显示chars作为ints没有显式强制转换 [英] Displaying chars as ints without explicit cast

问题描述

每次向cout对象发送一个字符时，除非将其转换为int，否则它将以ASCII字符显示。

Every time I send a char to the cout object, it displays in ASCII characters unless I cast it to an int.

>有没有一种方法来显示一个字符的数值没有显式转换？

Q: Is there a way to display the numerical value of a char without an explicit cast?

我读到某处，在代码中执行太多的转换可能会导致损失诚信（你的程序）。我猜，字符显示在ASCII为特定的原因，但我不知道为什么。

I read somewhere that doing too many casts in your code could lead to a loss of integrity (of your program). I am guessing that chars display in ASCII for a particular reason, but I'm not sure why.

我基本上创建一个游戏。我使用小数字（unsigned chars），我打算显示在控制台。我可能是偏执的，但是当我在我的代码中的任何地方发出垃圾邮件 static_cast< int> 时，我得到这种不安的感觉。

I am essentially creating a game. I am using small numbers (unsigned chars) that I plan to display to the console. I may be paranoid, but I get this uneasy feeling whenever I spam static_cast<int> everywhere in my code.

推荐答案

但是，类型转换没有什么问题，特别是如果你使用 static_cast 这就是你应该使用。

There is nothing wrong with type-casting, though, especially if you use static_cast to do it. That is what you should be using. It allows the compiler to validate the type-cast and make sure it is safe.

要改变<< 的行为， code>运算符，您必须覆盖 char 值的默认<< ：

To change the behavior of the << operator, you would have to override the default << operator for char values, eg:

std::ostream& operator <<(std::ostream &os, char c)
{
    os << static_cast<int>(c);
    return os;
}

char c = ...;
std::cout << c;

您可以创建一个自定义类型，它采用 char 作为输入，然后为该类型实现< 运算符，例如：

You could create a custom type that takes a char as input and then implement the << operator for that type, eg:

struct CharToInt
{
    int val;
    CharToInt(char c) : val(static_cast<int>(c)) {}
};

std::ostream& operator <<(std::ostream &os, const CharToInt &c)
{
    os << c.val;
    return os;
}

char c = ...;
std::cout << CharToInt(c);

你可以创建一个类似的函数，然后你不必重写< <$ / code>运算符，例如：

You could create a function that does something similar, then you don't have to override the << operator, eg:

int CharToInt(char c)
{
    return c;
}

char c = ...;
std::cout << CharToInt(c);

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