当获取max var时，使用函数和宏之间有什么区别

问题描述

阅读combase.cpp代码，找到以下内容：

reading combase.cpp code, I find following:

/* We have to ensure that we DON'T use a max macro, since these will typically   */
/* lead to one of the parameters being evaluated twice.  Since we are worried    */
/* about concurrency, we can't afford to access the m_cRef twice since we can't  */
/* afford to run the risk that its value having changed between accesses.        */

    template<class T> inline static T ourmax( const T & a, const T & b )
    {
        return a > b ? a : b;
    }

我不明白为什么max macro导致一个参数评估两次？

I don't understand why "max macro leads to one of the parameters being evaluated twice"?

推荐答案

请考虑 ：

Consider an usage like in this code sample:

#define max(a,b) (a>b?a:b)

int main()
{

  int a = 0;
  int b = 1;

  int c = max(a++, b++);

  cout << a << endl << b << endl;
  return 0;

}

打算打印 1 和 2 ，但宏扩展为：

The intention probably was to print 1 and 2, but macro expands to:

int c = a++ > b++ ? a++ : b++;

b 1 和 3 。

因此，

在某些情况下，作为参数传递给宏的表达式可以被多次计算。

Hence,
In some cases, expressions passed as arguments to macros can be evaluated more than once.

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