#define与运算符一起使用 [英] #define used with operators

问题描述

我知道 #define 具有以下语法: #define SYMBOL字符串例如，如果我写

I know that #define has the following syntax: #define SYMBOL string If I write, for example

#define ALPHA 2-1
#define BETA ALPHA*2

然后 ALPHA = 1 但 BETA = 0 .(为什么?)

但是如果我写这样的话

#define ALPHA (2-1)
#define BETA ALPHA*2

然后 ALPHA = 1 和 BETA = 2 .

有人可以解释一下这两者之间有什么区别吗?

Can someone explain me what's the difference between those two ?

推荐答案

使用 #define 创建的预处理器宏是 text 替换.

Pre-processor macros created using #define are textual substitutions.

这两个示例并不等效.第一个将 BETA 设置为 2-1 * 2 .第二个将 BETA 设置为(2-1)* 2 .像您一样声称 ALPHA == 1 是不正确的，因为 ALPHA 不是数字-它是免费的伙计！这只是一个字符序列.

The two examples are not equivalent. The first sets BETA to 2-1*2. The second sets BETA to (2-1)*2. It is not correct to claim that ALPHA == 1 as you do, because ALPHA is not a number - it is a free man! it's just a sequence of characters.

当解析为C或C ++时，这两个表达式是不同的(第一个与 2-(1 * 2)相同)

When parsed as C or C++, those two expressions are different (the first is the same as 2 - (1*2)).

我们可以通过打印 BETA 的字符串扩展并将其评估为表达式来显示差异:

We can show the difference, by printing the string expansion of BETA as well as evaluating it as an expression:

#ifdef PARENS
#define ALPHA (2-1)
#else
#define ALPHA 2-1
#endif

#define BETA ALPHA*2

#define str(x) str2(x)
#define str2(x) #x

#include <stdio.h>
int main()
{
    printf("%s = %d\n", str(BETA), BETA);
    return 0;
}

在定义和不定义 PARENS 的情况下编译以上内容，以查看两者的区别:

Compile the above with and without PARENS defined to see the difference:

(2-1)*2 = 2
2-1*2 = 0

---

其结果是，当使用 #define 创建扩展为表达式的宏时，通常最好使用比通常所需更多的括号，因为您不知道您的价值观将得到扩展的背景.例如:

---

The consequence of this is that when using #define to create macros that expand to expressions, it's generally a good idea to use many more parentheses than you would normally need, as you don't know the context in which your values will be expanded. For example:

#define ALPHA (2-1)
#define BETA ((ALPHA)*2)

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