宏的问题 [英] Problem with Macros

问题描述

HI，

有人可以帮助我理解为什么SQUARE（x）的值是49？

Can some one help me in understanding why the value of SQUARE(x) is 49 ?

我使用Visual C ++ 6.0。

I am using Visual C++ 6.0 .

#define SQUARE(X) X * X

int main(int argc, char* argv[])
{
    int y = 5;

    printf("%d\n",SQUARE(++y));
    return 0;
}

推荐答案

Neil Butterworth，Mark和Pavel

Neil Butterworth, Mark and Pavel are right.

SQUARE（++ y）扩展为++ y * ++ y，它的值增加两倍。

SQUARE(++y) expands to ++y * ++y, which increments twice the value of y.

您可能遇到的另一个问题：SQUARE（a + b）扩展为不是（a + b）*（a + b）而是a +（b * a）的+ b * + b。在定义宏时，您需要考虑在元素周围添加括号：#define SQUARE（X）（（X）*（X））的风险要小一些。 （Ian Kemp在他的评论中首先写道）

Another problem you could encounter: SQUARE(a + b) expands to a + b * a + b which is not (a+b)*(a+b) but a + (b * a) + b. You should take care of adding parentheses around elements when needed while defining macros: #define SQUARE(X) ((X) * (X)) is a bit less risky. (Ian Kemp wrote it first in his comment)

您可以使用内联模板函数（在运行时效率不低），如下所示：

You could instead use an inline template function (no less efficient at runtime) like this one:

template <class T>
inline T square(T value)
{
    return value*value;
}

您可以检查它是否正常：

You can check it works:

int i = 2;
std::cout << square(++i) << " should be 9" << std::endl;
std::cout << square(++i) << " should be 16" << std::endl;

（无需写入

square<int>(++i)

是隐含的i）

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