#define SQR（x）x * x。意外回答 [英] #define SQR(x) x*x. Unexpected Answer

问题描述

为什么这个宏会输出144而不是121？

Why this macro gives output 144, instead of 121?

#include<iostream>
#define SQR(x) x*x

int main()
{
    int p=10;
    std::cout<<SQR(++p);
}

推荐答案

宏有两个问题：

首先，对于参数 ++ p ，执行增量操作两次。这当然不是为了。 （作为一般的经验法则，只是不要在一行中做几件事，将它们分成更多的语句。它甚至不会在递增两次时停止：这些增量的顺序没有定义，所以没有这个操作的保证结果！

First, for the argument ++p, the increment operation is performed twice. That's certainly not intended. (As a general rule of thumb, just don't do several things in "one line". Separate them into more statements.). It doesn't even stop at incrementing twice: The order of these increments isn't defined, so there is no guaranteed outcome of this operation!

第二，即使你没有 ++ p 作为参数，您的宏中仍然有一个错误！考虑输入 1 + 1 。预期输出为 4 。 1 + 1 没有副作用，所以应该很好，应该不是吗？不，因为 SQR（1 + 1）转换为 1 + 1 * 1 + 1 $ c> 3 。

Second, even if you don't have ++p as the argument, there is still a bug in your macro! Consider the input 1 + 1. Expected output is 4. 1+1 has no side-effect, so it should be fine, shouldn't it? No, because SQR(1 + 1) translates to 1 + 1 * 1 + 1 which evaluates to 3.

要至少部分修复此宏，请使用括号：

To at least partially fix this macro, use parentheses:

#define SQR(x) (x) * (x)

$ b b

总之，你应该简单地用一个函数替换它（添加类型安全！）

Altogether, you should simply replace it by a function (to add type-safety!)

int sqr(int x)
{
    return x * x;
}

您可以将其设为模板

template <typename Type>
Type sqr(Type x)
{
   return x * x; // will only work on types for which there is the * operator.
}

您可以添加 constexpr （C ++ 11），如果你计划在模板中使用一个正方形，这是有用的：

and you may add a constexpr (C++11), which is useful if you ever plan on using a square in a template:

constexpr int sqr(int x)
{
    return x * x;
}

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