[英]handling operations with macros in C
I'm new to dealing with Macros and I've just stumbled into an exercise that I can't figure it out. 我刚开始处理Macros而且我偶然发现了一个我无法理解的练习。 Could someone explain to me what's happening here?
有人可以向我解释这里发生了什么吗? If I compile i can see what the output is but i can't get it there myself.
如果我编译我可以看到输出是什么,但我自己无法得到它。 Thank you in advance!
先感谢您!
#define M(varname, index) ( ( (unsigned char*) & varname )[index] )
int main(void) {
int a = 0x12345678;
printf( "%x %x\n", M(a,0), M(a,3) );
printf( "%x %x\n", M(a,1), M(a,2) );
}
Each Macro usage M(x,y)
is replaced with ( (unsigned char*) & x )[y]
每个宏用法
M(x,y)
替换为( (unsigned char*) & x )[y]
So your code looks like this after preprocessing: 所以你的代码在预处理后看起来像这样:
int main(void) {
int a = 0x12345678;
printf( "%x %x\n", ( (unsigned char*) & a )[0], ( (unsigned char*) & a )[3] );
printf( "%x %x\n", ( (unsigned char*) & a )[1], ( (unsigned char*) & a )[2] );
}
Like Thomas B Preusser added in the OP Question comments, most C compiler suites allow to get the pre-processed code with certain compiler flags or tools like fe with GCC as mentioned here. 就像Thomas B Preusser在OP问题评论中添加的那样,大多数C编译器套件允许使用某些编译器标志或工具来获取预处理代码,例如此处提到的带有GCC的 fe 。
Macros work by replacing one thing with another before attempting to compile the code (preprocessing). 在尝试编译代码(预处理)之前,宏通过将一个东西替换为另一个东西来工作。 So in the example you gave:
所以在你给出的例子中:
#define M(varname, index) ( ( (unsigned char*) & varname )[index] )
Each time a M(varname, index)
occurs, it will be replaced by ( (unsigned char*) & varname )[index]
每次出现
M(varname, index)
时,它将替换为( (unsigned char*) & varname )[index]
In terms of what the code is doing, as an example (reformatted slightly for readability): 就代码的作用而言,作为示例(为了可读性而略微重新格式化):
printf("%x %x\n", ((unsigned char*) &a)[0], ((unsigned char*) &a)[3]);
This: 这个:
a
a
的地址 unsigned char*
unsigned char*
%x
, as hex ( %x
is for hex) %x
替换为十六进制( %x
表示十六进制) It repeats this for the 3rd element with the second %x
它用第二个
%x
重复第三个元素
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