使用C ++ hex和cin [英] Using C++ hex and cin

问题描述

如果您有以下代码：

  cout< hex<< 10; 
  

输出是'a'，这意味着十进制10被转换为十六进制值。 p>

但是，在下面的代码中...

  int n; 
 cin>> hex>> n; 
 cout<< n<< endl 
  

当输入为12时，输出为18.任何人都可以解释转换的详细信息吗？它是如何成为一个十进制值？

我感兴趣的地方，它成为一个int。如果分解，它将是：

 （（cin>> hex）>> n）; 
  

这是否正确？

方案

六进制操纵器仅控制值的读取方式 - 它始终使用相同的内部二进制表示存储。变量没有办法记住它是以十六进制输入的。

If you have the following code:

cout << hex << 10;

The output is 'a', which means the decimal 10 is converted into its hexadecimal value.

However, in the code below...

int n;
cin >> hex >> n;
cout << n << endl;

When input is 12, the output becomes 18. Can anyone explain the details of the conversion? How did it became a decimal value?

I'm interested in the point where it became an int. If broken down, it would be:

(( cin >> hex ) >> n);

Is this correct?

解决方案

The hex manipulator only controls how a value is read - it is always stored using the same internal binary representation. There is no way for a variable to "remember" that it was input in hex.

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