似于函数的宏，这不是更好吗?

Why bother with the function-like macro if there are no input parameters, isn't this better?

#define PRINT_MSG printf("This is the message\n")

非常感谢您的任何建议，

Many thanks for any suggestions,

推荐答案

类对象宏和类函数宏之间有一个区别:

There's one difference between the object-like and function-like macros:

#define OBJECT     char *msg1 = NULL
#define FUNCTION() char *msg2 = NULL

void somefunc(void)
{
    int OBJECT = 5;
    int FUNCTION = 10;
    ...
}

OBJECT的声明被宏替换(因此代码不会编译)，但是对FUNCTION的引用不是宏调用，因为它后面没有括号.

The declaration for OBJECT is replaced by the macro (so the code won't compile), but the reference to FUNCTION is not a macro invocation because it is not followed by an open parenthesis.

这很少重要.但是，确实如此很重要.

This is seldom important. However, when it is, it really matters.

更典型的情况是可以实现为宏的函数.为了讨论起见(因为它很容易理解，而不是因为它是一个很好的例子):

A more typical case might be a function that can be implemented as a macro. For sake of discussion (because it is easily understood rather than because it is a good example):

extern int isdigit(int c);
#define isdigit(c) ((c) >= '0' && (c) <= '9')

并在实现文件中:

int (isdigit)(int c)
{
    assert((c >= 0 && c <= UCHAR_MAX) || c == EOF);
    return isdigit(c);
}

忽略这样的小细节并不是实现isdigit()的方式，并且isdigit()的宏实现不允许多次评估其自变量，并且您不应该重新定义那些在标准C库中，函数定义未进行宏扩展，因为名称isdigit后没有(，但是函数内部的宏已扩展.至少该功能是在宏方面实现的，几乎可以保证相同的行为.

Ignoring little details like that isn't how isdigit() is likely to be implemented, and a macro implementation of isdigit() is not allowed to evaluate its argument more than once, and you aren't supposed to redefine things that are in the standard C library, the function definition is not macro expanded because the name isdigit is not followed by (, but the macro inside the function is expanded. At least the function is implemented in terms of the macro, which pretty much guarantees the same behaviour.

这篇关于函数式宏与宏的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！