在 c 中对宏 SQR 求平方感到困惑
这个问题是在一次模拟面试中被问到的......真的很惊讶地发现了尴尬的答案......
This question was asked to me in a mock interview...Really got surprised to find awkward answers...
考虑一个宏:
#define SQR(x) (x*x)
示例 1:
SQR(2) //prints 4
示例 2:
如果给定 SQR(1+1)，它不会将 (1+1) 与 2 相加，而是...
If SQR(1+1) is given it doesn't sum (1+1) to 2 but rather ...
SQR(1+1) //prints 3
尴尬吧?是什么原因?这段代码是如何工作的?
Awkward right? What is the reason? How does this code work?
注意:我搜索了 SO，但找不到任何相关问题.如果有什么好心的请分享！
NOTE: I searched SO but couldn't find any relevant questions. If there are any kindly please share it!
推荐答案
SQR(1+1) 展开为 1+1*1+1 即 3，不是4，对吗?
SQR(1+1) expands to 1+1*1+1 which is 3, not 4, correct?
宏的正确定义是
#define SQR(x) ((x)*(x))
扩展为 (1+1)*(1+1) 并且，更重要的是，向您展示了不应使用宏的原因之一不需要.以下更好:
which expands to (1+1)*(1+1) and, more important, shows you one of the reasons you shouldn't use macros where they aren't needed. The following is better:
inline int SQR(int x)
{
return x*x;
}
此外:如果 SQR 是宏，则 SQR(i++) 将是 未定义的行为，如果 SQR 则完全正确code> 是一个函数.
Furthermore: SQR(i++) would be undefined behavior if SQR is a macro, and completely correct if SQR is a function.
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