如何在使用C预处理器宏时增加变量？ [英] How to increment a variable when using C preprocessor macros ?

问题描述

假设我想编写一个从x到y的简单循环，每次循环运行时，x都会增加1.



这是代码我试过了。

Say I want to write a simple loop that goes from x to y and every time this loop runs, x gets incremented by 1.

This is the code I've tried.

#define LOOP(x,y) 	while(x<=y)\
					{\
						printf("%d", x);\
						x++;\
					}

但这会在x ++行给出一个错误，说 [错误]左值作为递增操作数。



同样的错误即使我用++ x替换x ++，x = x + 1，x + = 1，也会出现。 使用宏编写函数时递增变​​量的正确方法是什么？



我尝试过：

But this gives an error at the x++ line saying "[Error] lvalue required as increment operand".

The same error appears even if I replaced x++ with ++x, x=x+1, x+=1. What is the correct way of incrementing a variable when writing a function using macros?

What I have tried:

#define LOOP(x,y) 	while(x<=y)\
					{\
						printf("%d", x);\
						x++;\
					}

<pre>#define LOOP(x,y) 	while(x<=y)\
					{\
						printf("%d", x);\
						++x;\
					}

<pre>#define LOOP(x,y) 	while(x<=y)\
					{\
						printf("%d", x);\
						x=x+1;\
					}

<pre>#define LOOP(x,y) 	while(x<=y)\
					{\
						printf("%d", x);\
						x+=1;\
					}

推荐答案

lvalue 是一个在内存中具有可识别位置的对象。只有那些可以修改。您可能正在将整数文字传递给您的宏：

An lvalue is an object that has an identifiable location in memory. Only those can be modified. You are probably passing integer literals to your macro:

LOOP(1,3);

这是行不通的，因为文字 1 和 3 不是左值。



只需像预处理器那样扩展代码：

That won't work because the literals 1 and 3 are not lvalues.

Just expand the code as done by the preprocessor:

while(1<=3)
{
    printf("%d", 1);
    // This is not allowed!
    // The left side must be an lvalue
    1++;
}

完全避免使用这些宏。它们容易产生不良副作用。例如：

Avoid such macros at all. They are prone to unwanted side effects. An example:

int x = 1;
int y = 3;
LOOP(x--, ++y);

虽然上面的编译没有错误，但未定义会发生什么，并且您将使用不同的编译器执行不同数量的循环。 br />


解决方案是在使用最新的C编译器时使用内联函数：

While the above would compile without errors, it is undefined what happens and you will have a different number of loops executed with different compilers.

The solution is to use inline functions when using a recent C compiler:

static inline void loop(int x, int y)
{
    while (x<=y)
    {
        printf("%d", x);
        x++;
    }
}

然而，这只允许传递 int （具有类型检查的优势）并且可能被编译器忽略（他仍然可以创建并调用该函数）。

However, this allows only ints to be passed (with the advantage of type checking) and may be ignored by the compiler (he may still create and call the function).

我只能假设你正在做类似的事情。

I can only assume you are doing something like

LOOP(1, 7)

评估为

which evaluates to

while(1 <= 7)
{
    printf("%d", 1);
    1++;
}

这种情况​​永远不会奏效。 x 字段必须是左值，即可以更改的非常量数字变量。

And that can never work. The x field must be an lvalue, i.e. a non-const numeric variable that can be changed.

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