类似函数的宏的错误扩展 [英] Incorrect expansion of a function-like macro

问题描述

晚上好，
我对宏有问题.我应该拿出一个宏ENTRY，它将一个值放入数组(给出了scanf("%d",&ENTRY(x,i))).

Good evening,
I have a problem with macros. I am supposed to come up with a macro ENTRY, that puts a value into an array ( scanf("%d",&ENTRY(x,i)) was given).

我尝试过:#define ENTRY (a,b) (a[b-1])，但这没用.
它创建了一个编译器错误，指出未声明a和b.
但是我认为我不必声明在宏中使用的变量，尤其是因为，例如:#define min (a,b) ((a)<(b)?(a):(b))在另一个程序中工作过.

I tried: #define ENTRY (a,b) (a[b-1]), but that didn't work.
It created a compiler error that says, that a and b are undeclared.
But I thought that I don't have to declare variables used in macros, especially because, for example: #define min (a,b) ((a)<(b)?(a):(b)) worked in another program.

那我在做什么错了?

#include <stdio.h>
#define N 3
#define ENTRY (a,b) (a[b-1])

int main(void)
{

    int x[N],i;
    float y[N];

    for(i=1;i<=N;i++){ printf("x_%d = ",i);scanf("%d",&ENTRY(x,i));}
    for(i=1;i<=N;i++){ printf("y_%d = ",i);scanf("%lf",&ENTRY(y,i));}

    return 0
}

推荐答案

类似于函数的宏定义不能具有

Function-like macro definitions can't have whitespace after the macro name

#define ENTRY (a,b) (a[b-1])              // wrong

=>

#define ENTRY(a,b) ((a)[(b)-1])               // correct

6.10 –预处理指令:

...

...

控制线:
...

control-line:
...

# define 标识符 lparen 标识符列表选择 ) 替换列表换行
# define 标识符 lparen ... ) 替换列表 换行符
# define 标识符 lparen 标识符列表 , ... ) 替换列表 换行符

# define identifier lparen identifier-listopt ) replacement-list new-line
# define identifier lparen ... ) replacement-list new-line
# define identifier lparen identifier-list , ... ) replacement-list new-line

...

lparen:
(字符，不能紧跟在空格后面

lparen:
a ( character not immediately preceded by white-space

有了空格，您将得到一个类似于对象的宏，该宏将扩展为(a,b) (a[b-1]).

With the space, you get an object-like macro that expands to (a,b) (a[b-1]).

(为增强鲁棒性，还建议在参数中加上括号，以便在传递更复杂的表达式时也可以使用.)

(For additional robustness, it's also recommended to parenthesize the parameters so that it also works if you pass in more complex expressions.)

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