C ++，传递引用变量不会在调用函数的同一行上更新 [英] C++, pass by reference variable not being update on the same line on which function is called

问题描述

在我的c ++程序中，我有这个函数，

char MostFrequentCharacter（ifstream& ifs，int& numOccurances） code>

并且在main（）中，是此代码，

  ifstream in（file.htm）; 
 int maxOccurances = 0; 
 cout<<最多频率字符是<<< MostFrequentCharacter（in，maxOccurances）<：<< maxOccurances; 
  

但这不工作，虽然我得到正确的字符，maxOccurance保持为零。
但是如果我用main替​​换上面的代码，

  ifstream in（file.htm）; 
 int maxOccurances = 0; 
 char maxFreq = MostFrequentCharacter（in，maxOccurances）; 
 cout<<最多频率char是<< maxFreq<<：<< maxOccurances; 
  

然后，它正常工作。我的问题是为什么它不能在第一种情况下工作。

解决方案

在C ++中，

  cout<< a<< b 
  

By Associativity评估为：

 （cout<< a）< b 
  

但编译器可以按任意顺序自由评估。

即，编译器首先可以评估 b ，然后 a 操作，而第二 <<$ 操作。这是因为没有与<

相关联的顺序点

---

为了简单起见，让我们考虑下面的代码，它是等价的：

 ＃include< iostream> 
 int main（）
 {
 int i = 0; 
 std :: cout<< i<< i ++; 
 return 0; 
} 
  

在上述源代码中：

  std :: cout<< i<< i ++; 
  

计算函数调用：

  operator<<（operator< <（std :: cout，i），i ++）; 
 

在此函数中调用是否 operator<<<（std :: cout， i）或 i ++ 首先被评估为未指定：

运算符<<（std :: cout，i）可能先计算或

i ++ 可以先计算或者

编译器执行一些魔术排序

鉴于上面的情况，没有办法定义这个排序

---

C ++ 03标准的相关引用：

第1.9节

抽象机器的其他方面和操作在本国际标准中被描述为未指定（例如，对函数的参数的求值顺序）。在可能的情况下，本国际标准定义了一组允许的行为。这些定义了抽象机的非确定性方面。

In my c++ program, I have this function,

char MostFrequentCharacter(ifstream &ifs, int &numOccurances);

and in main(), is this code,

ifstream in("file.htm");
int maxOccurances = 0;
cout <<"Most freq char is "<<MostFrequentCharacter(in, maxOccurances)<<" : "<<maxOccurances;

But this is not working, though I am getting the correct char, the maxOccurance remains zero. But if I replace the above code in main with this,

ifstream in("file.htm");
int maxOccurances = 0;
char maxFreq = MostFrequentCharacter(in, maxOccurances);
cout <<"Most freq char is "<<maxFreq<<" : "<<maxOccurances;

Then, it is working correctly. My question is why is it not working in first case.

解决方案

In C++,

cout << a << b 

By Associativity evaluates to:

(cout << a) << b 

but the compiler is free to evaluate them in any order.

i.e, the compiler can evaluate b first, then a, then the first << operation and the the second << operation. This because there is no sequence point associated with <<

---

For the sake of simplicity let us consider the following code, which is equivalent:

 #include<iostream>
    int main()
    {
       int i = 0;
       std::cout<<i<<i++;
       return 0;
    }

In the above source code:

std::cout<<i<<i++;

evaluates to the function call:

operator<<(operator<<(std::cout,i),i++);

In this function call whether operator<<(std::cout,i) or i++ gets evaluated first is Unspecified. i.e:

operator<<(std::cout,i) maybe evaluated first Or
i++ maybe evaluated first Or
Some Magic Ordering implemented by the compiler

Given the above, that there is no way to define this ordering and hence no explanation is possible either.

---

Relevant Quote from the C++03 Standard:
Section 1.9

Certain other aspects and operations of the abstract machine are described in this International Standard as unspecified (for example, order of evaluation of arguments to a function). Where possible, this International Standard defines a set of allowable behaviors. These define the nondeterministic aspects of the abstract machine.

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